Samacheer Kalvi Class 11 Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4

Find the derivatives of the following:

Question 1. \(y = x^{\cos x}\)
Answer: To find the derivative of \( y = x^{\cos x} \), we use logarithmic differentiation. First, take the natural logarithm on both sides to simplify the exponent. \[ \log y = \log (x^{\cos x}) \] This simplifies to: \[ \log y = \cos x \cdot \log x \] Now, differentiate both sides with respect to \( x \). Use the product rule for the right side. \[ \frac{1}{y} \frac{dy}{dx} = (-\sin x)(\log x) + (\cos x)\left(\frac{1}{x}\right) \] Rearrange to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y \left( \frac{\cos x}{x} - \sin x \log x \right) \] Substitute \( y = x^{\cos x} \) back into the equation: \[ \frac{dy}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \log x \right) \]In simple words: To differentiate this complex function, we first take the logarithm of both sides. This brings the exponent down, making it easier to differentiate using basic rules like the product rule. Then, we solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating \( \log y \) and use the product rule for terms like \( \cos x \cdot \log x \).

Question 2. \(y = x^{\log x} + (\log x)^x\)
Answer: We need to find the derivative of \( y = x^{\log x} + (\log x)^x \). Since it's a sum of two complex functions, we'll differentiate each part separately. Let \( u = x^{\log x} \) and \( v = (\log x)^x \), so \( y = u + v \). Then \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \). First, let's find \( \frac{du}{dx} \) for \( u = x^{\log x} \). Take natural log on both sides: \[ \log u = \log (x^{\log x}) \] \[ \log u = (\log x)(\log x) \] \[ \log u = (\log x)^2 \] Differentiate with respect to \( x \): \[ \frac{1}{u} \frac{du}{dx} = 2 (\log x) \left(\frac{1}{x}\right) \] So, \[ \frac{du}{dx} = u \left( \frac{2 \log x}{x} \right) \] Substitute \( u = x^{\log x} \) back: \[ \frac{du}{dx} = x^{\log x} \left( \frac{2 \log x}{x} \right) \quad \text{--- (1)} \] Next, let's find \( \frac{dv}{dx} \) for \( v = (\log x)^x \). Take natural log on both sides: \[ \log v = \log ((\log x)^x) \] \[ \log v = x \log (\log x) \] Differentiate with respect to \( x \), using the product rule for the right side: \[ \frac{1}{v} \frac{dv}{dx} = (1) \log (\log x) + x \left( \frac{1}{\log x} \cdot \frac{1}{x} \right) \] \[ \frac{1}{v} \frac{dv}{dx} = \log (\log x) + \frac{1}{\log x} \] So, \[ \frac{dv}{dx} = v \left( \log (\log x) + \frac{1}{\log x} \right) \] Substitute \( v = (\log x)^x \) back: \[ \frac{dv}{dx} = (\log x)^x \left( \log (\log x) + \frac{1}{\log x} \right) \quad \text{--- (2)} \] Finally, combine \( \frac{du}{dx} \) and \( \frac{dv}{dx} \): \[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \] \[ \frac{dy}{dx} = x^{\log x} \left( \frac{2 \log x}{x} \right) + (\log x)^x \left( \log (\log x) + \frac{1}{\log x} \right) \]In simple words: When you have a sum of terms where both the base and exponent are functions of x, you differentiate each term separately using logarithmic differentiation. This method involves taking logs, differentiating, and then substituting back the original function.

🎯 Exam Tip: Remember that if \( y \) is a sum of terms \( u+v \), its derivative is \( \frac{du}{dx} + \frac{dv}{dx} \). This allows you to tackle each complex term independently.

Question 3. \(\sqrt{x y} = e^{(x - y)}\)
Answer: We need to find the derivative of \( \sqrt{x y} = e^{(x - y)} \). First, let's simplify the equation by squaring both sides: \[ (\sqrt{x y})^2 = (e^{(x - y)})^2 \] \[ x y = e^{2(x - y)} \] Now, differentiate both sides with respect to \( x \). Use the product rule on the left side and the chain rule on the right side. For the left side, \( \frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx} \). For the right side, \( \frac{d}{dx}(e^{2(x - y)}) = e^{2(x - y)} \cdot \frac{d}{dx}(2(x - y)) \). \( \frac{d}{dx}(2(x - y)) = 2 \left( \frac{d}{dx}(x) - \frac{d}{dx}(y) \right) = 2 \left( 1 - \frac{dy}{dx} \right) \). So, the differentiation gives: \[ y + x \frac{dy}{dx} = e^{2(x - y)} \cdot 2 \left( 1 - \frac{dy}{dx} \right) \] Substitute back \( xy = e^{2(x - y)} \) from our simplified equation: \[ y + x \frac{dy}{dx} = xy \cdot 2 \left( 1 - \frac{dy}{dx} \right) \] \[ y + x \frac{dy}{dx} = 2xy - 2xy \frac{dy}{dx} \] Now, group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side: \[ x \frac{dy}{dx} + 2xy \frac{dy}{dx} = 2xy - y \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} (x + 2xy) = y (2x - 1) \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y(2x - 1)}{x(1 + 2y)} \]In simple words: To differentiate this equation, we first removed the square root by squaring both sides. Then, we differentiated both sides, keeping in mind the product rule for \( xy \) and the chain rule for the exponential term. We then collected all terms with \( \frac{dy}{dx} \) to one side to find the final derivative.

🎯 Exam Tip: When dealing with implicit functions, especially those involving roots or exponentials, simplifying the expression before differentiating often makes the process much easier.

Question 4. \(x^y = y^x\)
Answer: We need to find the derivative of \( x^y = y^x \). This equation has variables in both the base and the exponent, so we use logarithmic differentiation. Take the natural logarithm on both sides: \[ \log (x^y) = \log (y^x) \] Using the logarithm property \( \log (a^b) = b \log a \), we get: \[ y \log x = x \log y \] Now, differentiate both sides with respect to \( x \). Use the product rule on both sides. For the left side, \( \frac{d}{dx}(y \log x) = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \). For the right side, \( \frac{d}{dx}(x \log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \). So, the differentiation yields: \[ \frac{dy}{dx} \log x + \frac{y}{x} = \log y + \frac{x}{y} \frac{dy}{dx} \] Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side: \[ \frac{dy}{dx} \log x - \frac{x}{y} \frac{dy}{dx} = \log y - \frac{y}{x} \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x} \] To simplify the terms in the parentheses, find a common denominator: \[ \frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x} \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y(x \log y - y)}{x(y \log x - x)} \]In simple words: For equations where variables are in both the base and exponent, taking the logarithm first helps turn the exponents into products. Then, we differentiate using the product rule and rearrange to find \( \frac{dy}{dx} \). Remember to group terms with the derivative before isolating it.

🎯 Exam Tip: Logarithmic differentiation is crucial for functions of the form \( f(x)^{g(x)} \) or implicit equations involving such terms. Don't forget to simplify the expressions after grouping \( \frac{dy}{dx} \) terms.

Question 5. \(y = (\cos x)^{\log x}\)
Answer: We need to find the derivative of \( y = (\cos x)^{\log x} \). This is a function of the form \( f(x)^{g(x)} \), so we use logarithmic differentiation. Take the natural logarithm on both sides: \[ \log y = \log ((\cos x)^{\log x}) \] Using the logarithm property \( \log (a^b) = b \log a \), we get: \[ \log y = (\log x) (\log (\cos x)) \] Now, differentiate both sides with respect to \( x \). Use the product rule on the right side. For the left side, \( \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \). For the right side, use the product rule \( (uv)' = u'v + uv' \), where \( u = \log x \) and \( v = \log (\cos x) \). \( u' = \frac{1}{x} \) \( v' = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \) So, \( \frac{d}{dx}((\log x) (\log (\cos x))) = \left(\frac{1}{x}\right)(\log (\cos x)) + (\log x)(-\tan x) \). Putting it together: \[ \frac{1}{y} \frac{dy}{dx} = \frac{\log (\cos x)}{x} - \tan x \log x \] Now, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y \left( \frac{\log (\cos x)}{x} - \tan x \log x \right) \] Substitute \( y = (\cos x)^{\log x} \) back into the equation: \[ \frac{dy}{dx} = (\cos x)^{\log x} \left( \frac{\log (\cos x)}{x} - \tan x \log x \right) \]In simple words: When a function has another function in its exponent, we use logarithms to bring the exponent down. After that, we differentiate using the product rule for the logarithm terms. Finally, we put the original function back to get the derivative.

🎯 Exam Tip: Pay close attention to the chain rule when differentiating composite functions like \( \log(\cos x) \). A common mistake is forgetting to multiply by the derivative of the inner function.

Question 6. \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer: We need to find the derivative of the implicit equation \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \). Differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left( \frac{x^{2}}{a^{2}} \right) + \frac{d}{dx} \left( \frac{y^{2}}{b^{2}} \right) = \frac{d}{dx} (1) \] For the first term: \( \frac{d}{dx} \left( \frac{x^{2}}{a^{2}} \right) = \frac{1}{a^{2}} \cdot 2x = \frac{2x}{a^{2}} \). For the second term, remember that \( y \) is a function of \( x \), so use the chain rule: \( \frac{d}{dx} \left( \frac{y^{2}}{b^{2}} \right) = \frac{1}{b^{2}} \cdot 2y \cdot \frac{dy}{dx} = \frac{2y}{b^{2}} \frac{dy}{dx} \). The derivative of a constant is 0: \( \frac{d}{dx} (1) = 0 \). So, the differentiated equation is: \[ \frac{2x}{a^{2}} + \frac{2y}{b^{2}} \frac{dy}{dx} = 0 \] Now, solve for \( \frac{dy}{dx} \). First, move the term without \( \frac{dy}{dx} \) to the right side: \[ \frac{2y}{b^{2}} \frac{dy}{dx} = - \frac{2x}{a^{2}} \] Multiply both sides by \( \frac{b^{2}}{2y} \) to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = - \frac{2x}{a^{2}} \cdot \frac{b^{2}}{2y} \] Cancel out the 2s: \[ \frac{dy}{dx} = - \frac{b^{2}x}{a^{2}y} \] This is a classic derivative for an ellipse equation.In simple words: To find the derivative of an implicit equation like this, we differentiate each term with respect to \( x \). For terms with \( y \), we use the chain rule and multiply by \( \frac{dy}{dx} \). Then, we rearrange the equation to solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Always remember to apply the chain rule (multiplying by \( \frac{dy}{dx} \)) when differentiating terms involving \( y \) in implicit differentiation problems.

Question 7. \(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Answer: We need to find the derivative of \( \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right) \). This is an implicit equation. Differentiate both sides with respect to \( x \). For the left side, \( \frac{d}{dx}(\sqrt{x^2+y^2}) \). Using the chain rule, \( \frac{1}{2\sqrt{x^2+y^2}} \cdot \frac{d}{dx}(x^2+y^2) = \frac{1}{2\sqrt{x^2+y^2}} \cdot \left(2x + 2y \frac{dy}{dx}\right) = \frac{x + y \frac{dy}{dx}}{\sqrt{x^2+y^2}} \). For the right side, \( \frac{d}{dx}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) \). Using the chain rule, \( \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) \). The derivative of \( \frac{y}{x} \) using the quotient rule is \( \frac{\frac{dy}{dx} \cdot x - y \cdot 1}{x^2} \). So, the right side becomes \( \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{x \frac{dy}{dx} - y}{x^2} = \frac{x^2}{x^2+y^2} \cdot \frac{x \frac{dy}{dx} - y}{x^2} = \frac{x \frac{dy}{dx} - y}{x^2+y^2} \). Equating the derivatives of both sides: \[ \frac{x + y \frac{dy}{dx}}{\sqrt{x^2+y^2}} = \frac{x \frac{dy}{dx} - y}{x^2+y^2} \] We know that \( \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) \). This original equation can be rewritten as \( x^2+y^2 = (\tan^{-1}\left(\frac{y}{x}\right))^2 \). Let's use the \( \sqrt{x^2+y^2} \) term. Multiply both sides by \( \sqrt{x^2+y^2} \): \[ x + y \frac{dy}{dx} = \frac{\sqrt{x^2+y^2} (x \frac{dy}{dx} - y)}{x^2+y^2} \] Since \( \frac{\sqrt{x^2+y^2}}{x^2+y^2} = \frac{1}{\sqrt{x^2+y^2}} \): \[ x + y \frac{dy}{dx} = \frac{1}{\sqrt{x^2+y^2}} (x \frac{dy}{dx} - y) \] Now, substitute \( \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) \) back: \[ x + y \frac{dy}{dx} = \frac{x \frac{dy}{dx} - y}{\tan^{-1}\left(\frac{y}{x}\right)} \] This looks a bit complicated. Let's try an alternative simplification using the original equation: From the original equation, \( \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right) \). Let \( \tan^{-1}\left(\frac{y}{x}\right) = \theta \). Then \( \frac{y}{x} = \tan \theta \), and \( y = x \tan \theta \). Also, \( \sqrt{x^2+y^2} = \sqrt{x^2 + (x \tan \theta)^2} = \sqrt{x^2(1+\tan^2\theta)} = \sqrt{x^2 \sec^2\theta} = |x \sec \theta| \). So, \( |x \sec \theta| = \theta \). Differentiating \( y=x \tan \theta \) implicitly for \( \frac{dy}{dx} \). From \( \sqrt{x^2+y^2} = \theta \), if we assume \( x>0 \), then \( x \sec \theta = \theta \). This problem involves quite a bit of algebra. Let's return to the direct differentiation result and collect \( \frac{dy}{dx} \) terms. \[ x + y \frac{dy}{dx} = \frac{x \frac{dy}{dx} - y}{\sqrt{x^2+y^2}} \cdot \frac{1}{\sqrt{x^2+y^2}} \text{ (This step was wrong, the } \frac{1}{x^2+y^2} \text{ already included } x^2 \text{ in denominator for } \tan^{-1} ) \] Let's restart from: \[ \frac{x + y \frac{dy}{dx}}{\sqrt{x^2+y^2}} = \frac{x \frac{dy}{dx} - y}{x^2+y^2} \] Multiply both sides by \( x^2+y^2 \): \[ \sqrt{x^2+y^2} \left( x + y \frac{dy}{dx} \right) = x \frac{dy}{dx} - y \] Now, substitute \( \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) \): \[ \tan^{-1}\left(\frac{y}{x}\right) \left( x + y \frac{dy}{dx} \right) = x \frac{dy}{dx} - y \] Expand the left side: \[ x \tan^{-1}\left(\frac{y}{x}\right) + y \tan^{-1}\left(\frac{y}{x}\right) \frac{dy}{dx} = x \frac{dy}{dx} - y \] Group terms with \( \frac{dy}{dx} \): \[ y \tan^{-1}\left(\frac{y}{x}\right) \frac{dy}{dx} - x \frac{dy}{dx} = -y - x \tan^{-1}\left(\frac{y}{x}\right) \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( y \tan^{-1}\left(\frac{y}{x}\right) - x \right) = - \left( y + x \tan^{-1}\left(\frac{y}{x}\right) \right) \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = - \frac{y + x \tan^{-1}\left(\frac{y}{x}\right)}{y \tan^{-1}\left(\frac{y}{x}\right) - x} \] We can swap the terms in the denominator and change the sign: \[ \frac{dy}{dx} = \frac{y + x \tan^{-1}\left(\frac{y}{x}\right)}{x - y \tan^{-1}\left(\frac{y}{x}\right)} \] This looks correct. It's a complex differentiation, so it's good to be thorough.In simple words: We differentiate both sides of the equation with respect to \( x \), carefully applying the chain rule for each function. After differentiation, we group all terms containing \( \frac{dy}{dx} \) on one side and move the other terms to the other side. Finally, we factor out \( \frac{dy}{dx} \) and simplify the expression to get the final answer.

🎯 Exam Tip: For complex implicit differentiation, be very careful with the chain rule and algebraic manipulation. It's often helpful to simplify fractions as you go to avoid errors.

Question 8. \( \tan (x + y) + \tan (x - y) = x \)
Answer: We need to find the derivative of \( \tan (x + y) + \tan (x - y) = x \). Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(\tan (x + y)) + \frac{d}{dx}(\tan (x - y)) = \frac{d}{dx}(x) \] Using the chain rule, \( \frac{d}{dx}(\tan u) = \sec^2 u \cdot u' \). For the first term: \( \frac{d}{dx}(\tan (x + y)) = \sec^2 (x + y) \cdot \frac{d}{dx}(x + y) = \sec^2 (x + y) \left( 1 + \frac{dy}{dx} \right) \). For the second term: \( \frac{d}{dx}(\tan (x - y)) = \sec^2 (x - y) \cdot \frac{d}{dx}(x - y) = \sec^2 (x - y) \left( 1 - \frac{dy}{dx} \right) \). For the right side: \( \frac{d}{dx}(x) = 1 \). So, the differentiated equation is: \[ \sec^2 (x + y) \left( 1 + \frac{dy}{dx} \right) + \sec^2 (x - y) \left( 1 - \frac{dy}{dx} \right) = 1 \] Expand the terms: \[ \sec^2 (x + y) + \sec^2 (x + y) \frac{dy}{dx} + \sec^2 (x - y) - \sec^2 (x - y) \frac{dy}{dx} = 1 \] Group terms with \( \frac{dy}{dx} \): \[ \left( \sec^2 (x + y) - \sec^2 (x - y) \right) \frac{dy}{dx} = 1 - \sec^2 (x + y) - \sec^2 (x - y) \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1 - \sec^2 (x + y) - \sec^2 (x - y)}{\sec^2 (x + y) - \sec^2 (x - y)} \]In simple words: We differentiate each tangent term using the chain rule, remembering to include \( \frac{dy}{dx} \) whenever we differentiate \( y \). Then, we gather all \( \frac{dy}{dx} \) terms to one side and move the rest to the other side. This helps us isolate \( \frac{dy}{dx} \) to get the final derivative.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(\tan u) = \sec^2 u \cdot \frac{du}{dx} \). Be careful with the signs when differentiating \( (x-y) \) as \( 1-\frac{dy}{dx} \).

Question 9. If \( \cos(xy) = x \), show that \( \frac{dy}{dx}=\frac{-(1+y \sin (x y))}{x \sin x y} \)
Answer: We are given \( \cos(xy) = x \). We need to show the given derivative. Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(x) \] For the left side, use the chain rule and the product rule (for \( xy \)): \( \frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dx}(xy) \). \( \frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx} \). So, the left side is \( -\sin(xy) \left( y + x \frac{dy}{dx} \right) \). For the right side, \( \frac{d}{dx}(x) = 1 \). Equating the derivatives: \[ -\sin(xy) \left( y + x \frac{dy}{dx} \right) = 1 \] Distribute \( -\sin(xy) \): \[ -y \sin(xy) - x \sin(xy) \frac{dy}{dx} = 1 \] Move the term without \( \frac{dy}{dx} \) to the right side: \[ - x \sin(xy) \frac{dy}{dx} = 1 + y \sin(xy) \] Finally, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1 + y \sin(xy)}{-x \sin(xy)} \] \[ \frac{dy}{dx} = - \frac{1 + y \sin(xy)}{x \sin(xy)} \] This matches the expression we needed to show. Implicit differentiation is very helpful for these kinds of problems.In simple words: We differentiate both sides of the equation. On the left side, we use the chain rule for cosine and the product rule for \( xy \). Then, we rearrange the equation to gather all terms with \( \frac{dy}{dx} \) and solve for it.

🎯 Exam Tip: When differentiating \( \cos(xy) \), remember the chain rule for the outer function (cosine) and the product rule for the inner function (\( xy \)). Watch out for negative signs from differentiating cosine.

Question 10. \( \tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \)
Answer: We need to find the derivative of \( y = \tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \). First, simplify the expression inside the tangent inverse using trigonometric identities: We know that \( 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \) and \( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \). Substitute these into the expression: \[ \sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{2 \sin^2 \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2}\right)}} \] \[ = \sqrt{\tan^2 \left(\frac{x}{2}\right)} \] \[ = \left| \tan \left(\frac{x}{2}\right) \right| \] Assuming \( x \) is in a range where \( \tan \left(\frac{x}{2}\right) \) is positive (e.g., \( 0 < x < \pi \)), then \( \sqrt{\tan^2 \left(\frac{x}{2}\right)} = \tan \left(\frac{x}{2}\right) \). So, \( y = \tan^{-1} \left( \tan \left(\frac{x}{2}\right) \right) \). This simplifies to: \[ y = \frac{x}{2} \] Now, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} \right) \] \[ \frac{dy}{dx} = \frac{1}{2} \] This problem becomes very simple once the trigonometric identities are applied.In simple words: First, we use special trigonometry rules to make the expression inside the square root much simpler. The terms with \( 1-\cos x \) and \( 1+\cos x \) become \( \tan^2 \left(\frac{x}{2}\right) \). After taking the square root and applying the inverse tangent, the whole function becomes just \( \frac{x}{2} \). Then, finding the derivative is easy.

🎯 Exam Tip: Always look for trigonometric identities to simplify expressions before differentiation, especially when dealing with inverse trigonometric functions. Identities like \( 1-\cos x \) and \( 1+\cos x \) are very useful here.

Question 11. Find the derivative of \( y = \tan^{-1} \left(\frac{6x}{1-9x^2}\right) \)
Answer: We need to find the derivative of \( y = \tan^{-1} \left(\frac{6x}{1-9x^2}\right) \). This expression resembles the tangent double angle formula, \( \tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} \). Let \( 3x = \tan\theta \). Then \( \theta = \tan^{-1}(3x) \). Substitute \( 3x = \tan\theta \) into the expression for \( y \): \[ y = \tan^{-1} \left(\frac{2 \cdot (3x)}{1-(3x)^2}\right) \] \[ y = \tan^{-1} \left(\frac{2 \tan\theta}{1-\tan^2\theta}\right) \] Using the double angle identity, \( \frac{2 \tan\theta}{1-\tan^2\theta} = \tan(2\theta) \). So, \( y = \tan^{-1} (\tan(2\theta)) \). This simplifies to: \[ y = 2\theta \] Now, substitute \( \theta = \tan^{-1}(3x) \) back into the equation: \[ y = 2 \tan^{-1}(3x) \] Finally, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (2 \tan^{-1}(3x)) \] \[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x)^2} \cdot \frac{d}{dx}(3x) \] \[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+9x^2} \cdot 3 \] \[ \frac{dy}{dx} = \frac{6}{1+9x^2} \] This simplifies the differentiation significantly.In simple words: We saw that the given function looked like a known trigonometry formula. By letting \( 3x = \tan\theta \), we changed the expression into \( \tan^{-1}(\tan(2\theta)) \), which simplifies to \( 2\theta \). Then, we put \( \theta = \tan^{-1}(3x) \) back and easily differentiated the simpler form.

🎯 Exam Tip: Recognizing inverse trigonometric formulas (like for \( \tan(2\theta) \), \( \sin(2\theta) \), \( \cos(2\theta) \)) can dramatically simplify differentiation. Always try substitution to transform the expression into a simpler form.

Question 12. Find the derivative of \( y = \cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right) \)
Answer: We need to find the derivative of \( y = \cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right) \). This expression can be simplified using a substitution. Let \( x = \cos 2\theta \). Then, \( 2\theta = \cos^{-1} x \), so \( \theta = \frac{1}{2} \cos^{-1} x \). Substitute \( x = \cos 2\theta \) into the term \( \sqrt{\frac{1-x}{1+x}} \): \[ \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} \] Using the half-angle identities: \( 1-\cos 2\theta = 2\sin^2\theta \) and \( 1+\cos 2\theta = 2\cos^2\theta \). \[ = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \sqrt{\tan^2\theta} = |\tan\theta| \] Assuming \( x \) is in a range where \( \tan\theta \) is positive, we have \( |\tan\theta| = \tan\theta \). So, the expression inside \( 2 \tan^{-1} \) becomes \( \tan\theta \). \[ y = \cos (2 \tan^{-1} (\tan\theta)) \] \[ y = \cos (2\theta) \] Now, substitute back \( 2\theta = \cos^{-1} x \): \[ y = \cos (\cos^{-1} x) \] This simplifies to: \[ y = x \] Finally, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(x) \] \[ \frac{dy}{dx} = 1 \] This intricate problem becomes quite straightforward after the right trigonometric substitution.In simple words: We used a special trick by replacing \( x \) with \( \cos 2\theta \). This helped simplify the inside part of the function using trigonometry rules. After many steps, the whole complicated expression turned into just \( x \). Then, finding the derivative was very simple.

🎯 Exam Tip: When you see expressions like \( \sqrt{\frac{1-x}{1+x}} \), immediately think of the substitution \( x = \cos 2\theta \) (or \( x = \cos\theta \)) to simplify it using half-angle formulas.

Question 13. If \( x = a \cos^3 t \) and \( y = a \sin^3 t \), find \( \frac{dy}{dx} \)
Answer: We have parametric equations: \( x = a \cos^3 t \) and \( y = a \sin^3 t \). We need to find \( \frac{dy}{dx} \). First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = a \cos^3 t \): \[ \frac{dx}{dt} = a \cdot 3 \cos^2 t \cdot \frac{d}{dt}(\cos t) \] \[ \frac{dx}{dt} = 3a \cos^2 t (-\sin t) \] \[ \frac{dx}{dt} = -3a \cos^2 t \sin t \] For \( y = a \sin^3 t \): \[ \frac{dy}{dt} = a \cdot 3 \sin^2 t \cdot \frac{d}{dt}(\sin t) \] \[ \frac{dy}{dt} = 3a \sin^2 t (\cos t) \] Now, use the formula for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). \[ \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \] Cancel out common terms \( 3a, \sin t, \cos t \): \[ \frac{dy}{dx} = - \frac{\sin t}{\cos t} \] \[ \frac{dy}{dx} = - \tan t \] This method of using parametric derivatives is very efficient.In simple words: To find \( \frac{dy}{dx} \) when \( x \) and \( y \) are given in terms of a third variable (\( t \) here), we first find how \( x \) changes with \( t \) (\( \frac{dx}{dt} \)) and how \( y \) changes with \( t \) (\( \frac{dy}{dt} \)). Then, we divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to get \( \frac{dy}{dx} \).

🎯 Exam Tip: For parametric differentiation, remember the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Make sure to apply the chain rule correctly when differentiating terms like \( \cos^3 t \) or \( \sin^3 t \).

Question 14. If \( x = a (\cos t + t \sin t) \) and \( y = a (\sin t - t \cos t) \), find \( \frac{dy}{dx} \)
Answer: We have parametric equations: \( x = a (\cos t + t \sin t) \) and \( y = a (\sin t - t \cos t) \). We need to find \( \frac{dy}{dx} \). First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = a (\cos t + t \sin t) \): \[ \frac{dx}{dt} = a \left( \frac{d}{dt}(\cos t) + \frac{d}{dt}(t \sin t) \right) \] \[ \frac{dx}{dt} = a \left( -\sin t + (1 \cdot \sin t + t \cdot \cos t) \right) \] \[ \frac{dx}{dt} = a (-\sin t + \sin t + t \cos t) \] \[ \frac{dx}{dt} = a (t \cos t) \] For \( y = a (\sin t - t \cos t) \): \[ \frac{dy}{dt} = a \left( \frac{d}{dt}(\sin t) - \frac{d}{dt}(t \cos t) \right) \] \[ \frac{dy}{dt} = a \left( \cos t - (1 \cdot \cos t + t \cdot (-\sin t)) \right) \] \[ \frac{dy}{dt} = a (\cos t - \cos t + t \sin t) \] \[ \frac{dy}{dt} = a (t \sin t) \] Now, use the formula for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). \[ \frac{dy}{dx} = \frac{a (t \sin t)}{a (t \cos t)} \] Cancel out common terms \( a \) and \( t \): \[ \frac{dy}{dx} = \frac{\sin t}{\cos t} \] \[ \frac{dy}{dx} = \tan t \] This shows how parametric differentiation simplifies complex problems.In simple words: To find \( \frac{dy}{dx} \), we first calculated how \( x \) changes with \( t \) (\( \frac{dx}{dt} \)) and how \( y \) changes with \( t \) (\( \frac{dy}{dt} \)), using the product rule where needed. Then, we simply divided \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to get the final answer.

🎯 Exam Tip: Be very careful when applying the product rule for terms like \( t \sin t \) or \( t \cos t \), and double-check your signs, especially after distributing a negative sign.

Question 15. If \( x = \frac{1-t^2}{1+t^2} \) and \( y = \frac{2t}{1+t^2} \), find \( \frac{dy}{dx} \)
Answer: We have parametric equations: \( x = \frac{1-t^2}{1+t^2} \) and \( y = \frac{2t}{1+t^2} \). We need to find \( \frac{dy}{dx} \). These expressions look like trigonometric identities. Let \( t = \tan\theta \). Then, \( x = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta) \). And, \( y = \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta) \). Now we have simpler parametric equations: \( x = \cos(2\theta) \) and \( y = \sin(2\theta) \). Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). For \( x = \cos(2\theta) \): \[ \frac{dx}{d\theta} = - \sin(2\theta) \cdot \frac{d}{d\theta}(2\theta) = -2 \sin(2\theta) \] For \( y = \sin(2\theta) \): \[ \frac{dy}{d\theta} = \cos(2\theta) \cdot \frac{d}{d\theta}(2\theta) = 2 \cos(2\theta) \] Now, use the formula for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). \[ \frac{dy}{dx} = \frac{2 \cos(2\theta)}{-2 \sin(2\theta)} \] \[ \frac{dy}{dx} = - \frac{\cos(2\theta)}{\sin(2\theta)} \] \[ \frac{dy}{dx} = - \cot(2\theta) \] We can also express this in terms of \( t \) by substituting back \( \cos(2\theta) = \frac{1-t^2}{1+t^2} \) and \( \sin(2\theta) = \frac{2t}{1+t^2} \): \[ \frac{dy}{dx} = - \frac{(1-t^2)/(1+t^2)}{(2t)/(1+t^2)} \] \[ \frac{dy}{dx} = - \frac{1-t^2}{2t} \] Alternatively, since \( \cot(2\theta) = \frac{1}{\tan(2\theta)} = \frac{1-\tan^2\theta}{2\tan\theta} = \frac{1-t^2}{2t} \). So, \( \frac{dy}{dx} = - \frac{1-t^2}{2t} = \frac{t^2-1}{2t} \). This substitution simplifies the problem from quotient rule to basic derivatives.In simple words: By replacing \( t \) with \( \tan\theta \), the equations for \( x \) and \( y \) become simple trigonometric identities (\( \cos(2\theta) \) and \( \sin(2\theta) \)). Then, we find the derivatives with respect to \( \theta \) and divide them. The final answer can be given in terms of \( \theta \) or back in terms of \( t \).

🎯 Exam Tip: When \( x \) and \( y \) are given in terms of \( \frac{1-t^2}{1+t^2} \) and \( \frac{2t}{1+t^2} \), always think of the substitution \( t = \tan\theta \) to convert them into \( \cos 2\theta \) and \( \sin 2\theta \), simplifying the problem greatly.

Question 16. If \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \), find \( \frac{dy}{dx} \)
Answer: We need to find the derivative of \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \). This expression resembles the cosine double angle formula, \( \cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta} \). Let \( x = \tan\theta \). Then \( \theta = \tan^{-1}x \). Substitute \( x = \tan\theta \) into the expression for \( y \): \[ y = \cos^{-1} \left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \] Using the double angle identity, \( \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta) \). So, \( y = \cos^{-1} (\cos(2\theta)) \). This simplifies to: \[ y = 2\theta \] Now, substitute \( \theta = \tan^{-1}x \) back into the equation: \[ y = 2 \tan^{-1}x \] Finally, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (2 \tan^{-1}x) \] \[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \] \[ \frac{dy}{dx} = \frac{2}{1+x^2} \] This substitution technique is very powerful for simplifying inverse trigonometric derivatives.In simple words: We changed \( x \) to \( \tan\theta \) because the expression looked like a trigonometry formula for \( \cos(2\theta) \). This made the function much simpler, turning it into just \( 2\theta \). After replacing \( \theta \) with \( \tan^{-1}x \), we could easily find the derivative.

🎯 Exam Tip: Expressions involving \( \frac{1-x^2}{1+x^2} \) are strong indicators for the substitution \( x = \tan\theta \). This transformation helps reduce inverse trigonometric functions to simpler algebraic terms before differentiation.

Question 17. Find the derivative of \( y = \sin^{-1} (3x - 4x^3) \)
Answer: We need to find the derivative of \( y = \sin^{-1} (3x - 4x^3) \). This expression inside the inverse sine function resembles the sine triple angle formula, \( \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \). Let \( x = \sin\theta \). Then \( \theta = \sin^{-1}x \). Substitute \( x = \sin\theta \) into the expression for \( y \): \[ y = \sin^{-1} (3\sin\theta - 4\sin^3\theta) \] Using the triple angle identity, \( 3\sin\theta - 4\sin^3\theta = \sin(3\theta) \). So, \( y = \sin^{-1} (\sin(3\theta)) \). This simplifies to: \[ y = 3\theta \] Now, substitute \( \theta = \sin^{-1}x \) back into the equation: \[ y = 3 \sin^{-1}x \] Finally, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (3 \sin^{-1}x) \] \[ \frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}} \] \[ \frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}} \] Recognizing the trigonometric identity greatly simplifies this problem.In simple words: We noticed the pattern inside the inverse sine looked like the formula for \( \sin(3\theta) \). By letting \( x = \sin\theta \), the function simplified to \( y = 3\theta \). After replacing \( \theta \) with \( \sin^{-1}x \), we could easily find the derivative.

🎯 Exam Tip: When you see expressions like \( (3x - 4x^3) \) inside an inverse sine, consider the substitution \( x = \sin\theta \) to use the triple angle formula \( \sin(3\theta) \).

Question 18. Find the derivative of \( y = \tan^{-1} \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) \)
Answer: We need to find the derivative of \( y = \tan^{-1} \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) \). First, simplify the expression inside the inverse tangent. Divide the numerator and denominator by \( \cos x \): \[ \frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} = \frac{1 + \tan x}{1 - \tan x} \] This expression resembles the tangent sum formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). We know that \( \tan \left(\frac{\pi}{4}\right) = 1 \). So, we can write the expression as: \[ \frac{\tan \left(\frac{\pi}{4}\right) + \tan x}{1 - \tan \left(\frac{\pi}{4}\right) \tan x} \] This simplifies to \( \tan \left(\frac{\pi}{4} + x\right) \). So, \( y = \tan^{-1} \left(\tan \left(\frac{\pi}{4} + x\right)\right) \). This simplifies to: \[ y = \frac{\pi}{4} + x \] Finally, differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} + x \right) \] \[ \frac{dy}{dx} = 0 + 1 \] \[ \frac{dy}{dx} = 1 \] This simplification using trigonometric identities makes the differentiation trivial.In simple words: We simplified the fraction inside the inverse tangent by dividing everything by \( \cos x \). This turned the fraction into a form that matches the tangent sum formula. So, the whole function simplified to \( \frac{\pi}{4} + x \), making its derivative very easy to find.

🎯 Exam Tip: When you see expressions like \( \frac{\cos x \pm \sin x}{\cos x \mp \sin x} \) inside an inverse tangent, immediately divide by \( \cos x \) to convert it into the form \( \frac{1 \pm \tan x}{1 \mp \tan x} \) which simplifies to \( \tan\left(\frac{\pi}{4} \pm x\right) \).

Question 19. Find the derivative of \( \sin x^2 \) with respect to \( x^2 \).
Answer: Let \( u = \sin x^2 \) and \( v = x^2 \). We need to find \( \frac{\mathrm{d} u}{\mathrm{d} v} \).
First, we find the derivative of \( u \) with respect to \( x \):
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} (\sin x^2) \)
Using the chain rule, this becomes:
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \cos (x^2) \cdot \frac{\mathrm{d}}{\mathrm{d} x} (x^2) \)
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \cos (x^2) \cdot 2x \)
\( \frac{\mathrm{d} u}{\mathrm{d} x} = 2x \cos (x^2) \)
Next, we find the derivative of \( v \) with respect to \( x \):
\( \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} (x^2) \)
\( \frac{\mathrm{d} v}{\mathrm{d} x} = 2x \)
Now, to find \( \frac{\mathrm{d} u}{\mathrm{d} v} \), we use the formula \( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{\frac{\mathrm{d} u}{\mathrm{d} x}}{\frac{\mathrm{d} v}{\mathrm{d} x}} \):
\( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{2x \cos (x^2)}{2x} \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} v} = \cos (x^2) \)
In simple words: To find how one function changes with respect to another, we find how each function changes with respect to a common variable and then divide them. Here, \( \cos(x^2) \) is the final rate of change.

🎯 Exam Tip: When finding the derivative of one function with respect to another, remember to use the chain rule method by finding derivatives with respect to a common variable (often \( x \)) and then dividing them.

Question 20. Find the derivative of \( \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \) with respect to \( \tan^{-1} x \).
Answer: Let the first function be \( u = \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \) and the second function be \( v = \tan^{-1} x \). We need to find \( \frac{\mathrm{d} u}{\mathrm{d} v} \).
For \( u = \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \), let's substitute \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
So, \( u = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^{2} \theta}\right) \)
We know the trigonometric identity \( \frac{2 \tan \theta}{1+\tan^{2} \theta} = \sin 2\theta \).
\( \implies \) \( u = \sin^{-1}(\sin 2\theta) \)
\( \implies \) \( u = 2\theta \)
Now, substitute \( \theta = \tan^{-1} x \) back into the equation:
\( u = 2 \tan^{-1} x \)
Now, find the derivative of \( u \) with respect to \( x \):
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} (2 \tan^{-1} x) \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} x} = 2 \cdot \frac{1}{1+x^2} \)
Next, find the derivative of \( v = \tan^{-1} x \) with respect to \( x \):
\( \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} (\tan^{-1} x) \)
\( \implies \) \( \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{1}{1+x^2} \)
Finally, we calculate \( \frac{\mathrm{d} u}{\mathrm{d} v} \):
\( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{\frac{\mathrm{d} u}{\mathrm{d} x}}{\frac{\mathrm{d} v}{\mathrm{d} x}} = \frac{\frac{2}{1+x^2}}{\frac{1}{1+x^2}} \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} v} = 2 \)
In simple words: We changed the first function using a trigonometric substitution to make it simpler, which turned out to be two times the second function. When you differentiate a function with respect to itself, the answer is 1, so differentiating twice the second function with respect to the second function gives 2.

🎯 Exam Tip: For inverse trigonometric functions, always look for suitable trigonometric substitutions like \( x=\tan \theta \) or \( x=\sin \theta \) to simplify the expression before differentiating. This makes the process much faster and reduces error.

Question 21. If \( u = \tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x} \) and \( v = \tan^{-1}x \), find \( \frac{\mathrm{d}u}{\mathrm{d}v} \).
Answer: We are given two functions: \( u = \tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) \) and \( v = \tan^{-1}x \). We need to find \( \frac{\mathrm{d} u}{\mathrm{d} v} \).
Let's simplify \( u \) first. Substitute \( x = \tan \theta \), so \( \theta = \tan^{-1} x \).
\( u = \tan^{-1}\left(\frac{\sqrt{1+\tan^{2}\theta}-1}{\tan\theta}\right) \)
We know \( 1+\tan^{2}\theta = \sec^{2}\theta \), so \( \sqrt{1+\tan^{2}\theta} = \sec\theta \).
\( \implies \) \( u = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right) \)
Now, express \( \sec\theta \) and \( \tan\theta \) in terms of \( \sin\theta \) and \( \cos\theta \):
\( u = \tan^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) \)
\( \implies \) \( u = \tan^{-1}\left(\frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}\right) \)
\( \implies \) \( u = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right) \)
Using the half-angle identities \( 1-\cos\theta = 2\sin^2\frac{\theta}{2} \) and \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \):
\( u = \tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) \)
\( \implies \) \( u = \tan^{-1}\left(\tan\frac{\theta}{2}\right) \)
\( \implies \) \( u = \frac{\theta}{2} \)
Substitute \( \theta = \tan^{-1} x \) back into \( u \):
\( u = \frac{1}{2} \tan^{-1} x \)
Now, we find the derivative of \( u \) with respect to \( x \):
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} \left(\frac{1}{2} \tan^{-1} x\right) \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{1}{2} \cdot \frac{1}{1+x^2} \)
Next, we find the derivative of \( v \) with respect to \( x \):
\( v = \tan^{-1} x \)
\( \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{1}{1+x^2} \)
Finally, calculate \( \frac{\mathrm{d} u}{\mathrm{d} v} \):
\( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{\frac{\mathrm{d} u}{\mathrm{d} x}}{\frac{\mathrm{d} v}{\mathrm{d} x}} = \frac{\frac{1}{2(1+x^2)}}{\frac{1}{1+x^2}} \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{1}{2} \)
In simple words: We simplified the first function by using a trick called substitution and half-angle formulas. It turned out to be half of the second function. So, when we find the rate of change of the first function compared to the second, the answer is just one-half.

🎯 Exam Tip: When dealing with complex inverse trigonometric functions, always try to simplify them using substitutions and trigonometric identities before differentiating. This often reduces the problem to a much simpler form.

Question 22. Find the derivative with \( \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) \) with respect to \( \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \).
Answer: Let \( u = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) \) and \( v = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \). We need to find \( \frac{\mathrm{d} u}{\mathrm{d} v} \).
First, simplify \( u \):
\( u = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) \)
Using half-angle identities: \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \) and \( 1+\cos x = 2\cos^2\frac{x}{2} \)
\( \implies \) \( u = \tan^{-1}\left(\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\right) \)
\( \implies \) \( u = \tan^{-1}\left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\right) \)
\( \implies \) \( u = \tan^{-1}\left(\tan\frac{x}{2}\right) \)
\( \implies \) \( u = \frac{x}{2} \)
Now, find the derivative of \( u \) with respect to \( x \):
\( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} \left(\frac{x}{2}\right) \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{1}{2} \)
Next, simplify \( v \):
\( v = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
We can rewrite \( \cos x \) as \( \sin(\frac{\pi}{2}-x) \) and \( 1+\sin x \) as \( 1+\cos(\frac{\pi}{2}-x) \). This helps convert it into half-angle form for sine and cosine.
\( v = \tan^{-1}\left(\frac{\sin(\frac{\pi}{2}-x)}{1+\cos(\frac{\pi}{2}-x)}\right) \)
Using the same half-angle identities as for \( u \):
\( v = \tan^{-1}\left(\tan\left(\frac{\frac{\pi}{2}-x}{2}\right)\right) \)
\( \implies \) \( v = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \)
\( \implies \) \( v = \frac{\pi}{4}-\frac{x}{2} \)
Now, find the derivative of \( v \) with respect to \( x \):
\( \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{\mathrm{d}}{\mathrm{d} x} \left(\frac{\pi}{4}-\frac{x}{2}\right) \)
\( \implies \) \( \frac{\mathrm{d} v}{\mathrm{d} x} = 0 - \frac{1}{2} \)
\( \implies \) \( \frac{\mathrm{d} v}{\mathrm{d} x} = -\frac{1}{2} \)
Finally, calculate \( \frac{\mathrm{d} u}{\mathrm{d} v} \):
\( \frac{\mathrm{d} u}{\mathrm{d} v} = \frac{\frac{\mathrm{d} u}{\mathrm{d} x}}{\frac{\mathrm{d} v}{\mathrm{d} x}} = \frac{\frac{1}{2}}{-\frac{1}{2}} \)
\( \implies \) \( \frac{\mathrm{d} u}{\mathrm{d} v} = -1 \)
In simple words: We used special trigonometric identities to simplify both functions into much simpler forms involving \( x/2 \). Then we found how each simplified function changes with respect to \( x \). Dividing these rates of change gave us the final answer, which means the first function changes in the opposite direction and at the same rate as the second one.

🎯 Exam Tip: Simplifying complex inverse trigonometric expressions using identities is crucial. Remember that \( \frac{\cos x}{1+\sin x} \) can be transformed by noting \( \cos x = \sin(\frac{\pi}{2}-x) \) and \( \sin x = \cos(\frac{\pi}{2}-x) \) to apply half-angle formulas effectively.

Question 23. If \( y = \sin^{-1}x \) then find \( y'' \).
Answer: We are given \( y = \sin^{-1}x \). We need to find the second derivative, \( y'' \).
First, find the first derivative, \( y' \):
\( y' = \frac{\mathrm{d}}{\mathrm{d} x} (\sin^{-1}x) \)
\( \implies \) \( y' = \frac{1}{\sqrt{1-x^2}} \)
To make differentiation easier for the second derivative, rewrite \( y' \) using negative exponents:
\( y' = (1-x^2)^{-1/2} \)
Now, find the second derivative, \( y'' \), by differentiating \( y' \) with respect to \( x \):
\( y'' = \frac{\mathrm{d}}{\mathrm{d} x} ((1-x^2)^{-1/2}) \)
Using the chain rule (differentiate the outer power, then the inner function):
\( y'' = -\frac{1}{2} (1-x^2)^{(-1/2)-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x} (1-x^2) \)
\( y'' = -\frac{1}{2} (1-x^2)^{-3/2} \cdot (-2x) \)
Multiply the constants and \( x \):
\( y'' = x (1-x^2)^{-3/2} \)
To express it without negative exponents, move \( (1-x^2)^{-3/2} \) to the denominator:
\( \implies \) \( y'' = \frac{x}{(1-x^2)^{3/2}} \)
In simple words: We first found the usual rate of change of \( \sin^{-1}x \). Then, we found how that rate of change itself changes, which is the second derivative. This involves using the chain rule twice because the function has layers.

🎯 Exam Tip: When finding higher-order derivatives, rewrite expressions with square roots as fractional exponents (e.g., \( \frac{1}{\sqrt{A}} = A^{-1/2} \)) to simplify the application of the power rule and chain rule.

Question 24. If \( y = e^{\tan^{-1}x} \), show that \( (1 + x^2) y'' + (2x – 1) y' = 0 \).
Answer: We are given \( y = e^{\tan^{-1}x} \). We need to show the given differential equation.
First, find the first derivative, \( y' \):
\( y' = \frac{\mathrm{d}}{\mathrm{d} x} (e^{\tan^{-1}x}) \)
Using the chain rule, \( \frac{\mathrm{d}}{\mathrm{d} x} (e^f) = e^f \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \):
\( y' = e^{\tan^{-1}x} \cdot \frac{\mathrm{d}}{\mathrm{d} x} (\tan^{-1}x) \)
\( y' = e^{\tan^{-1}x} \cdot \frac{1}{1+x^2} \)
Since \( y = e^{\tan^{-1}x} \), we can substitute \( y \) back into the expression:
\( y' = \frac{y}{1+x^2} \)
Multiply both sides by \( (1+x^2) \) to clear the denominator:
\( (1+x^2)y' = y \) ---- (Equation 1)
Now, differentiate Equation 1 again with respect to \( x \) to find \( y'' \). Use the product rule on the left side and the chain rule on the right side.
\( \frac{\mathrm{d}}{\mathrm{d} x} ((1+x^2)y') = \frac{\mathrm{d}}{\mathrm{d} x} (y) \)
\( (1+x^2) \cdot y'' + y' \cdot \frac{\mathrm{d}}{\mathrm{d} x} (1+x^2) = y' \)
\( (1+x^2)y'' + y' \cdot (2x) = y' \)
Move \( y' \) from the right side to the left side:
\( (1+x^2)y'' + 2xy' - y' = 0 \)
Factor out \( y' \) from the last two terms:
\( \implies \) \( (1+x^2)y'' + (2x-1)y' = 0 \)
This matches the equation we needed to show. This type of problem often leads to a compact form by substituting the original function back during differentiation.
In simple words: We found the first rate of change for the given function. Then, we made it simpler by moving the denominator. We differentiated this new equation a second time. After rearranging the terms, we got the exact equation we needed to prove.

🎯 Exam Tip: When proving differential equations, it's often helpful to clear denominators after the first differentiation. Also, remember to substitute the original function back into the expression for \( y' \) if it simplifies the next differentiation step.

Question 25. If \( y = \frac{\sin^{-1} x}{\sqrt{1-x^{2}}} \), show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \).
Answer: We are given \( y = \frac{\sin^{-1} x}{\sqrt{1-x^{2}}} \). We need to show the given differential equation.
First, rewrite the equation by multiplying \( \sqrt{1-x^2} \) to the left side:
\( y\sqrt{1-x^2} = \sin^{-1} x \) ---- (Equation 1)
Now, differentiate Equation 1 with respect to \( x \). Use the product rule on the left side and the standard derivative for \( \sin^{-1} x \) on the right.
\( \frac{\mathrm{d}}{\mathrm{d} x} (y\sqrt{1-x^2}) = \frac{\mathrm{d}}{\mathrm{d} x} (\sin^{-1} x) \)
\( y_1\sqrt{1-x^2} + y \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{1}{\sqrt{1-x^2}} \)
Simplify the second term on the left side:
\( y_1\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
To clear the denominators, multiply the entire equation by \( \sqrt{1-x^2} \):
\( y_1(1-x^2) - xy = 1 \) ---- (Equation 2)
Now, differentiate Equation 2 again with respect to \( x \). Use the product rule on \( y_1(1-x^2) \) and \( xy \).
\( \frac{\mathrm{d}}{\mathrm{d} x} (y_1(1-x^2)) - \frac{\mathrm{d}}{\mathrm{d} x} (xy) = \frac{\mathrm{d}}{\mathrm{d} x} (1) \)
For the first term: \( y_2(1-x^2) + y_1(-2x) \)
For the second term: \( - (1 \cdot y + x \cdot y_1) \)
For the right term: \( 0 \)
So, combining these, we get:
\( y_2(1-x^2) - 2xy_1 - y - xy_1 = 0 \)
Combine the \( y_1 \) terms:
\( \implies \) \( (1-x^2)y_2 - 3xy_1 - y = 0 \)
This is the required differential equation. It demonstrates how to derive a higher-order differential equation from a given implicit function.
In simple words: We took the first function, rearranged it to remove the fraction, and then found its rate of change. We cleaned up this rate of change by removing more fractions. Then, we found the rate of change of that new equation. By simplifying and combining terms, we arrived at the final equation we needed to show.

🎯 Exam Tip: When proving differential equations involving square roots in the denominator, always multiply by the square root term (or square both sides) before differentiating to simplify calculations and avoid complex quotient rules.

Question 26. If \( x = a (\theta + \sin \theta) \), \( y = a (1 - \cos \theta) \) then prove that at \( \theta = \frac{\pi}{2} \), \( y'' = \frac{1}{a} \).
Answer: We are given parametric equations: \( x = a (\theta + \sin \theta) \) and \( y = a (1 - \cos \theta) \). We need to prove \( y'' = \frac{1}{a} \) at \( \theta = \frac{\pi}{2} \).
First, find \( \frac{\mathrm{d} x}{\mathrm{d} \theta} \) and \( \frac{\mathrm{d} y}{\mathrm{d} \theta} \):
\( \frac{\mathrm{d} x}{\mathrm{d} \theta} = \frac{\mathrm{d}}{\mathrm{d} \theta} (a(\theta + \sin \theta)) = a(1 + \cos \theta) \)
\( \frac{\mathrm{d} y}{\mathrm{d} \theta} = \frac{\mathrm{d}}{\mathrm{d} \theta} (a(1 - \cos \theta)) = a(0 - (-\sin \theta)) = a\sin \theta \)
Now, find the first derivative \( y' = \frac{\mathrm{d} y}{\mathrm{d} x} \):
\( y' = \frac{\frac{\mathrm{d} y}{\mathrm{d} \theta}}{\frac{\mathrm{d} x}{\mathrm{d} \theta}} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta} \)
Using half-angle identities \( \sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \) and \( 1 + \cos \theta = 2\cos^2\frac{\theta}{2} \):
\( y' = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} \)
Next, find the second derivative \( y'' = \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \). We differentiate \( y' \) with respect to \( \theta \) and then multiply by \( \frac{\mathrm{d} \theta}{\mathrm{d} x} \):
\( y'' = \frac{\mathrm{d}}{\mathrm{d} x} (y') = \frac{\mathrm{d}}{\mathrm{d} \theta} (y') \cdot \frac{\mathrm{d} \theta}{\mathrm{d} x} \)
We know \( y' = \tan\frac{\theta}{2} \). So, \( \frac{\mathrm{d}}{\mathrm{d} \theta} (\tan\frac{\theta}{2}) = \sec^2\frac{\theta}{2} \cdot \frac{1}{2} \)
And \( \frac{\mathrm{d} \theta}{\mathrm{d} x} = \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} \theta}} = \frac{1}{a(1+\cos \theta)} \)
\( \implies \) \( y'' = \left(\frac{1}{2}\sec^2\frac{\theta}{2}\right) \cdot \left(\frac{1}{a(1+\cos \theta)}\right) \)
Again, use the identity \( 1+\cos \theta = 2\cos^2\frac{\theta}{2} \):
\( y'' = \frac{1}{2}\sec^2\frac{\theta}{2} \cdot \frac{1}{a(2\cos^2\frac{\theta}{2})} \)
\( \implies \) \( y'' = \frac{1}{4a} \sec^2\frac{\theta}{2} \cdot \sec^2\frac{\theta}{2} \)
\( \implies \) \( y'' = \frac{1}{4a} \sec^4\frac{\theta}{2} \)
We can also write \( \sec^2 \frac{\theta}{2} = 1 + \tan^2 \frac{\theta}{2} \), so \( y'' = \frac{1}{4a} (1 + \tan^2 \frac{\theta}{2})^2 \).
Now, evaluate \( y'' \) at \( \theta = \frac{\pi}{2} \):
When \( \theta = \frac{\pi}{2} \), then \( \frac{\theta}{2} = \frac{\pi}{4} \).
\( \tan\frac{\pi}{4} = 1 \).
\( y'' = \frac{1}{4a} (1 + \tan^2 \frac{\pi}{4})^2 \)
\( y'' = \frac{1}{4a} (1 + 1^2)^2 \)
\( y'' = \frac{1}{4a} (1 + 1)^2 \)
\( y'' = \frac{1}{4a} (2)^2 \)
\( y'' = \frac{1}{4a} \cdot 4 \)
\( \implies \) \( y'' = \frac{1}{a} \)
This proves the required statement. The use of half-angle formulas greatly simplifies the expression for \( y' \).
In simple words: We found how \( x \) and \( y \) change with respect to \( \theta \). Then, we used these to find the first rate of change of \( y \) with respect to \( x \). To get the second rate of change, we differentiated again. After plugging in the given angle \( \theta = \frac{\pi}{2} \) and simplifying, we confirmed the result.

🎯 Exam Tip: For parametric differentiation, carefully compute \( \frac{\mathrm{d} y}{\mathrm{d} x} \) first. When finding \( \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \), remember it's \( \frac{\mathrm{d}}{\mathrm{d} \theta} \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} x} \), not just \( \frac{\mathrm{d}}{\mathrm{d} \theta} \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right) \).

Question 27. If \( \sin y = x \sin (a + y) \), prove that \( \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin ^{2}(a+y)}{\sin a} \), \( a \neq n\pi \).
Answer: We are given \( \sin y = x \sin (a + y) \). We need to prove the given expression for \( \frac{\mathrm{d} y}{\mathrm{d} x} \).
First, isolate \( x \):
\( x = \frac{\sin y}{\sin (a + y)} \)
Now, differentiate \( x \) with respect to \( y \). This is often easier than differentiating \( y \) directly with respect to \( x \). Use the quotient rule: \( \frac{\mathrm{d}}{\mathrm{d} y} \left(\frac{f(y)}{g(y)}\right) = \frac{f'(y)g(y) - f(y)g'(y)}{(g(y))^2} \).
\( \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{\frac{\mathrm{d}}{\mathrm{d} y}(\sin y) \cdot \sin(a+y) - \sin y \cdot \frac{\mathrm{d}}{\mathrm{d} y}(\sin(a+y))}{(\sin(a+y))^2} \)
\( \implies \) \( \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{\cos y \sin(a+y) - \sin y \cos(a+y) \cdot 1}{\sin^2(a+y)} \)
The numerator is in the form of the trigonometric identity \( \sin A \cos B - \cos A \sin B = \sin (A-B) \). Here, \( A = a+y \) and \( B = y \).
\( \implies \) \( \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{\sin((a+y)-y)}{\sin^2(a+y)} \)
\( \implies \) \( \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{\sin a}{\sin^2(a+y)} \)
Now, to find \( \frac{\mathrm{d} y}{\mathrm{d} x} \), we take the reciprocal of \( \frac{\mathrm{d} x}{\mathrm{d} y} \):
\( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} y}} \)
\( \implies \) \( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\sin^2(a+y)}{\sin a} \)
The condition \( a \neq n\pi \) ensures that \( \sin a \neq 0 \), so the denominator is not zero. This completes the proof. This method is often preferred for implicit differentiation involving products of trigonometric functions.
In simple words: We first put the equation so that \( x \) is alone on one side. Then, we found how \( x \) changes as \( y \) changes, using a special division rule. The top part of this rule simplified nicely with a trigonometry trick. Finally, to find how \( y \) changes as \( x \) changes, we just flipped our answer upside down.

🎯 Exam Tip: When dealing with implicit differentiation, especially for expressions like \( \sin y = x \sin (a+y) \), it's often simpler to express \( x \) in terms of \( y \) and then find \( \frac{\mathrm{d} x}{\mathrm{d} y} \). Then, simply take the reciprocal to get \( \frac{\mathrm{d} y}{\mathrm{d} x} \).

Question 28. If \( y = \frac{\sin^{-1} x}{\sqrt{1-x^{2}}} \), show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \). Hence find \( y_2 \) when \( x = 0 \).
Answer: We are given \( y = \frac{\sin^{-1} x}{\sqrt{1-x^{2}}} \). We need to show the differential equation and then find \( y_2 \) at \( x=0 \).
First, rewrite the equation by multiplying \( \sqrt{1-x^2} \) to the left side:
\( y\sqrt{1-x^2} = \sin^{-1} x \) ---- (Equation A)
Now, differentiate Equation A with respect to \( x \). Use the product rule on the left side.
\( \frac{\mathrm{d}}{\mathrm{d} x} (y\sqrt{1-x^2}) = \frac{\mathrm{d}}{\mathrm{d} x} (\sin^{-1} x) \)
\( y_1\sqrt{1-x^2} + y \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{1}{\sqrt{1-x^2}} \)
Simplify the second term on the left side:
\( y_1\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
To clear the denominators, multiply the entire equation by \( \sqrt{1-x^2} \):
\( y_1(1-x^2) - xy = 1 \) ---- (Equation B)
Now, differentiate Equation B again with respect to \( x \). Use the product rule on \( y_1(1-x^2) \) and \( xy \).
\( \frac{\mathrm{d}}{\mathrm{d} x} (y_1(1-x^2)) - \frac{\mathrm{d}}{\mathrm{d} x} (xy) = \frac{\mathrm{d}}{\mathrm{d} x} (1) \)
For the first term: \( y_2(1-x^2) + y_1(-2x) \)
For the second term: \( - (1 \cdot y + x \cdot y_1) \)
For the right term: \( 0 \)
So, combining these, we get:
\( y_2(1-x^2) - 2xy_1 - y - xy_1 = 0 \)
Combine the \( y_1 \) terms:
\( \implies \) \( (1-x^2)y_2 - 3xy_1 - y = 0 \)
This proves the first part of the question.

Now, we need to find \( y_2 \) when \( x = 0 \).
From the original expression \( y = \frac{\sin^{-1} x}{\sqrt{1-x^2}} \):
When \( x = 0 \), \( y = \frac{\sin^{-1} 0}{\sqrt{1-0^2}} = \frac{0}{1} = 0 \).
From Equation B: \( y_1(1-x^2) - xy = 1 \):
When \( x = 0 \), \( y_1(1-0^2) - 0 \cdot y = 1 \)
\( \implies \) \( y_1(1) - 0 = 1 \)
\( \implies \) \( y_1 = 1 \)
Now, substitute \( x=0 \), \( y=0 \), and \( y_1=1 \) into the derived differential equation \( (1-x^2)y_2 - 3xy_1 - y = 0 \):
\( (1-0^2)y_2 - 3(0)(1) - 0 = 0 \)
\( (1)y_2 - 0 - 0 = 0 \)
\( \implies \) \( y_2 = 0 \)
Thus, the value of \( y_2 \) when \( x=0 \) is 0.
In simple words: We first found how the function changes twice, following the same steps as a previous question. Once we got the main equation, we used it to find the value of the second rate of change when \( x \) is zero. We did this by first finding the values of \( y \) and its first rate of change at \( x=0 \), then plugging those numbers into our main equation.

🎯 Exam Tip: When a question asks for a specific value of a higher-order derivative, substitute the value of \( x \) and previously calculated derivatives (like \( y \) and \( y_1 \)) into the derived differential equation to simplify the calculation significantly.

Question 1. Find the derivative of \(y = x^{\cos x}\).
Answer:
To find the derivative of \(y = x^{\cos x}\), we first take the natural logarithm on both sides to handle the variable exponent:
\( \log y = \log x^{\cos x} \)
Using logarithm properties, we bring the exponent down:
\( \log y = (\cos x) (\log x) \)
Next, we differentiate both sides with respect to \(x\). We use the product rule for the right side \((\cos x \cdot \log x)\) and the chain rule for the left side \((\log y)\):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \log x + \cos x \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{y} \frac{dy}{dx} = (-\sin x) \log x + \cos x \cdot \frac{1}{x} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x} - (\sin x) (\log x) \)
Finally, we multiply both sides by \(y\) to solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{\cos x}{x} - (\sin x) (\log x) \right] \)
Substitute back \(y = x^{\cos x}\):
\( \frac{dy}{dx} = x^{\cos x} \left[ \frac{\cos x}{x} - (\sin x) (\log x) \right] \)
This method makes it easier to differentiate functions where both the base and exponent are variables.
In simple words: To find the derivative, we use a trick called logarithmic differentiation. We take the log on both sides, differentiate it using product rule, and then solve for dy/dx.

🎯 Exam Tip: Remember to use logarithmic differentiation when the function has a variable in both its base and exponent, like \(f(x)^{g(x)}\). Always apply the product rule carefully after taking the logarithm.

Question 2. Find the derivative of \(y = x^{\log x} + (\log x)^x\).
Answer:
We need to find the derivative of \(y = x^{\log x} + (\log x)^x\). Since this is a sum of two functions, each with a variable base and exponent, we will differentiate each term separately.
Let \(u = x^{\log x}\) and \(v = (\log x)^x\). Then \(y = u + v\), and \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).

**Step 1: Differentiate \(u = x^{\log x}\)**
Take the natural logarithm on both sides:
\( \log u = \log (x^{\log x}) \)
\( \log u = (\log x) (\log x) \)
\( \log u = (\log x)^2 \)
Differentiate both sides with respect to \(x\) using the chain rule on the right:
\( \frac{1}{u} \frac{du}{dx} = 2 (\log x) \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{u} \frac{du}{dx} = 2 (\log x) \cdot \frac{1}{x} \)
\( \frac{du}{dx} = u \left( \frac{2 \log x}{x} \right) \)
Substitute back \(u = x^{\log x}\):
\( \frac{du}{dx} = x^{\log x} \left( \frac{2 \log x}{x} \right) \) (Equation 1)

**Step 2: Differentiate \(v = (\log x)^x\)**
Take the natural logarithm on both sides:
\( \log v = \log ((\log x)^x) \)
\( \log v = x \log (\log x) \)
Differentiate both sides with respect to \(x\) using the product rule on the right:
\( \frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(x) \cdot \log (\log x) + x \cdot \frac{d}{dx}(\log (\log x)) \)
\( \frac{1}{v} \frac{dv}{dx} = 1 \cdot \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{v} \frac{dv}{dx} = \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \)
\( \frac{1}{v} \frac{dv}{dx} = \log (\log x) + \frac{1}{\log x} \)
\( \frac{dv}{dx} = v \left[ \log (\log x) + \frac{1}{\log x} \right] \)
Substitute back \(v = (\log x)^x\):
\( \frac{dv}{dx} = (\log x)^x \left[ \log (\log x) + \frac{1}{\log x} \right] \) (Equation 2)

**Step 3: Combine the derivatives**
\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \)
\( \frac{dy}{dx} = x^{\log x} \left( \frac{2 \log x}{x} \right) + (\log x)^x \left[ \log (\log x) + \frac{1}{\log x} \right] \)
This method breaks down complex differentiation problems into manageable steps.
In simple words: We split the problem into two parts because of the plus sign. For each part, we take the logarithm, differentiate it, and then combine the results.

🎯 Exam Tip: When differentiating a sum of functions, each of which requires logarithmic differentiation, remember to treat them as separate terms (e.g., \(u\) and \(v\)) and then add their individual derivatives. This prevents errors from trying to differentiate the sum directly.

Question 3. If \( \sqrt{x y} = e^{(x - y)} \), find \( \frac{dy}{dx} \).
Answer:
Given the equation \( \sqrt{x y} = e^{(x - y)} \).
First, we square both sides to remove the square root:
\( (\sqrt{x y})^2 = (e^{(x - y)})^2 \)
\( x y = e^{2(x - y)} \) (Equation 1)
Now, we differentiate both sides with respect to \(x\). We use the product rule on the left side and the chain rule on the right side.
For the left side \(xy\): \( \frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} \)
For the right side \(e^{2(x-y)}\): \( \frac{d}{dx}(e^{2(x-y)}) = e^{2(x-y)} \cdot \frac{d}{dx}(2(x-y)) = e^{2(x-y)} \cdot 2 \left( 1 - \frac{dy}{dx} \right) \)
So, putting these together:
\( y + x \frac{dy}{dx} = 2 e^{2(x - y)} \left( 1 - \frac{dy}{dx} \right) \)
Distribute \(2 e^{2(x - y)}\) on the right side:
\( y + x \frac{dy}{dx} = 2 e^{2(x - y)} - 2 e^{2(x - y)} \frac{dy}{dx} \)
Collect all terms involving \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( x \frac{dy}{dx} + 2 e^{2(x - y)} \frac{dy}{dx} = 2 e^{2(x - y)} - y \)
Factor out \( \frac{dy}{dx} \) from the left side:
\( \frac{dy}{dx} (x + 2 e^{2(x - y)}) = 2 e^{2(x - y)} - y \)
Now, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{2 e^{2(x - y)} - y}{x + 2 e^{2(x - y)}} \)
From Equation 1, we know \(e^{2(x-y)} = xy\). Substitute this back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{2 (xy) - y}{x + 2 (xy)} \)
Factor out \(y\) from the numerator and \(x\) from the denominator:
\( \frac{dy}{dx} = \frac{y (2x - 1)}{x (1 + 2y)} \)
This method uses implicit differentiation and substitution to find the derivative.
In simple words: First, we remove the square root by squaring. Then we differentiate both sides carefully, gathering all terms with dy/dx. Finally, we replace \(e^{2(x-y)}\) with \(xy\) to get a simpler answer.

🎯 Exam Tip: When dealing with implicit functions, always remember to apply the chain rule correctly when differentiating terms involving \(y\) with respect to \(x\), treating \(y\) as a function of \(x\).

Question 4. If \(x^y = y^x\), find \( \frac{dy}{dx} \).
Answer:
Given the equation \(x^y = y^x\). Since both sides have variable bases and exponents, we use logarithmic differentiation.
Take the natural logarithm on both sides:
\( \log (x^y) = \log (y^x) \)
Using logarithm properties \( \log (a^b) = b \log a \):
\( y \log x = x \log y \)
Now, differentiate both sides with respect to \(x\). We will use the product rule on both sides:
For the left side \(y \log x\): \( \frac{d}{dx}(y \log x) = \frac{dy}{dx} \cdot \log x + y \cdot \frac{d}{dx}(\log x) = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
For the right side \(x \log y\): \( \frac{d}{dx}(x \log y) = \frac{d}{dx}(x) \cdot \log y + x \cdot \frac{d}{dx}(\log y) = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \)
Equating the derivatives:
\( \frac{dy}{dx} \log x + \frac{y}{x} = \log y + \frac{x}{y} \frac{dy}{dx} \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( \frac{dy}{dx} \log x - \frac{x}{y} \frac{dy}{dx} = \log y - \frac{y}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x} \)
To simplify the terms in the parenthesis and on the right side, find common denominators:
\( \frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y}{x} \frac{(x \log y - y)}{(y \log x - x)} \)
This method is crucial for functions where both base and exponent contain variables.
In simple words: Since we have variables in both the base and power, we take the logarithm on both sides. Then we differentiate each side using the product rule and rearrange the equation to find dy/dx.

🎯 Exam Tip: When performing logarithmic differentiation, always remember to differentiate \( \log y \) as \( \frac{1}{y} \frac{dy}{dx} \) using the chain rule. Pay close attention to collecting terms with \( \frac{dy}{dx} \) to one side.

Question 5. Find the derivative of \(y = (\cos x)^{\log x}\).
Answer:
Given \(y = (\cos x)^{\log x}\). This function has a variable base and a variable exponent, so we use logarithmic differentiation.
Take the natural logarithm on both sides:
\( \log y = \log ((\cos x)^{\log x}) \)
Using the logarithm property \( \log (a^b) = b \log a \):
\( \log y = (\log x) (\log (\cos x)) \)
Now, differentiate both sides with respect to \(x\). We use the product rule on the right side:
For the left side \( \log y \): \( \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \)
For the right side \( (\log x) (\log (\cos x)) \):
\( \frac{d}{dx}((\log x) (\log (\cos x))) = \frac{d}{dx}(\log x) \cdot \log (\cos x) + \log x \cdot \frac{d}{dx}(\log (\cos x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \cdot \log (\cos x) + \log x \cdot \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log (\cos x)}{x} + \log x \cdot \frac{1}{\cos x} \cdot (-\sin x) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log (\cos x)}{x} - \frac{\sin x}{\cos x} \log x \)
Since \( \frac{\sin x}{\cos x} = \tan x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log (\cos x)}{x} - \tan x \log x \)
Finally, multiply both sides by \(y\) to solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{\log (\cos x)}{x} - \tan x \log x \right] \)
Substitute back \(y = (\cos x)^{\log x}\):
\( \frac{dy}{dx} = (\cos x)^{\log x} \left[ \frac{\log (\cos x)}{x} - \tan x \log x \right] \)
This process applies the rules of logarithms and differentiation to find the derivative.
In simple words: To differentiate this, we take the natural log of both sides. This lets us use the product rule. Then we find dy/dx by isolating it.

🎯 Exam Tip: Be careful with the chain rule when differentiating \( \log (\cos x) \). It's \( \frac{1}{\cos x} \cdot (-\sin x) \), which simplifies to \( -\tan x \).

Question 6. Find the derivative of \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \).
Answer:
Given the equation \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \). We need to find \( \frac{dy}{dx} \) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\):
\( \frac{d}{dx} \left( \frac{x^{2}}{a^{2}} \right) + \frac{d}{dx} \left( \frac{y^{2}}{b^{2}} \right) = \frac{d}{dx} (1) \)
For the first term, \( \frac{1}{a^{2}} \frac{d}{dx}(x^{2}) = \frac{1}{a^{2}} (2x) = \frac{2x}{a^{2}} \).
For the second term, treat \(y\) as a function of \(x\) and use the chain rule:
\( \frac{1}{b^{2}} \frac{d}{dx}(y^{2}) = \frac{1}{b^{2}} (2y) \frac{dy}{dx} = \frac{2y}{b^{2}} \frac{dy}{dx} \).
The derivative of a constant (1) is 0.
So, the differentiated equation is:
\( \frac{2x}{a^{2}} + \frac{2y}{b^{2}} \frac{dy}{dx} = 0 \)
Now, we isolate the term containing \( \frac{dy}{dx} \):
\( \frac{2y}{b^{2}} \frac{dy}{dx} = - \frac{2x}{a^{2}} \)
Multiply both sides by \( \frac{b^{2}}{2y} \) to solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = - \frac{2x}{a^{2}} \cdot \frac{b^{2}}{2y} \)
\( \frac{dy}{dx} = - \frac{b^{2}x}{a^{2}y} \)
This derivative represents the slope of the tangent to an ellipse at any point \((x, y)\).
In simple words: We find the derivative of each part of the equation. Remember that when you differentiate a term with \(y\), you also multiply by dy/dx. Then, we rearrange the equation to find dy/dx.

🎯 Exam Tip: When implicitly differentiating, remember to apply the chain rule to any term involving \(y\). For example, \( \frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx} \).

Question 7. If \( \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right) \), find \( \frac{dy}{dx} \).
Answer:
Given the equation \( \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right) \). We will differentiate both sides with respect to \(x\) using implicit differentiation.
**Left side differentiation:** \( \frac{d}{dx} (\sqrt{x^2+y^2}) \)
Using the chain rule, \( \frac{d}{dx} (\sqrt{u}) = \frac{1}{2\sqrt{u}} \frac{du}{dx} \), where \(u = x^2+y^2\):
\( \frac{1}{2\sqrt{x^2+y^2}} \cdot \frac{d}{dx}(x^2+y^2) = \frac{1}{2\sqrt{x^2+y^2}} \cdot \left( 2x + 2y \frac{dy}{dx} \right) \)
\( = \frac{x + y \frac{dy}{dx}}{\sqrt{x^2+y^2}} \)

**Right side differentiation:** \( \frac{d}{dx} \left(\tan ^{-1}\left(\frac{y}{x}\right)\right) \)
Using the chain rule, \( \frac{d}{dx} (\tan^{-1}(u)) = \frac{1}{1+u^2} \frac{du}{dx} \), where \(u = \frac{y}{x}\):
\( \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) \)
For \( \frac{d}{dx}\left(\frac{y}{x}\right) \), use the quotient rule: \( \frac{x \frac{dy}{dx} - y \cdot 1}{x^2} \).
So, the right side becomes:
\( \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{\frac{x^2+y^2}{x^2}} \cdot \frac{x \frac{dy}{dx} - y}{x^2} \)
\( = \frac{x^2}{x^2+y^2} \cdot \frac{x \frac{dy}{dx} - y}{x^2} = \frac{x \frac{dy}{dx} - y}{x^2+y^2} \)

**Equating both sides:**
\( \frac{x + y \frac{dy}{dx}}{\sqrt{x^2+y^2}} = \frac{x \frac{dy}{dx} - y}{x^2+y^2} \)
Multiply both sides by \( \sqrt{x^2+y^2} \):
\( x + y \frac{dy}{dx} = \frac{x \frac{dy}{dx} - y}{\sqrt{x^2+y^2}} \)
Let's use the given equation \( \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right) \). We could substitute this back but it might complicate. Let's rearrange algebraically first.
Multiply by \( \sqrt{x^2+y^2} \):
\( \sqrt{x^2+y^2} (x + y \frac{dy}{dx}) = x \frac{dy}{dx} - y \)
\( x \sqrt{x^2+y^2} + y \sqrt{x^2+y^2} \frac{dy}{dx} = x \frac{dy}{dx} - y \)
Collect terms with \( \frac{dy}{dx} \) on one side:
\( y \sqrt{x^2+y^2} \frac{dy}{dx} - x \frac{dy}{dx} = -y - x \sqrt{x^2+y^2} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} (y \sqrt{x^2+y^2} - x) = - (y + x \sqrt{x^2+y^2}) \)
Therefore,
\( \frac{dy}{dx} = - \frac{y + x \sqrt{x^2+y^2}}{y \sqrt{x^2+y^2} - x} \)
This uses the chain rule multiple times for complex functions. The negative sign can be absorbed into the denominator if preferred: \( \frac{dy}{dx} = \frac{y + x \sqrt{x^2+y^2}}{x - y \sqrt{x^2+y^2}} \).
In simple words: We differentiate both sides of the equation using the chain rule and quotient rule. Then, we group all terms that have dy/dx and solve for it.

🎯 Exam Tip: For functions like \( \tan^{-1}(u) \) or \( \sqrt{u} \), remember the chain rule is crucial. When differentiating fractions like \( \frac{y}{x} \), use the quotient rule carefully. Simplifying common factors (like \(x^2\)) after differentiating helps reduce complexity.

Question 8. If \( \tan (x + y) + \tan (x - y) = x \), find \( \frac{dy}{dx} \).
Answer:
Given the equation \( \tan (x + y) + \tan (x - y) = x \). We will find \( \frac{dy}{dx} \) using implicit differentiation.
Differentiate both sides with respect to \(x\):
\( \frac{d}{dx}(\tan (x + y)) + \frac{d}{dx}(\tan (x - y)) = \frac{d}{dx}(x) \)
Using the chain rule for \( \tan(u) \), which is \( \sec^2(u) \frac{du}{dx} \):
For \( \tan(x+y) \): \( \sec^2(x+y) \cdot \frac{d}{dx}(x+y) = \sec^2(x+y) \left( 1 + \frac{dy}{dx} \right) \)
For \( \tan(x-y) \): \( \sec^2(x-y) \cdot \frac{d}{dx}(x-y) = \sec^2(x-y) \left( 1 - \frac{dy}{dx} \right) \)
The derivative of \(x\) with respect to \(x\) is 1.
So, the equation becomes:
\( \sec^2(x+y) \left( 1 + \frac{dy}{dx} \right) + \sec^2(x-y) \left( 1 - \frac{dy}{dx} \right) = 1 \)
Expand the terms:
\( \sec^2(x+y) + \sec^2(x+y) \frac{dy}{dx} + \sec^2(x-y) - \sec^2(x-y) \frac{dy}{dx} = 1 \)
Group terms with \( \frac{dy}{dx} \) on one side and other terms on the other:
\( \sec^2(x+y) \frac{dy}{dx} - \sec^2(x-y) \frac{dy}{dx} = 1 - \sec^2(x+y) - \sec^2(x-y) \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} (\sec^2(x+y) - \sec^2(x-y)) = 1 - \sec^2(x+y) - \sec^2(x-y) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1 - \sec^2(x+y) - \sec^2(x-y)}{\sec^2(x+y) - \sec^2(x-y)} \)
This uses the chain rule for trigonometric functions.
In simple words: We differentiate each tan term using the chain rule. This means multiplying by sec squared of the angle, then by the derivative of the angle. Then we collect all dy/dx terms and solve.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(\tan u) = \sec^2 u \cdot \frac{du}{dx} \). Be careful with the signs when differentiating \( (x-y) \), which yields \( (1 - \frac{dy}{dx}) \).

Question 9. If \( \cos(xy) = x \), show that \( \frac{d y}{d x}=\frac{-(1+y \sin (x y))}{x \sin x y} \).
Answer:
Given the equation \( \cos(xy) = x \). We need to show the required derivative using implicit differentiation.
Differentiate both sides with respect to \(x\):
\( \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(x) \)
For the left side, use the chain rule: \( \frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx} \), where \(u = xy\).
So, \( \frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = 1 \cdot y + x \frac{dy}{dx} \).
The derivative of \(x\) with respect to \(x\) is 1.
Putting these together:
\( -\sin(xy) \left( y + x \frac{dy}{dx} \right) = 1 \)
Distribute \(-\sin(xy)\) on the left side:
\( -y \sin(xy) - x \sin(xy) \frac{dy}{dx} = 1 \)
Move the term \( -y \sin(xy) \) to the right side:
\( -x \sin(xy) \frac{dy}{dx} = 1 + y \sin(xy) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1 + y \sin(xy)}{-x \sin(xy)} \)
This can be written as:
\( \frac{dy}{dx} = - \frac{1 + y \sin(xy)}{x \sin(xy)} \)
Thus, the given statement is proved. This method combines implicit differentiation with the chain rule.
In simple words: We differentiate both sides of the equation. On the left, we use the chain rule for cos(xy). Then we collect terms with dy/dx and solve to show the final expression.

🎯 Exam Tip: Pay close attention to negative signs when differentiating trigonometric functions, especially \( \frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx} \). Ensure all terms are correctly multiplied out before isolating \( \frac{dy}{dx} \).

Question 10. Find the derivative of \( \tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \).
Answer:
Let \(y = \tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\).
We can simplify the expression inside the square root using the half-angle trigonometric identities:
\( 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \)
\( 1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right) \)
Substitute these identities into the expression for \(y\):
\( y = \tan ^{-1} \sqrt{\frac{2 \sin^2 \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2}\right)}} \)
\( y = \tan ^{-1} \sqrt{\tan^2 \left(\frac{x}{2}\right)} \)
\( y = \tan ^{-1} \left( \tan \left(\frac{x}{2}\right) \right) \)
For a valid range, \( \tan^{-1}(\tan \theta) = \theta \). Assuming \(x\) is in a suitable range:
\( y = \frac{x}{2} \)
Now, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} \right) \)
\( \frac{dy}{dx} = \frac{1}{2} \cdot 1 \)
\( \frac{dy}{dx} = \frac{1}{2} \)
This simplification using trigonometric identities makes the differentiation straightforward.
In simple words: First, we simplify the expression inside the inverse tan using special math rules for cos x. This makes the whole expression much simpler, leaving us with just x/2. Then, finding the derivative is easy.

🎯 Exam Tip: Always look for opportunities to simplify expressions using trigonometric identities (like half-angle formulas for \(1 \pm \cos x\)) before differentiating. This can turn a complex problem into a very simple one.

Question 11. Find the derivative of \( \tan ^{-1} \left(\frac{6x}{1-9 x^{2}}\right) \).
Answer:
Let \(y = \tan ^{-1} \left(\frac{6x}{1-9 x^{2}}\right)\).
We can simplify this using a trigonometric substitution. Recall the identity \( \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \).
Let \(3x = \tan \theta\). Then \(x = \frac{1}{3} \tan \theta\).
Substitute \(3x\) into the expression for \(y\):
\( y = \tan ^{-1} \left(\frac{2 \cdot (3x)}{1-(3x)^2}\right) \)
\( y = \tan ^{-1} \left(\frac{2 \tan \theta}{1-\tan^2 \theta}\right) \)
Using the identity for \( \tan(2\theta) \):
\( y = \tan ^{-1} (\tan (2\theta)) \)
Assuming \( \theta \) is in the principal value range for \( \tan^{-1} \):
\( y = 2\theta \)
Now, substitute back \( \theta = \tan^{-1}(3x) \):
\( y = 2 \tan^{-1}(3x) \)
Differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1}(3x)) \)
Using the chain rule \( \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1+u^2} \frac{du}{dx} \):
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x)^2} \cdot \frac{d}{dx}(3x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+9x^2} \cdot 3 \)
\( \frac{dy}{dx} = \frac{6}{1+9x^2} \)
This substitution method simplifies the problem significantly.
In simple words: We make a substitution \(3x = \tan \theta\). This changes the expression into a simpler form using a trigonometric identity, which helps us easily find the derivative.

🎯 Exam Tip: Look for expressions resembling \( \frac{2A}{1-A^2} \) (for \( \tan(2\theta) \)) or \( \frac{2A}{1+A^2} \) (for \( \sin(2\theta) \)) or \( \frac{1-A^2}{1+A^2} \) (for \( \cos(2\theta) \)) when dealing with inverse trigonometric functions. These forms often suggest a substitution like \( A = \tan \theta \).

Question 12. Find the derivative of \( y = \cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right) \).
Answer:
Let \(y = \cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\).
We can simplify the expression inside the inverse tangent using a trigonometric substitution. Let \(x = \cos(2\theta)\). Then \( \theta = \frac{1}{2} \cos^{-1}(x) \).
Substitute \(x = \cos(2\theta)\) into the expression:
\( y = \cos \left(2 \tan ^{-1} \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\right) \)
Use the double-angle identities: \( 1-\cos(2\theta) = 2\sin^2\theta \) and \( 1+\cos(2\theta) = 2\cos^2\theta \).
\( y = \cos \left(2 \tan ^{-1} \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\right) \)
\( y = \cos \left(2 \tan ^{-1} \sqrt{\tan^2\theta}\right) \)
\( y = \cos \left(2 \tan ^{-1} (\tan\theta)\right) \)
Assuming \( \theta \) is in the principal value range:
\( y = \cos (2\theta) \)
We started with \(x = \cos(2\theta)\), so now we have \(y = x\).
Differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(x) \)
\( \frac{dy}{dx} = 1 \)
This trigonometric substitution greatly simplifies the function before differentiation.
In simple words: We make a substitution for \(x\) using a cosine double angle. This simplifies the inverse tangent part of the equation, making the whole function just \(y=x\). Then, finding the derivative is very easy.

🎯 Exam Tip: For expressions involving \( \sqrt{\frac{1-x}{1+x}} \), a common and effective substitution is \( x = \cos(2\theta) \). This helps simplify the expression inside the square root using double-angle identities.

Question 13. If \( x = a \cos^3 t \) and \( y = a \sin^3 t \), find \( \frac{dy}{dx} \).
Answer:
Given the parametric equations \( x = a \cos^3 t \) and \( y = a \sin^3 t \). We need to find \( \frac{dy}{dx} \).
First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
Differentiate \(x\) with respect to \(t\):
\( \frac{dx}{dt} = \frac{d}{dt}(a \cos^3 t) = a \cdot 3 \cos^2 t \cdot \frac{d}{dt}(\cos t) = 3a \cos^2 t (-\sin t) \)
\( \frac{dx}{dt} = -3a \cos^2 t \sin t \)
Differentiate \(y\) with respect to \(t\):
\( \frac{dy}{dt} = \frac{d}{dt}(a \sin^3 t) = a \cdot 3 \sin^2 t \cdot \frac{d}{dt}(\sin t) = 3a \sin^2 t (\cos t) \)
\( \frac{dy}{dt} = 3a \sin^2 t \cos t \)
Now, use the chain rule for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
\( \frac{dy}{dx} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \)
Cancel out common terms \(3a\), \( \sin t \), and \( \cos t \):
\( \frac{dy}{dx} = - \frac{\sin t}{\cos t} \)
\( \frac{dy}{dx} = - \tan t \)
This method is used when \(x\) and \(y\) are both defined by a third variable, \(t\).
In simple words: We find the derivative of \(x\) with respect to \(t\) and \(y\) with respect to \(t\). Then, we divide \(dy/dt\) by \(dx/dt\) to get \(dy/dx\).

🎯 Exam Tip: For parametric equations, remember the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Be careful when differentiating powers of trigonometric functions (e.g., \( \cos^3 t \)), as you need to use the chain rule twice.

Question 14. If \( x = a (\cos t + t \sin t) \) and \( y = a (\sin t - t \cos t) \), find \( \frac{dy}{dx} \).
Answer:
Given the parametric equations \( x = a (\cos t + t \sin t) \) and \( y = a (\sin t - t \cos t) \). We need to find \( \frac{dy}{dx} \).
First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
Differentiate \(x\) with respect to \(t\):
\( \frac{dx}{dt} = \frac{d}{dt} [a (\cos t + t \sin t)] \)
\( \frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(t \sin t) \right] \)
Using the product rule for \(t \sin t\): \( \frac{d}{dt}(t \sin t) = 1 \cdot \sin t + t \cdot \cos t \).
\( \frac{dx}{dt} = a [-\sin t + \sin t + t \cos t] \)
\( \frac{dx}{dt} = a (t \cos t) \) (Equation 1)
Differentiate \(y\) with respect to \(t\):
\( \frac{dy}{dt} = \frac{d}{dt} [a (\sin t - t \cos t)] \)
\( \frac{dy}{dt} = a \left[ \frac{d}{dt}(\sin t) - \frac{d}{dt}(t \cos t) \right] \)
Using the product rule for \(t \cos t\): \( \frac{d}{dt}(t \cos t) = 1 \cdot \cos t + t \cdot (-\sin t) = \cos t - t \sin t \).
\( \frac{dy}{dt} = a [\cos t - (\cos t - t \sin t)] \)
\( \frac{dy}{dt} = a [\cos t - \cos t + t \sin t] \)
\( \frac{dy}{dt} = a (t \sin t) \) (Equation 2)
Now, use the chain rule for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
\( \frac{dy}{dx} = \frac{a (t \sin t)}{a (t \cos t)} \)
Cancel out common terms \(a\) and \(t\):
\( \frac{dy}{dx} = \frac{\sin t}{\cos t} \)
\( \frac{dy}{dx} = \tan t \)
This problem demonstrates careful application of the product rule within parametric differentiation.
In simple words: We find the derivative of \(x\) with respect to \(t\) and \(y\) with respect to \(t\). For terms like \(t \sin t\), we use the product rule. Then, we divide \(dy/dt\) by \(dx/dt\) to find \(dy/dx\).

🎯 Exam Tip: When using the product rule, especially with subtraction, be careful with signs. For example, \( \frac{d}{dt}(t \cos t) = \cos t - t \sin t \), so \( -\frac{d}{dt}(t \cos t) \) becomes \( -(\cos t - t \sin t) = -\cos t + t \sin t \).

Question 15. If \( x = \frac{1-t^{2}}{1+t^{2}} \) and \( y = \frac{2t}{1+t^{2}} \), find \( \frac{dy}{dx} \).
Answer:
Given the parametric equations \( x = \frac{1-t^{2}}{1+t^{2}} \) and \( y = \frac{2t}{1+t^{2}} \). We need to find \( \frac{dy}{dx} \).
We can use a trigonometric substitution to simplify these expressions. Let \(t = \tan \theta\).
Then \( x = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \). This is the identity for \( \cos(2\theta) \). So, \( x = \cos(2\theta) \).
And \( y = \frac{2\tan \theta}{1+\tan^2 \theta} \). This is the identity for \( \sin(2\theta) \). So, \( y = \sin(2\theta) \).
Now we have simpler parametric equations: \( x = \cos(2\theta) \) and \( y = \sin(2\theta) \).
First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
Differentiate \(x\) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(\cos(2\theta)) = -\sin(2\theta) \cdot \frac{d}{d\theta}(2\theta) = -\sin(2\theta) \cdot 2 \)
\( \frac{dx}{d\theta} = -2\sin(2\theta) \)
Differentiate \(y\) with respect to \( \theta \):
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin(2\theta)) = \cos(2\theta) \cdot \frac{d}{d\theta}(2\theta) = \cos(2\theta) \cdot 2 \)
\( \frac{dy}{d\theta} = 2\cos(2\theta) \)
Now, use the chain rule for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
\( \frac{dy}{dx} = \frac{2\cos(2\theta)}{-2\sin(2\theta)} \)
Cancel out 2:
\( \frac{dy}{dx} = - \frac{\cos(2\theta)}{\sin(2\theta)} \)
\( \frac{dy}{dx} = - \cot(2\theta) \)
Substitute back \( x = \cos(2\theta) \). Since \( \cot(2\theta) = \frac{\cos(2\theta)}{\sin(2\theta)} \):
\( \frac{dy}{dx} = - \frac{x}{\sqrt{1-x^2}} \)
Alternatively, since \( t = \tan \theta \), we know \( 2\theta = \tan^{-1} \left( \frac{2t}{1-t^2} \right) \). No, better to keep it in terms of \( \cos(2\theta) \).
We have \( \cos(2\theta) = x \) and \( \sin(2\theta) = y \).
So, \( \frac{dy}{dx} = - \frac{x}{y} \). This is another simple form.
In simple words: We substitute \(t = \tan \theta\) to make \(x\) and \(y\) become simpler trigonometric functions of \(2\theta\). Then, we find their derivatives with respect to \( \theta \) and divide them to get \(dy/dx\).

🎯 Exam Tip: Recognizing the trigonometric forms for \( \cos(2\theta) \) and \( \sin(2\theta) \) in terms of \( \tan \theta \) is key for simplifying these types of parametric equations quickly. This reduces calculation steps significantly.

Question 16. Find the derivative of \( y = \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \).
Answer:
Let \(y = \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\).
We use a trigonometric substitution to simplify the expression inside the inverse cosine. Recall the identity \( \cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \).
Let \(x = \tan \theta\). Then \( \theta = \tan^{-1}(x) \).
Substitute \(x = \tan \theta\) into the expression for \(y\):
\( y = \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \)
Using the identity for \( \cos(2\theta) \):
\( y = \cos^{-1}(\cos (2\theta)) \)
Assuming \( \theta \) is in the principal value range for \( \cos^{-1} \):
\( y = 2\theta \)
Now, substitute back \( \theta = \tan^{-1}(x) \):
\( y = 2 \tan^{-1}(x) \)
Differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1}(x)) \)
Using the derivative rule \( \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \):
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
This substitution greatly simplifies the differentiation process.
In simple words: We replace \(x\) with \( \tan \theta \). This transforms the expression inside the inverse cosine into a simpler trigonometric form. The whole function becomes much simpler to differentiate.

🎯 Exam Tip: When you see \( \frac{1-x^2}{1+x^2} \) inside an inverse trigonometric function, immediately think of the substitution \( x = \tan \theta \) because it simplifies to \( \cos(2\theta) \).

Question 17. Find the derivative of \( \sin^{-1} (3x – 4x^3) \).
Answer:
Let \(y = \sin^{-1} (3x – 4x^3)\).
We use a trigonometric substitution to simplify the expression inside the inverse sine. Recall the identity \( \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \).
Let \(x = \sin \theta\). Then \( \theta = \sin^{-1}(x) \).
Substitute \(x = \sin \theta\) into the expression for \(y\):
\( y = \sin^{-1} (3\sin\theta - 4\sin^3\theta) \)
Using the identity for \( \sin(3\theta) \):
\( y = \sin^{-1}(\sin (3\theta)) \)
Assuming \( \theta \) is in the principal value range for \( \sin^{-1} \):
\( y = 3\theta \)
Now, substitute back \( \theta = \sin^{-1}(x) \):
\( y = 3 \sin^{-1}(x) \)
Differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(3 \sin^{-1}(x)) \)
Using the derivative rule \( \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}} \):
\( \frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}} \)
This substitution streamlines the differentiation process significantly.
In simple words: We replace \(x\) with \( \sin \theta \). This changes the expression inside the inverse sine into a simpler form using a triple angle identity. The whole function becomes much simpler to differentiate.

🎯 Exam Tip: Recognize the form \(3x - 4x^3\) as a hint for the identity \( \sin(3\theta) = 3\sin\theta - 4\sin^3\theta \). This substitution is commonly used to simplify inverse sine functions.

Question 18. Find the derivative of \( \tan^{-1} \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) \).
Answer:
Let \(y = \tan^{-1} \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\).
We can simplify the expression inside the inverse tangent by dividing both the numerator and the denominator by \( \cos x \):
\( y = \tan^{-1} \left(\frac{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right) \)
\( y = \tan^{-1} \left(\frac{1+\tan x}{1-\tan x}\right) \)
Recall the tangent addition formula: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).
We know that \( \tan(\frac{\pi}{4}) = 1 \). So, we can write the expression as:
\( y = \tan^{-1} \left(\frac{\tan(\frac{\pi}{4})+\tan x}{1-\tan(\frac{\pi}{4})\tan x}\right) \)
Using the tangent addition formula:
\( y = \tan^{-1} \left( \tan \left(\frac{\pi}{4} + x\right) \right) \)
Assuming \( (\frac{\pi}{4} + x) \) is in the principal value range for \( \tan^{-1} \):
\( y = \frac{\pi}{4} + x \)
Now, differentiate \(y\) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + x\right) \)
\( \frac{dy}{dx} = 0 + 1 \)
\( \frac{dy}{dx} = 1 \)
This simplification using trigonometric identities and division by \( \cos x \) makes the problem much easier.
In simple words: We divide the top and bottom of the fraction inside inverse tan by cos x. This transforms the expression into a standard tangent addition formula. After simplifying, the whole function becomes very easy to differentiate.

🎯 Exam Tip: When you encounter fractions like \( \frac{\cos x \pm \sin x}{\cos x \mp \sin x} \) inside an inverse tangent, divide the numerator and denominator by \( \cos x \) to simplify it into the form of \( \tan(\frac{\pi}{4} \pm x) \).

Question 19. Find the derivative of \( \sin (x^2) \) with respect to \( x^2 \).
Answer:
To find the derivative of \( \sin (x^2) \) with respect to \( x^2 \), we can use a substitution.
Let \(u = \sin (x^2)\) and \(v = x^2\). We need to find \( \frac{du}{dv} \).
First, find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx}(\sin (x^2)) \)
Using the chain rule, \( \frac{d}{dx}(\sin w) = \cos w \frac{dw}{dx} \), where \(w = x^2\):
\( \frac{du}{dx} = \cos (x^2) \cdot \frac{d}{dx}(x^2) \)
\( \frac{du}{dx} = \cos (x^2) \cdot (2x) \)
\( \frac{du}{dx} = 2x \cos (x^2) \)
Next, find \( \frac{dv}{dx} \):
\( \frac{dv}{dx} = \frac{d}{dx}(x^2) \)
\( \frac{dv}{dx} = 2x \)
Now, use the chain rule for finding \( \frac{du}{dv} \):
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} \)
\( \frac{du}{dv} = \frac{2x \cos (x^2)}{2x} \)
Cancel out \(2x\) (assuming \(x \neq 0\)):
\( \frac{du}{dv} = \cos (x^2) \)
This method makes it clear how to differentiate with respect to a function of \(x\) rather than \(x\) itself.
In simple words: We want to find how fast \( \sin(x^2) \) changes when \(x^2\) changes. We treat \(x^2\) as a new variable. We find the derivative of \( \sin(x^2) \) with respect to \(x\) and the derivative of \(x^2\) with respect to \(x\). Then we divide them to get the answer.

🎯 Exam Tip: When asked to find the derivative of \( f(x) \) with respect to \( g(x) \), let \(u = f(x)\) and \(v = g(x)\). Then apply the chain rule formula \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \). Ensure you differentiate both \(u\) and \(v\) correctly with respect to \(x\).

Question 20. Find the derivative of \( \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \) with respect to \( \tan^{-1} x \).
Answer:
We need to find the derivative of \( u = \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \) with respect to \( v = \tan^{-1} x \).
We will find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) separately, then use \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \).

**Step 1: Find \( \frac{du}{dx} \) for \( u = \sin^{-1}\left(\frac{2 x}{1+x^{2}}\right) \)**
Let \(x = \tan \theta\). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the expression for \(u\):
\( u = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^{2}\theta}\right) \)
Recall the identity \( \sin(2\theta) = \frac{2 \tan \theta}{1+\tan^{2}\theta} \):
\( u = \sin^{-1}(\sin (2\theta)) \)
Assuming \( \theta \) is in the principal value range:
\( u = 2\theta \)
Substitute back \( \theta = \tan^{-1} x \):
\( u = 2 \tan^{-1} x \)
Now, differentiate \(u\) with respect to \(x\):
\( \frac{du}{dx} = \frac{d}{dx}(2 \tan^{-1} x) \)
\( \frac{du}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{du}{dx} = \frac{2}{1+x^2} \)

**Step 2: Find \( \frac{dv}{dx} \) for \( v = \tan^{-1} x \)**
\( \frac{dv}{dx} = \frac{d}{dx}(\tan^{-1} x) \)
\( \frac{dv}{dx} = \frac{1}{1+x^2} \)

**Step 3: Find \( \frac{du}{dv} \)**
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{2}{1+x^2}}{\frac{1}{1+x^2}} \)
\( \frac{du}{dv} = 2 \)
This problem demonstrates a common simplification using trigonometric identities before applying the derivative rule.
In simple words: We let the first function be \(u\) and the second be \(v\). We use a trigonometric trick to simplify \(u\) first. Then, we find the derivative of \(u\) with respect to \(x\) and the derivative of \(v\) with respect to \(x\). Finally, we divide these two derivatives to get the answer.

🎯 Exam Tip: The form \( \frac{2x}{1+x^2} \) is a strong indicator for the substitution \( x = \tan \theta \), as it simplifies to \( \sin(2\theta) \). This makes the differentiation of the inverse sine function much simpler.

Question 21. If \( u = \tan ^{-1}\frac{\sqrt{1+x^{2}}-1}{x} \) and \( v = \tan^{-1}x \), find \( \frac{\mathrm{du}}{\mathrm{dv}} \).
Answer:
We need to find \( \frac{du}{dv} \), where \( u = \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) \) and \( v = \tan^{-1}x \).
We will find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) separately, then use \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \).

**Step 1: Simplify \(u\) and find \( \frac{du}{dx} \)**
Let \(x = \tan \theta\). Then \( \theta = \tan^{-1} x \).
Substitute \(x = \tan \theta\) into the expression for \(u\):
\( u = \tan ^{-1}\left(\frac{\sqrt{1+\tan^{2}\theta}-1}{\tan \theta}\right) \)
Recall \( 1+\tan^2\theta = \sec^2\theta \), so \( \sqrt{1+\tan^2\theta} = \sec\theta \).
\( u = \tan ^{-1}\left(\frac{\sec\theta-1}{\tan \theta}\right) \)
Express \( \sec\theta \) and \( \tan\theta \) in terms of \( \sin\theta \) and \( \cos\theta \):
\( u = \tan ^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) \)
Combine the terms in the numerator:
\( u = \tan ^{-1}\left(\frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}\right) \)
Cancel out \( \cos\theta \):
\( u = \tan ^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right) \)
Now, use the half-angle identities: \( 1-\cos\theta = 2\sin^2\left(\frac{\theta}{2}\right) \) and \( \sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \).
\( u = \tan ^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) \)
Cancel out \( 2\sin\left(\frac{\theta}{2}\right) \):
\( u = \tan ^{-1}\left(\frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\right) \)
\( u = \tan ^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) \)
Assuming \( \theta/2 \) is in the principal value range:
\( u = \frac{\theta}{2} \)
Substitute back \( \theta = \tan^{-1} x \):
\( u = \frac{1}{2} \tan^{-1} x \)
Now, differentiate \(u\) with respect to \(x\):
\( \frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{2} \tan^{-1} x\right) \)
\( \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \)
\( \frac{du}{dx} = \frac{1}{2(1+x^2)} \)

**Step 2: Find \( \frac{dv}{dx} \) for \( v = \tan^{-1} x \)**
\( \frac{dv}{dx} = \frac{d}{dx}(\tan^{-1} x) \)
\( \frac{dv}{dx} = \frac{1}{1+x^2} \)

**Step 3: Find \( \frac{du}{dv} \)**
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2(1+x^2)}}{\frac{1}{1+x^2}} \)
\( \frac{du}{dv} = \frac{1}{2} \)
This problem shows how extensive simplification using trigonometric identities can lead to a very simple derivative.
In simple words: We use a trigonometric substitution to simplify the expression for \(u\) greatly. After many steps, \(u\) becomes \( \frac{1}{2} \tan^{-1} x \). Then, we find the derivatives of \(u\) and \(v\) with respect to \(x\) and divide them.

🎯 Exam Tip: For expressions like \( \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) \), the substitution \( x = \tan \theta \) is very effective. Remember to simplify using identities for \( \sec \theta \), \( \tan \theta \), and then half-angle formulas for \( 1-\cos \theta \) and \( \sin \theta \).

Question 22. Find the derivative of \( \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) \) with respect to \( \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \).
Answer:
Let \(u = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)\) and \(v = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)\). We need to find \( \frac{du}{dv} \).
We will simplify \(u\) and \(v\) first, then find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

**Step 1: Simplify \(u\) and find \( \frac{du}{dx} \)**
\( u = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right) \)
Use half-angle identities: \( \sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \) and \( 1+\cos x = 2\cos^2\left(\frac{x}{2}\right) \).
\( u = \tan^{-1}\left(\frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)}\right) \)
\( u = \tan^{-1}\left(\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}\right) \)
\( u = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) \)
Assuming \( x/2 \) is in the principal value range:
\( u = \frac{x}{2} \)
Differentiate \(u\) with respect to \(x\):
\( \frac{du}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2} \)

**Step 2: Simplify \(v\) and find \( \frac{dv}{dx} \)**
\( v = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right) \)
Use half-angle identities for \( \cos x \) and \( \sin x \), and the Pythagorean identity. It's often easier to rewrite \( \cos x \) and \( \sin x \) using angle shifts:
\( \cos x = \sin\left(\frac{\pi}{2}-x\right) \)
\( 1+\sin x = 1+\cos\left(\frac{\pi}{2}-x\right) \)
Now apply half-angle identities to these shifted terms:
\( \sin\left(\frac{\pi}{2}-x\right) = 2\sin\left(\frac{\frac{\pi}{2}-x}{2}\right)\cos\left(\frac{\frac{\pi}{2}-x}{2}\right) = 2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right) \)
\( 1+\cos\left(\frac{\pi}{2}-x\right) = 2\cos^2\left(\frac{\frac{\pi}{2}-x}{2}\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right) \)
Substitute these into the expression for \(v\):
\( v = \tan^{-1}\left(\frac{2\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right) \)
\( v = \tan^{-1}\left(\frac{\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)}{\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right) \)
\( v = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \)
Assuming \( \left(\frac{\pi}{4}-\frac{x}{2}\right) \) is in the principal value range:
\( v = \frac{\pi}{4} - \frac{x}{2} \)
Differentiate \(v\) with respect to \(x\):
\( \frac{dv}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2} \)

**Step 3: Find \( \frac{du}{dv} \)**
\( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2}}{-\frac{1}{2}} \)
\( \frac{du}{dv} = -1 \)
This problem requires using half-angle identities and sometimes angle shifts to simplify expressions.
In simple words: We simplify both expressions, \(u\) and \(v\), using special trigonometric rules to make them much simpler forms of \(x\). Then we find their derivatives with respect to \(x\) and divide them to get the final answer.

🎯 Exam Tip: For expressions like \( \frac{\sin x}{1+\cos x} \) and \( \frac{\cos x}{1+\sin x} \), using half-angle formulas is key. For the latter, it's often helpful to first rewrite \( \cos x \) as \( \sin(\frac{\pi}{2}-x) \) and \( \sin x \) as \( \cos(\frac{\pi}{2}-x) \) to apply the half-angle identities in a familiar form.

Question 23. If \( y = \sin^{-1}x \) then find \(y''\).
Answer:
Given \( y = \sin^{-1}x \). We need to find the second derivative, \(y''\).
First, find the first derivative, \(y'\):
\( y' = \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} \)
To make it easier to differentiate again, rewrite this expression using exponents:
\( y' = (1-x^2)^{-1/2} \)
Now, find the second derivative, \(y''\), by differentiating \(y'\) with respect to \(x\). Use the chain rule:
\( y'' = \frac{d}{dx}((1-x^2)^{-1/2}) \)
\( y'' = -\frac{1}{2} (1-x^2)^{(-1/2)-1} \cdot \frac{d}{dx}(1-x^2) \)
\( y'' = -\frac{1}{2} (1-x^2)^{-3/2} \cdot (-2x) \)
Multiply the terms:
\( y'' = x (1-x^2)^{-3/2} \)
We can also write this with a positive exponent and a square root:
\( y'' = \frac{x}{(1-x^2)^{3/2}} \)
This derivative calculation involves two applications of the chain rule.
In simple words: First, we find the derivative of \( \sin^{-1}x \). Then we rewrite this derivative using powers to make it easier to differentiate a second time using the chain rule.

🎯 Exam Tip: When finding higher-order derivatives, converting radical expressions like \( \sqrt{1-x^2} \) to exponential form \( (1-x^2)^{1/2} \) often simplifies the application of the power rule and chain rule.

Question 24. If \( y = e^{\tan^{-1}x} \), show that \( (1 + x^2) y'' + (2x – 1) y' = 0 \).
Answer:
Given \( y = e^{\tan^{-1}x} \). We need to show that \( (1 + x^2) y'' + (2x – 1) y' = 0 \).
First, find the first derivative, \(y'\):
\( y' = \frac{d}{dx}(e^{\tan^{-1}x}) \)
Using the chain rule: \( \frac{d}{dx}(e^u) = e^u \frac{du}{dx} \), where \(u = \tan^{-1}x\).
\( y' = e^{\tan^{-1}x} \cdot \frac{d}{dx}(\tan^{-1}x) \)
\( y' = e^{\tan^{-1}x} \cdot \frac{1}{1+x^2} \)
From the original equation, \( e^{\tan^{-1}x} = y \). Substitute this into the expression for \(y'\):
\( y' = \frac{y}{1+x^2} \)
Rearrange this to get rid of the fraction, which often helps with the second derivative:
\( y'(1+x^2) = y \) (Equation 1)
Now, differentiate Equation 1 with respect to \(x\) to find \(y''\). Use the product rule on the left side:
\( \frac{d}{dx}(y'(1+x^2)) = \frac{d}{dx}(y) \)
\( y'' (1+x^2) + y' \cdot \frac{d}{dx}(1+x^2) = y' \)
\( y'' (1+x^2) + y' (2x) = y' \)
Move all terms to one side to match the target equation:
\( y'' (1+x^2) + 2xy' - y' = 0 \)
Factor out \(y'\) from the last two terms:
\( (1+x^2) y'' + (2x - 1) y' = 0 \)
This proves the given statement. The steps involve both the chain rule and the product rule in sequence.
In simple words: First, we find the derivative of \(y\). Then we rearrange it to remove fractions. Next, we differentiate this new equation again using the product rule. Finally, we simplify to show the given equation.

🎯 Exam Tip: When asked to prove a relationship involving \(y''\), it's often helpful to clear denominators after finding \(y'\) to simplify the process of calculating \(y''\). Keep the original function \(y\) in mind for potential substitutions.

Question 25. If \( y = \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \), show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \).
Answer:
Given \( y = \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \). We need to show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \), where \(y_1 = \frac{dy}{dx}\) and \(y_2 = \frac{d^2y}{dx^2}\).
First, rearrange the given equation to clear the denominator:
\( y\sqrt{1-x^2} = \sin^{-1} x \)
Now, differentiate both sides with respect to \(x\). Use the product rule for the left side and the chain rule for \( \sqrt{1-x^2} \):
\( y_1\sqrt{1-x^2} + y \cdot \frac{d}{dx}(\sqrt{1-x^2}) = \frac{d}{dx}(\sin^{-1} x) \)
\( y_1\sqrt{1-x^2} + y \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{1}{\sqrt{1-x^2}} \)
\( y_1\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
To remove the remaining denominators, multiply the entire equation by \( \sqrt{1-x^2} \):
\( y_1(1-x^2) - xy = 1 \) (Equation 1)
Now, differentiate Equation 1 with respect to \(x\) to find \(y_2\). Use the product rule for \( y_1(1-x^2) \) and \( xy \):
\( \frac{d}{dx}(y_1(1-x^2)) - \frac{d}{dx}(xy) = \frac{d}{dx}(1) \)
\( [y_2(1-x^2) + y_1(-2x)] - [1 \cdot y + x \cdot y_1] = 0 \)
\( y_2(1-x^2) - 2xy_1 - y - xy_1 = 0 \)
Combine the terms involving \(xy_1\):
\( (1-x^2)y_2 - 3xy_1 - y = 0 \)
This proves the given statement. This solution uses implicit differentiation and the product rule multiple times.
In simple words: First, we clear the square root from the denominator. Then, we differentiate both sides once. After simplifying, we differentiate the equation again. This gives us the second derivative, and then we arrange the terms to match the required equation.

🎯 Exam Tip: For problems involving \( \sin^{-1}x \) and \( \sqrt{1-x^2} \), always try to clear denominators by multiplying by \( \sqrt{1-x^2} \) as an initial step. This usually simplifies the subsequent differentiation process significantly.

Question 26. If \( x = a (\theta + \sin \theta) \) and \( y = a (1 – \cos \theta) \), then prove that at \( \theta = \frac{\pi}{2} \), \( y'' = \frac{1}{a} \).
Answer:
Given the parametric equations \( x = a (\theta + \sin \theta) \) and \( y = a (1 – \cos \theta) \). We need to prove that \( y'' = \frac{1}{a} \) at \( \theta = \frac{\pi}{2} \).

**Step 1: Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)**
Differentiate \(x\) with respect to \( \theta \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(a (\theta + \sin \theta)) = a (1 + \cos \theta) \)
Differentiate \(y\) with respect to \( \theta \):
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(a (1 – \cos \theta)) = a (0 - (-\sin \theta)) = a \sin \theta \)

**Step 2: Find \( \frac{dy}{dx} \)**
Using the chain rule: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \)
\( \frac{dy}{dx} = \frac{a \sin \theta}{a (1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta} \)
Use half-angle identities: \( \sin \theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \) and \( 1+\cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) \).
\( \frac{dy}{dx} = \frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \)
\( \frac{dy}{dx} = \tan\left(\frac{\theta}{2}\right) \)

**Step 3: Find \( \frac{d^2y}{dx^2} \) (which is \(y''\))**
Recall that \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx} \).
We know \( \frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{a(1+\cos \theta)} \).
First, find \( \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \):
\( \frac{d}{d\theta}\left(\tan\left(\frac{\theta}{2}\right)\right) = \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{d}{d\theta}\left(\frac{\theta}{2}\right) = \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} \)
Now, substitute this into the formula for \(y''\):
\( y'' = \frac{1}{2} \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{a(1+\cos \theta)} \)
Use the identity \( 1+\cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) \):
\( y'' = \frac{1}{2} \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{a \cdot 2\cos^2\left(\frac{\theta}{2}\right)} \)
Recall \( \sec\theta = \frac{1}{\cos\theta} \):
\( y'' = \frac{1}{2} \cdot \frac{1}{\cos^2\left(\frac{\theta}{2}\right)} \cdot \frac{1}{2a \cos^2\left(\frac{\theta}{2}\right)} \)
\( y'' = \frac{1}{4a} \cdot \frac{1}{\cos^4\left(\frac{\theta}{2}\right)} \)
Alternatively, using \( \sec^2\left(\frac{\theta}{2}\right) = 1 + \tan^2\left(\frac{\theta}{2}\right) \):
\( y'' = \frac{1}{2a(1+\cos\theta)} \sec^2\left(\frac{\theta}{2}\right) = \frac{1}{2a(2\cos^2(\theta/2))} (1+\tan^2(\theta/2)) \)
\( y'' = \frac{1}{4a \cos^2(\theta/2)} (1+\tan^2(\theta/2)) = \frac{1}{4a} \sec^2(\theta/2) (1+\tan^2(\theta/2)) \)
\( y'' = \frac{1}{4a} (1+\tan^2(\theta/2))(1+\tan^2(\theta/2)) = \frac{1}{4a} (1+\tan^2(\theta/2))^2 \).

**Step 4: Evaluate \(y''\) at \( \theta = \frac{\pi}{2} \)**
If \( \theta = \frac{\pi}{2} \), then \( \frac{\theta}{2} = \frac{\pi}{4} \).
\( \tan\left(\frac{\pi}{4}\right) = 1 \).
Substitute this into the expression for \(y''\):
\( y'' = \frac{1}{4a} (1 + 1^2)^2 \)
\( y'' = \frac{1}{4a} (1 + 1)^2 \)
\( y'' = \frac{1}{4a} (2)^2 \)
\( y'' = \frac{1}{4a} \cdot 4 \)
\( y'' = \frac{1}{a} \)
This proves the given statement. This problem combines parametric differentiation for both first and second derivatives with trigonometric identities.
In simple words: First, we find the derivatives of \(x\) and \(y\) with respect to \( \theta \). Then, we find \(dy/dx\). Next, we find the second derivative \(y''\) using a specific formula for parametric equations. Finally, we plug in \( \theta = \pi/2 \) into the \(y''\) expression to get the required value.

🎯 Exam Tip: The formula for the second derivative of parametric equations is \( \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx} \). Remember to simplify \( \frac{dy}{dx} \) using half-angle identities before differentiating it again, as this often makes the process much simpler.

Question 27. If \( \sin y = x \sin (a + y) \), prove that \( \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a} \), where \( a \neq n\pi \).
Answer:
Given the equation \( \sin y = x \sin (a + y) \). We need to prove \( \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a} \).
First, isolate \(x\) from the equation:
\( x = \frac{\sin y}{\sin (a + y)} \)
Now, differentiate \(x\) with respect to \(y\) using the quotient rule: \( \frac{d}{dy}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).
Here, \(u = \sin y\) and \(v = \sin (a+y)\).
\( u' = \frac{d}{dy}(\sin y) = \cos y \)
\( v' = \frac{d}{dy}(\sin (a+y)) = \cos (a+y) \cdot \frac{d}{dy}(a+y) = \cos (a+y) \cdot 1 \)
So, \( \frac{dx}{dy} = \frac{\cos y \sin (a+y) - \sin y \cos (a+y)}{(\sin (a+y))^2} \)
Recall the trigonometric identity for \( \sin(A-B) = \sin A \cos B - \cos A \sin B \). The numerator matches this form with \(A = a+y\) and \(B = y\).
\( \frac{dx}{dy} = \frac{\sin ((a+y) - y)}{\sin^2 (a+y)} \)
\( \frac{dx}{dy} = \frac{\sin a}{\sin^2 (a+y)} \)
Since we need \( \frac{dy}{dx} \), we take the reciprocal of \( \frac{dx}{dy} \):
\( \frac{dy}{dx} = \frac{1}{dx/dy} \)
\( \frac{dy}{dx} = \frac{\sin^2 (a+y)}{\sin a} \)
This proves the given statement. This method cleverly differentiates with respect to \(y\) first, simplifying the process.
In simple words: We first rearrange the equation to get \(x\) by itself. Then we find the derivative of \(x\) with respect to \(y\) using the quotient rule. We simplify this derivative using a trigonometric identity, and then flip it to get \(dy/dx\).

🎯 Exam Tip: When \(x\) is easily expressed as a function of \(y\), it's often more efficient to calculate \( \frac{dx}{dy} \) first using implicit or direct differentiation, and then find \( \frac{dy}{dx} \) as its reciprocal. This is particularly useful when the expression for \( \frac{dx}{dy} \) simplifies via trigonometric identities.

Question 28. If \( y = \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \), show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \). Hence find \(y_2\) when \(x = 0\).
Answer:
**Part 1: Show that \( (1 – x^2)y_2 – 3xy_1 – y = 0 \)**
This part is identical to Question 25.
Given \( y = \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \).
First, rearrange the equation:
\( y\sqrt{1-x^2} = \sin^{-1} x \)
Differentiate both sides with respect to \(x\) (using product rule on left, chain rule for \(\sqrt{1-x^2}\)):
\( y_1\sqrt{1-x^2} + y \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{1}{\sqrt{1-x^2}} \)
\( y_1\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
Multiply by \( \sqrt{1-x^2} \) to clear denominators:
\( y_1(1-x^2) - xy = 1 \) (Equation 1)
Now, differentiate Equation 1 with respect to \(x\) (using product rule twice):
\( [y_2(1-x^2) + y_1(-2x)] - [1 \cdot y + x \cdot y_1] = 0 \)
\( y_2(1-x^2) - 2xy_1 - y - xy_1 = 0 \)
Combine terms:
\( (1-x^2)y_2 - 3xy_1 - y = 0 \)
This proves the first part.

**Part 2: Find \(y_2\) when \(x = 0\)**
Substitute \(x = 0\) into the proved differential equation \( (1-x^2)y_2 - 3xy_1 - y = 0 \):
\( (1-0^2)y_2 - 3(0)y_1 - y = 0 \)
\( (1)y_2 - 0 - y = 0 \)
\( y_2 = y \)
Now, we need to find the value of \(y\) when \(x=0\) using the original function \( y = \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} \):
\( y(0) = \frac{\sin ^{-1} (0)}{\sqrt{1-0^{2}}} = \frac{0}{\sqrt{1}} = \frac{0}{1} = 0 \)
Therefore, when \(x = 0\), \(y = 0\).
Substitute \(y=0\) back into \(y_2 = y\):
\( y_2 = 0 \)
So, when \(x = 0\), the value of \(y_2\) is 0. This involves using the derived differential equation to find a specific value.
In simple words: First, we follow the same steps as Question 25 to show the given equation. Then, to find \(y_2\) when \(x=0\), we plug \(x=0\) into that equation. We also find the value of \(y\) itself when \(x=0\) from the original function. Finally, we use these values to solve for \(y_2\).

🎯 Exam Tip: When asked to find the value of a derivative at a specific point, it is often easiest to derive the differential equation (if necessary) first, and then substitute the point's coordinates into the derived equation. This avoids finding complex higher-order derivatives and then evaluating them.