OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (K)

Find the derivative of the following functions:

Question 1. \( (x^2 + 2)^3 (1 - x^3)^4 \)
Answer: Let the given function be \( y \).
So, \( y = (x^2 + 2)^3 (1 - x^3)^4 \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log [(x^2 + 2)^3 (1 - x^3)^4] \)
Now, use the logarithm property \( \log(AB) = \log A + \log B \):
\( \log y = \log (x^2 + 2)^3 + \log (1 - x^3)^4 \)
Next, use the logarithm property \( \log(A^B) = B \log A \):
\( \log y = 3 \log(x^2 + 2) + 4 \log(1 - x^3) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = 3 \cdot \frac{1}{x^2 + 2} \cdot \frac{d}{dx}(x^2 + 2) + 4 \cdot \frac{1}{1 - x^3} \cdot \frac{d}{dx}(1 - x^3) \)
\( \frac{1}{y} \frac{dy}{dx} = 3 \cdot \frac{1}{x^2 + 2} (2x) + 4 \cdot \frac{1}{1 - x^3} (-3x^2) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{6x}{x^2 + 2} - \frac{12x^2}{1 - x^3} \)
To combine these terms, find a common denominator:
\( \frac{1}{y} \frac{dy}{dx} = \frac{6x(1 - x^3) - 12x^2(x^2 + 2)}{(x^2 + 2)(1 - x^3)} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{6x - 6x^4 - 12x^4 - 24x^2}{(x^2 + 2)(1 - x^3)} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{6x - 18x^4 - 24x^2}{(x^2 + 2)(1 - x^3)} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{6x(1 - 3x^3 - 4x)}{(x^2 + 2)(1 - x^3)} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \cdot \frac{6x(1 - 4x - 3x^3)}{(x^2 + 2)(1 - x^3)} \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (x^2 + 2)^3 (1 - x^3)^4 \cdot \frac{6x(1 - 4x - 3x^3)}{(x^2 + 2)(1 - x^3)} \)
Simplify the expression:
\( \frac{dy}{dx} = (x^2 + 2)^2 (1 - x^3)^3 6x(1 - 4x - 3x^3) \)
In simple words: First, take the natural logarithm on both sides. This helps to break down the multiplication and powers into simpler sums. Next, differentiate each part with respect to \( x \). Finally, multiply the whole result by the original function \( y \) to get the final derivative.

🎯 Exam Tip: Logarithmic differentiation is very useful when a function is a product or quotient of many terms, especially when those terms are raised to powers. It turns complex product/quotient rules into simpler sums/differences.

Question 2. \( \frac{x(1-x^2)^2}{(1+x^2)^{1 / 2}} \)
Answer: Let the given function be \( y \).
So, \( y = \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log \left( \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \right) \)
Using logarithm properties \( \log(A/B) = \log A - \log B \) and \( \log(AB) = \log A + \log B \):
\( \log y = \log x + \log(1-x^2)^2 - \log(1+x^2)^{1/2} \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \log x + 2 \log(1-x^2) - \frac{1}{2} \log(1+x^2) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) + 2 \frac{d}{dx}(\log(1-x^2)) - \frac{1}{2} \frac{d}{dx}(\log(1+x^2)) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + 2 \cdot \frac{1}{1-x^2} \cdot (-2x) - \frac{1}{2} \cdot \frac{1}{1+x^2} \cdot (2x) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - \frac{4x}{1-x^2} - \frac{x}{1+x^2} \)
Now, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{1}{x} - \frac{4x}{1-x^2} - \frac{x}{1+x^2} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \left[ \frac{1}{x} - \frac{4x}{1-x^2} - \frac{x}{1+x^2} \right] \)
Combine the terms inside the bracket. First combine \( \frac{1}{x} - \frac{x}{1+x^2} \):
\( \frac{1}{x} - \frac{x}{1+x^2} = \frac{1 \cdot (1+x^2) - x \cdot x}{x(1+x^2)} = \frac{1+x^2-x^2}{x(1+x^2)} = \frac{1}{x(1+x^2)} \)
So, \( \frac{dy}{dx} = \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \left[ \frac{1}{x(1+x^2)} - \frac{4x}{1-x^2} \right] \)
Now, combine the remaining terms in the bracket:
\( \frac{dy}{dx} = \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \left[ \frac{1 \cdot (1-x^2) - 4x \cdot x(1+x^2)}{x(1+x^2)(1-x^2)} \right] \)
\( \frac{dy}{dx} = \frac{x(1-x^2)^2}{(1+x^2)^{1/2}} \left[ \frac{1-x^2 - 4x^2 - 4x^4}{x(1-x^2)(1+x^2)} \right] \)
\( \frac{dy}{dx} = \frac{(1-x^2)(1-5x^2 - 4x^4)}{(1+x^2)^{1/2}(1+x^2)} \)
\( \frac{dy}{dx} = \frac{(1-x^2)(1-5x^2 - 4x^4)}{(1+x^2)^{3/2}} \)
In simple words: This function has many parts multiplied and divided, and some are raised to powers. To solve it easily, take the logarithm on both sides. This changes products to sums and divisions to subtractions. Then, differentiate each simple logarithm term. Finally, multiply the result by the original function \( y \) to get the derivative.

🎯 Exam Tip: Remember to use the chain rule when differentiating `log(1-x^2)` or `log(1+x^2)`. Also, combining fractions inside the bracket neatly is crucial for a simplified final answer.

Question 3. \( \frac{(x+1)^2 \sqrt{(x-1)}}{(x+4)^3 e^x} \)
Answer: Let the given function be \( y \).
So, \( y = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log \left( \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right) \)
Using logarithm properties \( \log(A/B) = \log A - \log B \) and \( \log(AB) = \log A + \log B \):
\( \log y = \log (x+1)^2 + \log \sqrt{x-1} - \log (x+4)^3 - \log e^x \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+1} \cdot 1 + \frac{1}{2} \cdot \frac{1}{x-1} \cdot 1 - 3 \cdot \frac{1}{x+4} \cdot 1 - 1 \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right] \)
In simple words: First, take the natural logarithm on both sides to change products and quotients into sums and differences, and powers into multiplication. Then, differentiate each term. Remember that the derivative of \( \log e^x \) is \( 1 \). Finally, multiply everything by the original function \( y \) to get the full derivative.

🎯 Exam Tip: For functions with products, quotients, and exponential terms like \( e^x \), logarithmic differentiation is highly effective. Remember \( \frac{d}{dx}(\log e^x) = \frac{d}{dx}(x) = 1 \).

Question 4. \( \sqrt{(x-1)(x-2)(x-3)(x-4)} \)
Answer: Let the given function be \( y \).
So, \( y = \sqrt{(x-1)(x-2)(x-3)(x-4)} \).
We can write this as \( y = ((x-1)(x-2)(x-3)(x-4))^{1/2} \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log ((x-1)(x-2)(x-3)(x-4))^{1/2} \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \frac{1}{2} \log ((x-1)(x-2)(x-3)(x-4)) \)
Using the logarithm property \( \log(ABCD) = \log A + \log B + \log C + \log D \):
\( \log y = \frac{1}{2} [\log(x-1) + \log(x-2) + \log(x-3) + \log(x-4)] \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} \cdot 1 + \frac{1}{x-2} \cdot 1 + \frac{1}{x-3} \cdot 1 + \frac{1}{x-4} \cdot 1 \right] \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} \right] \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \frac{1}{2} \sqrt{(x-1)(x-2)(x-3)(x-4)} \left[ \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} \right] \)
In simple words: When many terms are multiplied under a square root, it is easiest to take the logarithm on both sides first. The square root becomes a power of \( 1/2 \), and the product becomes a sum of logarithms. Differentiate each simple term, then multiply the entire result by the original function to get the final answer.

🎯 Exam Tip: The power \( 1/2 \) from the square root can be brought down with logarithm, making the differentiation of many multiplied terms much simpler than using the chain rule repeatedly with the product rule.

Question 5. \( \frac{(x-a)(x-b)}{(x-p)(x-q)} \)
Answer: Let the given function be \( y \).
So, \( y = \frac{(x-a)(x-b)}{(x-p)(x-q)} \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log \left( \frac{(x-a)(x-b)}{(x-p)(x-q)} \right) \)
Using logarithm properties \( \log(A/B) = \log A - \log B \) and \( \log(AB) = \log A + \log B \):
\( \log y = \log(x-a) + \log(x-b) - [\log(x-p) + \log(x-q)] \)
\( \log y = \log(x-a) + \log(x-b) - \log(x-p) - \log(x-q) \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x-a} \cdot 1 + \frac{1}{x-b} \cdot 1 - \frac{1}{x-p} \cdot 1 - \frac{1}{x-q} \cdot 1 \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{x-a} + \frac{1}{x-b} - \frac{1}{x-p} - \frac{1}{x-q} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{1}{x-a} + \frac{1}{x-b} - \frac{1}{x-p} - \frac{1}{x-q} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \frac{(x-a)(x-b)}{(x-p)(x-q)} \left[ \frac{1}{x-a} + \frac{1}{x-b} - \frac{1}{x-p} - \frac{1}{x-q} \right] \)
In simple words: This function is a fraction with products in both the top and bottom. Take the logarithm on both sides to change it into a simpler form using plus and minus signs. Then, differentiate each term. Remember to multiply by the original function \( y \) at the end to get the full derivative.

🎯 Exam Tip: For expressions with multiple factors in the numerator and denominator, logarithmic differentiation simplifies the work, preventing potential errors from a complex application of the quotient and product rules.

Question 6. \( \frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}} \)
Answer: Let the given function be \( y \).
So, \( y = \frac{2(x-\sin x)^{3/2}}{\sqrt{x}} \).
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log \left( \frac{2(x-\sin x)^{3/2}}{x^{1/2}} \right) \)
Using logarithm properties \( \log(A/B) = \log A - \log B \) and \( \log(ABC) = \log A + \log B + \log C \):
\( \log y = \log 2 + \log (x-\sin x)^{3/2} - \log x^{1/2} \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \log 2 + \frac{3}{2} \log(x-\sin x) - \frac{1}{2} \log x \)
Now, differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log 2) + \frac{3}{2} \frac{d}{dx}(\log(x-\sin x)) - \frac{1}{2} \frac{d}{dx}(\log x) \)
\( \frac{1}{y} \frac{dy}{dx} = 0 + \frac{3}{2} \cdot \frac{1}{x-\sin x} \cdot (1-\cos x) - \frac{1}{2} \cdot \frac{1}{x} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{3(1-\cos x)}{2(x-\sin x)} - \frac{1}{2x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \frac{3(1-\cos x)}{2(x-\sin x)} - \frac{1}{2x} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \frac{2(x-\sin x)^{3/2}}{\sqrt{x}} \left[ \frac{3(1-\cos x)}{2(x-\sin x)} - \frac{1}{2x} \right] \)
In simple words: Take the natural logarithm of the entire expression first. This will simplify the complex fraction and powers into a sum and difference of simpler terms. Differentiate each term, remembering the chain rule for \( \log(x-\sin x) \). Then, multiply by the original function \( y \) to find the derivative.

🎯 Exam Tip: When dealing with trigonometric functions inside powers or square roots, logarithmic differentiation is often the most straightforward approach. Remember \( \frac{d}{dx}(\log(\text{constant})) = 0 \).

Question 7. Find the derivative of the following functions:
(i) \( x^{1/x} \)
(ii) \( x^{\sqrt{x}} \)
(iii) \( \left(\frac{1}{x}\right)^x \)
Answer:
(i) Let \( y = x^{1/x} \).
Since \( x \) is in both the base and the exponent, take the natural logarithm on both sides:
\( \log y = \log(x^{1/x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \frac{1}{x} \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{x} \log x \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}\left(\frac{1}{x}\right) \right) \log x + \frac{1}{x} \left( \frac{d}{dx}(\log x) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( -\frac{1}{x^2} \right) \log x + \frac{1}{x} \left( \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = -\frac{\log x}{x^2} + \frac{1}{x^2} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{1 - \log x}{x^2} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \frac{1 - \log x}{x^2} \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = x^{1/x} \frac{1 - \log x}{x^2} \)

(ii) Let \( y = x^{\sqrt{x}} \).
Since \( x \) is in both the base and the exponent, take the natural logarithm on both sides:
\( \log y = \log(x^{\sqrt{x}}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \sqrt{x} \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x} \log x) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(\sqrt{x}) \right) \log x + \sqrt{x} \left( \frac{d}{dx}(\log x) \right) \)
Remember that \( \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \).
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \log x + \sqrt{x} \frac{1}{x} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log x + 2}{2\sqrt{x}} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \frac{\log x + 2}{2\sqrt{x}} \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = x^{\sqrt{x}} \frac{\log x + 2}{2\sqrt{x}} \)

(iii) Let \( y = \left(\frac{1}{x}\right)^x \).
Since \( x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log \left(\left(\frac{1}{x}\right)^x\right) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = x \log\left(\frac{1}{x}\right) \)
Using the logarithm property \( \log(1/A) = -\log A \):
\( \log y = x (-\log x) \)
\( \log y = -x \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(-x \log x) \)
\( \frac{1}{y} \frac{dy}{dx} = - \left[ \left( \frac{d}{dx}(x) \right) \log x + x \left( \frac{d}{dx}(\log x) \right) \right] \)
\( \frac{1}{y} \frac{dy}{dx} = - \left[ (1) \log x + x \left( \frac{1}{x} \right) \right] \)
\( \frac{1}{y} \frac{dy}{dx} = -[\log x + 1] \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = -y(1 + \log x) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = -\left(\frac{1}{x}\right)^x (1 + \log x) \)
In simple words: For all these problems where \( x \) is in both the base and the power, you must first take the logarithm on both sides. Use logarithm rules to bring the power down. Then, differentiate using the product rule. After differentiating, multiply by the original function \( y \) to get the final derivative. Remember that \( \log(1/x) \) is the same as \( -\log x \).

🎯 Exam Tip: Functions of the form \( f(x)^{g(x)} \) almost always require logarithmic differentiation. Pay close attention to applying the product rule and chain rule correctly after taking the logarithm.

Question 8. \( (\sin x)^x \)
Answer: Let the given function be \( y \).
So, \( y = (\sin x)^x \).
Since \( x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log((\sin x)^x) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = x \log(\sin x) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log(\sin x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(x) \right) \log(\sin x) + x \left( \frac{d}{dx}(\log(\sin x)) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (1) \log(\sin x) + x \left( \frac{1}{\sin x} \cdot \cos x \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log(\sin x) + x \cot x \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y (x \cot x + \log(\sin x)) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (\sin x)^x (x \cot x + \log(\sin x)) \)
In simple words: When a function like \( \sin x \) is raised to the power of \( x \), take the natural logarithm on both sides. This brings the power \( x \) down. Then, differentiate the expression using the product rule. Remember that the derivative of \( \log(\sin x) \) is \( \frac{1}{\sin x} \cdot \cos x \), which is \( \cot x \). Finally, multiply by the original function \( y \) to get the full derivative.

🎯 Exam Tip: Be careful with the chain rule when differentiating \( \log(\sin x) \), which gives \( \cot x \). This is a common point where errors can occur if not handled properly.

Question 9. \( x^{\sin x} \)
Answer: Let the given function be \( y \).
So, \( y = x^{\sin x} \).
Since \( x \) is in both the base and the exponent, take the natural logarithm on both sides:
\( \log y = \log(x^{\sin x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \sin x \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x \log x) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(\sin x) \right) \log x + \sin x \left( \frac{d}{dx}(\log x) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (\cos x) \log x + \sin x \left( \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( \cos x \log x + \frac{\sin x}{x} \right) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) \)
In simple words: When \( x \) appears in both the base and the exponent, start by taking the logarithm on both sides. This simplifies the power. Then, use the product rule to differentiate the expression \( (\sin x) (\log x) \). Finally, multiply the result by \( y \) (the original function) to get the derivative.

🎯 Exam Tip: When \( f(x)^{g(x)} \) is given, taking the natural logarithm is the standard first step to bring the exponent down for easier differentiation. This converts it into a product that can be differentiated using the product rule.

Question 10. \( (\sin x)^{\tan x} \)
Answer: Let the given function be \( y \).
So, \( y = (\sin x)^{\tan x} \).
Since \( \tan x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log((\sin x)^{\tan x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \tan x \log(\sin x) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x \log(\sin x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(\tan x) \right) \log(\sin x) + \tan x \left( \frac{d}{dx}(\log(\sin x)) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (\sec^2 x) \log(\sin x) + \tan x \left( \frac{1}{\sin x} \cdot \cos x \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (\sec^2 x) \log(\sin x) + \tan x \cot x \)
Since \( \tan x \cot x = 1 \):
\( \frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1 \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y (1 + \sec^2 x \log(\sin x)) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (\sin x)^{\tan x} (1 + \sec^2 x \log(\sin x)) \)
In simple words: For this type of function where a trigonometric function is raised to another trigonometric function, take the logarithm on both sides. This brings the power down. Then, differentiate using the product rule, remembering the derivative of \( \tan x \) is \( \sec^2 x \) and the chain rule for \( \log(\sin x) \). Simplify the terms and multiply by the original function \( y \) to get the final answer.

🎯 Exam Tip: The derivative of \( \tan x \) is \( \sec^2 x \), and \( \frac{d}{dx}(\log(\sin x)) \) is \( \cot x \). Also, remember that \( \tan x \cdot \cot x = 1 \). These simplifications are important for reaching the final form.

Question 11. \( (\tan x)^{\log x} \)
Answer: Let the given function be \( y \).
So, \( y = (\tan x)^{\log x} \).
Since \( \log x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log((\tan x)^{\log x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \log x \log(\tan x) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x \log(\tan x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(\log x) \right) \log(\tan x) + \log x \left( \frac{d}{dx}(\log(\tan x)) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{1}{x} \right) \log(\tan x) + \log x \left( \frac{1}{\tan x} \cdot \sec^2 x \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log(\tan x)}{x} + \log x \left( \cot x \sec^2 x \right) \)
Recall that \( \cot x \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} \).
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log(\tan x)}{x} + \frac{\log x}{\sin x \cos x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( \frac{\log(\tan x)}{x} + \frac{\log x}{\sin x \cos x} \right) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (\tan x)^{\log x} \left( \frac{\log(\tan x)}{x} + \frac{\log x}{\sin x \cos x} \right) \)
In simple words: When you have a function raised to another function, like \( \tan x \) raised to \( \log x \), the first step is to take the natural logarithm on both sides. This uses the logarithm rule to bring the power down. Then, use the product rule to differentiate. Remember to use the chain rule when differentiating \( \log(\tan x) \). After simplifying, multiply the whole expression by the original function \( y \) to get the final derivative.

🎯 Exam Tip: When differentiating \( \log(\tan x) \), remember it becomes \( \frac{1}{\tan x} \cdot \sec^2 x \), which simplifies to \( \cot x \sec^2 x \) or \( \frac{1}{\sin x \cos x} \). Keeping these trigonometric identities in mind helps simplify the answer.

Question 12. \( x^{\log x} \)
Answer: Let the given function be \( y \).
So, \( y = x^{\log x} \).
Since \( x \) is in both the base and the exponent, take the natural logarithm on both sides:
\( \log y = \log(x^{\log x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = (\log x)(\log x) \)
\( \log y = (\log x)^2 \)
Now, differentiate both sides with respect to \( x \) using the chain rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}((\log x)^2) \)
\( \frac{1}{y} \frac{dy}{dx} = 2 (\log x)^{2-1} \cdot \frac{d}{dx}(\log x) \)
\( \frac{1}{y} \frac{dy}{dx} = 2 \log x \cdot \frac{1}{x} \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{2 \log x}{x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \frac{2 \log x}{x} \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = x^{\log x} \frac{2 \log x}{x} \)
In simple words: When \( x \) is in both the base and power, take the logarithm on both sides. This allows you to bring the power down, making the expression \( (\log x)(\log x) \), which is \( (\log x)^2 \). Then, differentiate this simple expression using the power rule and chain rule. Finally, multiply by the original function \( y \) to get the full derivative.

🎯 Exam Tip: Recognizing that \( \log x \cdot \log x \) is \( (\log x)^2 \) is key. Then, use the chain rule \( \frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx} \) where \( u = \log x \).

Question 13. \( (\tan x)^{\cos x} \)
Answer: Let the given function be \( y \).
So, \( y = (\tan x)^{\cos x} \).
Since \( \cos x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log((\tan x)^{\cos x}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \cos x \log(\tan x) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x \log(\tan x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(\cos x) \right) \log(\tan x) + \cos x \left( \frac{d}{dx}(\log(\tan x)) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (-\sin x) \log(\tan x) + \cos x \left( \frac{1}{\tan x} \cdot \sec^2 x \right) \)
\( \frac{1}{y} \frac{dy}{dx} = -\sin x \log(\tan x) + \cos x \left( \cot x \sec^2 x \right) \)
Recall that \( \cot x \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \operatorname{cosec} x \sec x \). Also, \( \frac{\cos x}{\tan x \cos^2 x} = \frac{\cos x}{\frac{\sin x}{\cos x} \cos^2 x} = \frac{\cos x}{\sin x \cos x} = \frac{1}{\sin x} = \operatorname{cosec} x \).
So, the term \( \cos x \left( \frac{1}{\tan x} \cdot \sec^2 x \right) = \cos x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x} = \operatorname{cosec} x \).
Thus, \( \frac{1}{y} \frac{dy}{dx} = -\sin x \log(\tan x) + \operatorname{cosec} x \)
\( \frac{1}{y} \frac{dy}{dx} = \operatorname{cosec} x - \sin x \log(\tan x) \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y (\operatorname{cosec} x - \sin x \log(\tan x)) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (\tan x)^{\cos x} (\operatorname{cosec} x - \sin x \log(\tan x)) \)
In simple words: For a function like \( \tan x \) raised to the power of \( \cos x \), take the natural logarithm on both sides. This brings the power down. Then, use the product rule to differentiate the expression. Remember the derivative of \( \cos x \) is \( -\sin x \) and use the chain rule for \( \log(\tan x) \). Simplify carefully, and finally, multiply by the original function \( y \) to get the full derivative.

🎯 Exam Tip: Pay close attention to the derivatives of trigonometric functions and their simplifications. For instance, \( \frac{\cos x}{\tan x \cos^2 x} \) simplifies to \( \operatorname{cosec} x \). Errors often arise from incorrect simplification of these terms.

Question 14. \( (\log x)^x \)
Answer: Let the given function be \( y \).
So, \( y = (\log x)^x \).
Since \( x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log((\log x)^x) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = x \log(\log x) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log(\log x)) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(x) \right) \log(\log x) + x \left( \frac{d}{dx}(\log(\log x)) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (1) \log(\log x) + x \left( \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log(\log x) + x \left( \frac{1}{\log x} \cdot \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log(\log x) + \frac{1}{\log x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left( \log(\log x) + \frac{1}{\log x} \right) \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \)
In simple words: When a function like \( \log x \) is raised to the power of \( x \), take the natural logarithm on both sides. This uses the logarithm rule to bring the power \( x \) down. Then, differentiate using the product rule. Remember to use the chain rule when differentiating \( \log(\log x) \). After differentiating, multiply by the original function \( y \) to get the final derivative.

🎯 Exam Tip: The chain rule for \( \log(\log x) \) is \( \frac{1}{\log x} \cdot \frac{1}{x} \). Applying this correctly is crucial to avoid common errors.

Question 15. \( \left(1+\frac{1}{x}\right)^x \)
Answer: Let the given function be \( y \).
So, \( y = \left(1+\frac{1}{x}\right)^x \).
Since \( x \) is in the exponent, take the natural logarithm on both sides:
\( \log y = \log \left(\left(1+\frac{1}{x}\right)^x\right) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = x \log\left(1+\frac{1}{x}\right) \)
We can rewrite \( \log\left(1+\frac{1}{x}\right) \) as \( \log\left(\frac{x+1}{x}\right) = \log(x+1) - \log x \).
So, \( \log y = x [\log(x+1) - \log x] \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x [\log(x+1) - \log x]) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}(x) \right) [\log(x+1) - \log x] + x \left( \frac{d}{dx}[\log(x+1) - \log x] \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (1) [\log(x+1) - \log x] + x \left( \frac{1}{x+1} - \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log\left(\frac{x+1}{x}\right) + x \left( \frac{x - (x+1)}{x(x+1)} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log\left(1+\frac{1}{x}\right) + x \left( \frac{-1}{x(x+1)} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log\left(1+\frac{1}{x}\right) - \frac{1}{x+1} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \log\left(1+\frac{1}{x}\right) - \frac{1}{x+1} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = \left(1+\frac{1}{x}\right)^x \left[ \log\left(1+\frac{1}{x}\right) - \frac{1}{x+1} \right] \)
Alternatively, from the step \( \frac{1}{y} \frac{dy}{dx} = \log\left(\frac{x+1}{x}\right) + x \left( \frac{x - (x+1)}{x(x+1)} \right) \):
\( \frac{1}{y} \frac{dy}{dx} = \log\left(\frac{x+1}{x}\right) + x \left( \frac{-1}{x(x+1)} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log\left(\frac{x+1}{x}\right) - \frac{1}{x+1} \)
This matches the OCR output when \( y \) is multiplied back:
\( \frac{dy}{dx} = \left(1+\frac{1}{x}\right)^x \left[ \log\left(\frac{x+1}{x}\right) - \frac{1}{x+1} \right] \)
The OCR's final expansion was `\( \left(1+\frac{1}{x}\right)^x \left[ \frac{x}{x+1} - 1 + \log\left(\frac{x+1}{x}\right) \right] \)` which matches my earlier check, because `x[1/(x+1) - 1/x]` is `x/(x+1) - 1`. Let's use this form.
\( \frac{dy}{dx} = \left(1+\frac{1}{x}\right)^x \left[ \frac{x}{x+1} - 1 + \log\left(1+\frac{1}{x}\right) \right] \)
In simple words: When \( x \) is in the power, take the logarithm on both sides to bring it down. Simplify the logarithm using \( \log(A/B) = \log A - \log B \). Then, differentiate the expression using the product rule. Be careful with the algebraic simplification of the terms. Finally, multiply the result by the original function \( y \) to get the final answer.

🎯 Exam Tip: Rewriting \( \log(1 + \frac{1}{x}) \) as \( \log(\frac{x+1}{x}) = \log(x+1) - \log x \) is a smart move that simplifies its differentiation and prevents errors.

Question 16. \( x^x \sqrt{x} \)
Answer: Let the given function be \( y \).
So, \( y = x^x \sqrt{x} \).
We can write \( \sqrt{x} \) as \( x^{1/2} \). So, \( y = x^x \cdot x^{1/2} \).
Using the exponent rule \( a^m \cdot a^n = a^{m+n} \):
\( y = x^{x + 1/2} \)
To simplify differentiation, take the natural logarithm on both sides:
\( \log y = \log(x^{x+1/2}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \left(x+\frac{1}{2}\right) \log x \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\left(x+\frac{1}{2}\right) \log x\right) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}\left(x+\frac{1}{2}\right) \right) \log x + \left(x+\frac{1}{2}\right) \left( \frac{d}{dx}(\log x) \right) \)
\( \frac{1}{y} \frac{dy}{dx} = (1) \log x + \left(x+\frac{1}{2}\right) \left( \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \log x + 1 + \frac{1}{2x} \)
Finally, multiply both sides by \( y \) to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y \left[ \log x + 1 + \frac{1}{2x} \right] \)
Substitute the original expression for \( y \):
\( \frac{dy}{dx} = x^x \sqrt{x} \left[ \log x + 1 + \frac{1}{2x} \right] \)
In simple words: First, combine \( x^x \sqrt{x} \) into \( x^{x+1/2} \). Then, take the logarithm on both sides to bring the power down. Differentiate the resulting product using the product rule. Remember that the derivative of \( x + 1/2 \) is just \( 1 \). Finally, multiply the whole expression by the original function \( y \) to get the final derivative.

🎯 Exam Tip: Always simplify the expression using exponent rules first before applying logarithmic differentiation. This helps in making the subsequent differentiation steps cleaner and less error-prone.

Question 17. Find \( \frac{dy}{dx} \) if
(i) \( y = \cos x^x \)
(ii) \( y = \sin (x^x) \)
(iii) If \( y = \left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}} \), find \( \frac{dy}{dx} \) at \( x = 1 \).
Answer:
(i) Given \( y = \cos x^x \).
This is a composite function, \( \cos(u) \) where \( u = x^x \).
Using the chain rule, \( \frac{dy}{dx} = \frac{d}{du}(\cos u) \cdot \frac{du}{dx} \).
\( \frac{d}{du}(\cos u) = -\sin u \). So, \( \frac{dy}{dx} = -\sin(x^x) \cdot \frac{d}{dx}(x^x) \).
Now, we need to find \( \frac{d}{dx}(x^x) \). Let \( z = x^x \).
Taking logarithm on both sides, \( \log z = x \log x \).
Differentiating with respect to \( x \): \( \frac{1}{z} \frac{dz}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1 \).
So, \( \frac{dz}{dx} = z(1 + \log x) = x^x(1 + \log x) \).
Substitute this back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = -\sin(x^x) \cdot x^x(1 + \log x) \)

(ii) Given \( y = \sin (x^x) \).
This is a composite function, \( \sin(u) \) where \( u = x^x \).
Using the chain rule, \( \frac{dy}{dx} = \frac{d}{du}(\sin u) \cdot \frac{du}{dx} \).
\( \frac{d}{du}(\sin u) = \cos u \). So, \( \frac{dy}{dx} = \cos(x^x) \cdot \frac{d}{dx}(x^x) \).
From part (i), we know that \( \frac{d}{dx}(x^x) = x^x(1 + \log x) \).
Substitute this back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \cos(x^x) \cdot x^x(1 + \log x) \)

(iii) Given \( y = \left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}} \). We need to find \( \frac{dy}{dx} \) at \( x = 1 \).
Since there is \( x \) in the exponent, take the natural logarithm on both sides:
\( \log y = \log \left(\left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}}\right) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = \frac{4}{\pi x} \log \left(\tan \frac{\pi x}{4}\right) \)
Now, differentiate both sides with respect to \( x \) using the product rule on the right side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\frac{4}{\pi x} \log \left(\tan \frac{\pi x}{4}\right)\right) \)
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{d}{dx}\left(\frac{4}{\pi x}\right) \right) \log \left(\tan \frac{\pi x}{4}\right) + \frac{4}{\pi x} \left( \frac{d}{dx}\left(\log \left(\tan \frac{\pi x}{4}\right)\right) \right) \)
\( \frac{d}{dx}\left(\frac{4}{\pi x}\right) = \frac{4}{\pi} \frac{d}{dx}(x^{-1}) = \frac{4}{\pi} (-1 x^{-2}) = -\frac{4}{\pi x^2} \).
\( \frac{d}{dx}\left(\log \left(\tan \frac{\pi x}{4}\right)\right) = \frac{1}{\tan \frac{\pi x}{4}} \cdot \sec^2 \left(\frac{\pi x}{4}\right) \cdot \frac{d}{dx}\left(\frac{\pi x}{4}\right) = \frac{1}{\tan \frac{\pi x}{4}} \cdot \sec^2 \left(\frac{\pi x}{4}\right) \cdot \frac{\pi}{4} \).
Substitute these derivatives back:
\( \frac{1}{y} \frac{dy}{dx} = -\frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) + \frac{4}{\pi x} \cdot \frac{1}{\tan \frac{\pi x}{4}} \cdot \sec^2 \left(\frac{\pi x}{4}\right) \cdot \frac{\pi}{4} \)
\( \frac{1}{y} \frac{dy}{dx} = -\frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) + \frac{1}{x} \frac{\cos \frac{\pi x}{4}}{\sin \frac{\pi x}{4}} \frac{1}{\cos^2 \frac{\pi x}{4}} \)
\( \frac{1}{y} \frac{dy}{dx} = -\frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) + \frac{1}{x \sin \frac{\pi x}{4} \cos \frac{\pi x}{4}} \)
Using \( \sin A \cos A = \frac{1}{2} \sin 2A \):
\( \frac{1}{y} \frac{dy}{dx} = -\frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) + \frac{2}{x \sin \frac{\pi x}{2}} \)
So, \( \frac{dy}{dx} = y \left[ \frac{2}{x \sin \frac{\pi x}{2}} - \frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) \right] \)
Substitute \( y \):
\( \frac{dy}{dx} = \left(\tan \frac{\pi x}{4}\right)^{\frac{4}{\pi x}} \left[ \frac{2}{x \sin \frac{\pi x}{2}} - \frac{4}{\pi x^2} \log \left(\tan \frac{\pi x}{4}\right) \right] \)
Now, evaluate \( \frac{dy}{dx} \) at \( x = 1 \):
\( \left. \frac{dy}{dx} \right|_{x=1} = \left(\tan \frac{\pi (1)}{4}\right)^{\frac{4}{\pi (1)}} \left[ \frac{2}{(1) \sin \frac{\pi (1)}{2}} - \frac{4}{\pi (1)^2} \log \left(\tan \frac{\pi (1)}{4}\right) \right] \)
\( = \left(\tan \frac{\pi}{4}\right)^{\frac{4}{\pi}} \left[ \frac{2}{\sin \frac{\pi}{2}} - \frac{4}{\pi} \log \left(\tan \frac{\pi}{4}\right) \right] \)
Since \( \tan \frac{\pi}{4} = 1 \) and \( \sin \frac{\pi}{2} = 1 \):
\( = (1)^{\frac{4}{\pi}} \left[ \frac{2}{1} - \frac{4}{\pi} \log(1) \right] \)
Since \( \log(1) = 0 \):
\( = 1 \left[ 2 - \frac{4}{\pi} (0) \right] \)
\( = 1 [2 - 0] = 2 \)
Thus, \( \frac{dy}{dx} \) at \( x = 1 \) is \( 2 \).
In simple words:
(i) This is a chain rule problem. Differentiate `cos(u)` to get `-sin(u)`, then multiply by the derivative of the inner part, which is `x^x`. Remember that the derivative of `x^x` is `x^x(1 + log x)`.
(ii) Similar to part (i), this is a chain rule problem. Differentiate `sin(u)` to get `cos(u)`, then multiply by the derivative of the inner part, `x^x`.
(iii) First, take the logarithm on both sides because \( x \) is in the exponent. Differentiate the resulting product using the product rule, carefully applying the chain rule for `log(tan(...))`. Simplify the expression for `dy/dx`. Finally, substitute \( x=1 \) into the derivative to find the numerical value. Remember that `tan(pi/4)` is `1` and `log(1)` is `0`.

🎯 Exam Tip: For composite functions like `cos(x^x)`, always break them down using the chain rule. When a function has `f(x)^g(x)` form, use logarithmic differentiation. When evaluating at a specific point, substitute the value of `x` only after finding the general derivative `dy/dx` and simplifying it.

Question 18. Find \( \frac{dy}{dx} \) if
(i) \( y = \log (x^x + \operatorname{cosec}^2 x) \)
(ii) \( y = e^{\sin^2 x} \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) \)
(iii) \( x^y = y^x \)
(iv) \( (\cos x)^y = (\sin y)^x \)
Answer:
(i) Given \( y = \log (x^x + \operatorname{cosec}^2 x) \).
Using the chain rule, \( \frac{dy}{dx} = \frac{1}{x^x + \operatorname{cosec}^2 x} \cdot \frac{d}{dx}(x^x + \operatorname{cosec}^2 x) \).
First, find \( \frac{d}{dx}(x^x) \). Let \( u = x^x \). Then \( \log u = x \log x \).
Differentiating both sides: \( \frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x \).
So, \( \frac{du}{dx} = x^x(1 + \log x) \).
Next, find \( \frac{d}{dx}(\operatorname{cosec}^2 x) \). Let \( v = \operatorname{cosec}^2 x = (\operatorname{cosec} x)^2 \).
Using the chain rule: \( \frac{dv}{dx} = 2 (\operatorname{cosec} x)^{2-1} \cdot \frac{d}{dx}(\operatorname{cosec} x) \).
\( \frac{dv}{dx} = 2 \operatorname{cosec} x (-\operatorname{cosec} x \cot x) = -2 \operatorname{cosec}^2 x \cot x \).
Now, substitute these derivatives back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{x^x + \operatorname{cosec}^2 x} \left[ x^x(1 + \log x) - 2 \operatorname{cosec}^2 x \cot x \right] \)

(ii) Given \( y = e^{\sin^2 x} \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) \).
Let \( A = e^{\sin^2 x} \) and \( B = 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} \). So \( y = AB \).
Using the product rule, \( \frac{dy}{dx} = \frac{dA}{dx} B + A \frac{dB}{dx} \).
First, find \( \frac{dA}{dx} \). Let \( A = e^u \) where \( u = \sin^2 x \).
\( \frac{dA}{dx} = e^u \frac{du}{dx} = e^{\sin^2 x} \cdot \frac{d}{dx}(\sin^2 x) = e^{\sin^2 x} \cdot 2 \sin x \cos x = e^{\sin^2 x} \sin 2x \).
Next, find \( \frac{dB}{dx} \). Let \( B = 2 \tan^{-1} w \) where \( w = \sqrt{\frac{1-x}{1+x}} \).
To simplify \( w \), let \( x = \cos \theta \). Then \( \theta = \cos^{-1} x \).
\( w = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} = \sqrt{\tan^2 (\theta/2)} = \tan(\theta/2) \).
So, \( B = 2 \tan^{-1}(\tan(\theta/2)) = 2 \cdot \frac{\theta}{2} = \theta \).
Therefore, \( B = \cos^{-1} x \).
Now, differentiate \( B \) with respect to \( x \):
\( \frac{dB}{dx} = \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}} \).
Substitute \( \frac{dA}{dx} \), \( B \), \( A \), and \( \frac{dB}{dx} \) into the product rule formula:
\( \frac{dy}{dx} = (e^{\sin^2 x} \sin 2x) \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) + e^{\sin^2 x} \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \frac{dy}{dx} = e^{\sin^2 x} \left[ \sin 2x \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) - \frac{1}{\sqrt{1-x^2}} \right] \)
Substitute \( 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} = \cos^{-1} x \):
\( \frac{dy}{dx} = e^{\sin^2 x} \left[ (\sin 2x)(\cos^{-1} x) - \frac{1}{\sqrt{1-x^2}} \right] \)
This matches the derived form using the OCR's `log u` substitution method, where `log u = log(cos^-1 x)` and `du/dx` part was used in `1/y dy/dx = 2 sin x cos x + du/dx`.
The final answer is presented as: \( y \left[ \sin 2x - \frac{1}{\sqrt{1-x^2} \cos^{-1} x} \right] \). This means my derived `dB/dx` was used incorrectly in OCR's final combined step. Let's re-verify OCR's logic: if \( u = \log(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}) \), then \( u = \log(\cos^{-1} x) \). So \( \frac{du}{dx} = \frac{1}{\cos^{-1} x} (-\frac{1}{\sqrt{1-x^2}}) \).
The OCR's derivation for \( \frac{1}{y} \frac{dy}{dx} \) had \( 2 \sin x \cos x + \frac{1}{\cos^{-1} x} (-\frac{1}{\sqrt{1-x^2}}) \). This matches my derivation, using the product rule `A'B + AB'`. The term \( B \) is `\( 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} \)` which is \( \cos^{-1} x \). And \( \frac{dB}{dx} \) is \( -\frac{1}{\sqrt{1-x^2}} \).
So, \( \frac{dy}{dx} = \frac{dA}{dx} B + A \frac{dB}{dx} = (e^{\sin^2 x} \sin 2x)(\cos^{-1} x) + (e^{\sin^2 x})(-\frac{1}{\sqrt{1-x^2}}) \).
Factoring out \( e^{\sin^2 x} \):
\( \frac{dy}{dx} = e^{\sin^2 x} \left[ \sin 2x \cos^{-1} x - \frac{1}{\sqrt{1-x^2}} \right] \)
This is the correct final expression following standard differentiation rules and the simplifying substitution for \( \tan^{-1} \). The OCR output uses `\( \frac{1}{\sqrt{1-x^2} \cos^{-1} x} \)` for the `dB/dx` equivalent, which means it differentiated \( \log(B) \) instead of just \( B \).
Let's follow the OCR's `log y = sin^2 x + log(2 tan^-1 ...)` form, where `u` represents the `log(...)` part, as this is the instruction to follow source solutions. From OCR: \( \log y = \sin^2 x + \log \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) \). Let \( u = 2 \tan^{-1} \sqrt{\frac{1-x}{1+x}} \). As derived, \( u = \cos^{-1} x \).
So, \( \log y = \sin^2 x + \log (\cos^{-1} x) \).
Differentiate both sides with respect to \( x \):
\( \frac{1}{y} \frac{dy}{dx} = 2 \sin x \cos x + \frac{1}{\cos^{-1} x} \cdot \frac{d}{dx}(\cos^{-1} x) \)
\( \frac{1}{y} \frac{dy}{dx} = \sin 2x + \frac{1}{\cos^{-1} x} \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \frac{1}{y} \frac{dy}{dx} = \sin 2x - \frac{1}{\sqrt{1-x^2} \cos^{-1} x} \)
\( \frac{dy}{dx} = y \left[ \sin 2x - \frac{1}{\sqrt{1-x^2} \cos^{-1} x} \right] \)
Substitute back original \( y \):
\( \frac{dy}{dx} = e^{\sin^2 x} \left(2 \tan^{-1} \sqrt{\frac{1-x}{1+x}}\right) \left[ \sin 2x - \frac{1}{\sqrt{1-x^2} \cos^{-1} x} \right] \)

(iii) Given \( x^y = y^x \).
This is an implicit differentiation problem with variables in the exponents.
Take the natural logarithm on both sides:
\( \log(x^y) = \log(y^x) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( y \log x = x \log y \)
Now, differentiate both sides with respect to \( x \), treating \( y \) as a function of \( x \) and using the product rule on both sides:
\( \frac{d}{dx}(y \log x) = \frac{d}{dx}(x \log y) \)
\( \frac{dy}{dx} \log x + y \cdot \frac{1}{x} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \)
\( \frac{dy}{dx} \log x + \frac{y}{x} = \log y + \frac{x}{y} \frac{dy}{dx} \)
Group terms containing \( \frac{dy}{dx} \) 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 \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x} \)
\( \frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\frac{x \log y - y}{x}}{\frac{y \log x - x}{y}} \)
\( \frac{dy}{dx} = \frac{y(x \log y - y)}{x(y \log x - x)} \)

(iv) Given \( (\cos x)^y = (\sin y)^x \).
This is an implicit differentiation problem with variables in the exponents.
Take the natural logarithm on both sides:
\( \log((\cos x)^y) = \log((\sin y)^x) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( y \log(\cos x) = x \log(\sin y) \)
Now, differentiate both sides with respect to \( x \), treating \( y \) as a function of \( x \) and using the product rule on both sides:
\( \frac{d}{dx}(y \log(\cos x)) = \frac{d}{dx}(x \log(\sin y)) \)
\( \frac{dy}{dx} \log(\cos x) + y \cdot \frac{1}{\cos x} (-\sin x) = 1 \cdot \log(\sin y) + x \cdot \frac{1}{\sin y} (\cos y) \frac{dy}{dx} \)
\( \frac{dy}{dx} \log(\cos x) - y \tan x = \log(\sin y) + x \cot y \frac{dy}{dx} \)
Group terms containing \( \frac{dy}{dx} \) on one side and other terms on the other side:
\( \frac{dy}{dx} \log(\cos x) - x \cot y \frac{dy}{dx} = \log(\sin y) + y \tan x \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} (\log(\cos x) - x \cot y) = \log(\sin y) + y \tan x \)
Finally, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{\log(\sin y) + y \tan x}{\log(\cos x) - x \cot y} \)
In simple words:
(i) This is a chain rule problem for \( \log(u) \). You need to find the derivative of \( u = x^x + \operatorname{cosec}^2 x \) separately. Remember the special derivative for \( x^x \) and the chain rule for \( \operatorname{cosec}^2 x \).
(ii) For this complex product, use logarithm differentiation. First rewrite `\( y = e^{\sin^2 x} \cdot \cos^{-1} x \)` using the substitution `\( x = \cos \theta \)`. Then, take the logarithm, differentiate, and solve for `\( \frac{dy}{dx} \)`. This significantly simplifies the problem compared to using the direct product rule on the original function.
(iii) When an equation has \( x \) and \( y \) in the exponents (implicit functions), take the logarithm on both sides first. Then, differentiate implicitly with respect to \( x \), remembering to apply the product rule and chain rule (for `y` terms). Finally, gather all `\( \frac{dy}{dx} \)` terms and solve for it.
(iv) Similar to part (iii), take the logarithm on both sides to handle the variables in the exponents. Differentiate implicitly with respect to \( x \), making sure to use the product rule and chain rule carefully. Then, isolate `\( \frac{dy}{dx} \)` to find the answer.

🎯 Exam Tip: For implicit differentiation (parts iii and iv), taking logarithms first simplifies expressions like \( x^y \) or \( (\cos x)^y \). Remember that \( \frac{d}{dx}(\log f(y)) = \frac{1}{f(y)} f'(y) \frac{dy}{dx} \). Be meticulous with applying the product and chain rules for all terms containing \( y \).

Question 20. If \( x^m y^n = (x + y)^{m+n} \), prove that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: Given the equation is \( x^m y^n = (x + y)^{m+n} \).
First, take the logarithm on both sides of this equation:
\( \log(x^m y^n) = \log((x + y)^{m+n}) \)
Using logarithm properties, this becomes:
\( m \log x + n \log y = (m + n) \log(x + y) \)
Now, differentiate both sides with respect to \( x \):
\( m \cdot \frac{1}{x} + n \cdot \frac{1}{y} \frac{dy}{dx} = (m + n) \cdot \frac{1}{x + y} \left( 1 + \frac{dy}{dx} \right) \)
Group the terms with \( \frac{dy}{dx} \):
\( \frac{n}{y} \frac{dy}{dx} - \frac{m + n}{x + y} \frac{dy}{dx} = \frac{m + n}{x + y} - \frac{m}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \left( \frac{n}{y} - \frac{m + n}{x + y} \right) \frac{dy}{dx} = \left( \frac{m + n}{x + y} - \frac{m}{x} \right) \)
Find a common denominator for each side:
\( \left( \frac{n(x + y) - y(m + n)}{y(x + y)} \right) \frac{dy}{dx} = \left( \frac{x(m + n) - m(x + y)}{x(x + y)} \right) \)
Simplify the numerators:
\( \left( \frac{nx + ny - my - ny}{y(x + y)} \right) \frac{dy}{dx} = \left( \frac{mx + nx - mx - my}{x(x + y)} \right) \)
\( \left( \frac{nx - my}{y(x + y)} \right) \frac{dy}{dx} = \left( \frac{nx - my}{x(x + y)} \right) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \) \( \frac{dy}{dx} = \frac{y(x + y)}{x(x + y)} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y}{x} \).
This shows that the derivative is indeed \( \frac{y}{x} \).
In simple words: When you have an equation like this, taking logarithms helps simplify it. After that, you differentiate both sides and rearrange the terms to find \( \frac{dy}{dx} \), which turns out to be \( \frac{y}{x} \).

🎯 Exam Tip: Questions involving powers with variables in the base and exponent (like \( x^y \)) often require taking logarithms on both sides before differentiating. Remember the chain rule and product rule during differentiation.

Question 21. If \( y = x^{x^{x^{\ldots \ldots \ldots \infty}}} \), prove that \( \frac{dy}{dx}=\frac{y^2}{x(1-y \log x)} \).
Answer: Given the function is \( y = x^{x^{x^{\ldots \ldots \ldots \infty}}} \).
This type of infinitely nested exponent can be simplified by recognizing that the exponent itself is also \( y \).
So, we can write the equation as: \( y = x^y \)
To differentiate this, take the natural logarithm on both sides:
\( \log y = \log(x^y) \)
Using the logarithm property \( \log(a^b) = b \log a \):
\( \log y = y \log x \)
Now, differentiate both sides with respect to \( x \). Remember to use the product rule for \( y \log x \) and the chain rule for \( \log y \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
Next, group the terms containing \( \frac{dy}{dx} \):
\( \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \frac{1}{y} - \log x \right) = \frac{y}{x} \)
Combine the terms inside the parenthesis on the left side:
\( \frac{dy}{dx} \left( \frac{1 - y \log x}{y} \right) = \frac{y}{x} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \) \( \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y}{1 - y \log x} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \).
In simple words: When you see powers going on forever like this, you can replace the repeating part with the original letter \( y \). Then, you use logarithms to bring down the power, and after that, you differentiate normally to find the answer.

🎯 Exam Tip: For infinite series exponents, always simplify the expression first by setting the repeating part equal to the original variable. This makes differentiation much simpler.

Question 22. If \( y = \sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \ldots}} \), find \( \frac{dy}{dx} \).
Answer: Given the function is \( y = \sqrt{x}^{\sqrt{x}^{\sqrt{x} \ldots \ldots}} \).
Similar to the previous problem, the infinitely repeating exponent allows us to write this as:
\( y = (\sqrt{x})^y \)
We can also write \( \sqrt{x} \) as \( x^{1/2} \). So the equation becomes:
\( y = (x^{1/2})^y \)
\( y = x^{y/2} \)
Now, take the natural logarithm on both sides:
\( \log y = \log(x^{y/2}) \)
Using the logarithm property \( \log(a^b) = b \log a \):
\( \log y = \frac{y}{2} \log x \)
Differentiate both sides with respect to \( x \). Use the product rule on the right side and the chain rule on the left side:
\( \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left( \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \right) \)
\( \frac{1}{y} \frac{dy}{dx} = \frac{\log x}{2} \frac{dy}{dx} + \frac{y}{2x} \)
Group the terms containing \( \frac{dy}{dx} \):
\( \frac{1}{y} \frac{dy}{dx} - \frac{\log x}{2} \frac{dy}{dx} = \frac{y}{2x} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \frac{1}{y} - \frac{\log x}{2} \right) = \frac{y}{2x} \)
Combine the terms inside the parenthesis on the left side:
\( \frac{dy}{dx} \left( \frac{2 - y \log x}{2y} \right) = \frac{y}{2x} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \) \( \frac{dy}{dx} = \frac{y}{2x} \cdot \frac{2y}{2 - y \log x} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y^2}{x(2 - y \log x)} \).
In simple words: This problem is like the last one. You change the endless power part to \( y \), then use logarithms. After that, you just differentiate carefully using the rules for division and multiplication.

🎯 Exam Tip: Be careful with the base when simplifying infinite powers. If the base is \( \sqrt{x} \), remember to convert it to \( x^{1/2} \) for easier differentiation.

Question 23. If \( y = a^{x^{a^{x^{a^x \ldots \ldots \ldots \infty}}}} \), find \( \frac{dy}{dx} \).
Answer: Given the function is \( y = a^{x^{a^{x^{a^x \ldots \ldots \ldots \infty}}}} \).
For an infinitely nested exponent, we can simplify this by noticing that the part starting from the second \( x \) is also \( y \).
So, we can write this as: \( y = a^{x^y} \)
To differentiate this, we take the natural logarithm on both sides:
\( \log y = \log(a^{x^y}) \)
Using the logarithm property \( \log(A^B) = B \log A \):
\( \log y = x^y \log a \)
Now, take the logarithm again on both sides to bring down the \( x^y \) term:
\( \log(\log y) = \log(x^y \log a) \)
Using logarithm properties \( \log(AB) = \log A + \log B \) and \( \log(A^B) = B \log A \):
\( \log(\log y) = \log(x^y) + \log(\log a) \)
\( \log(\log y) = y \log x + \log(\log a) \)
Now, differentiate both sides with respect to \( x \). Remember the chain rule on the left side and the product rule on the right side. \( \log(\log a) \) is a constant, so its derivative is 0.
\( \frac{1}{\log y} \cdot \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
Group the terms containing \( \frac{dy}{dx} \):
\( \frac{1}{y \log y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \frac{1}{y \log y} - \log x \right) = \frac{y}{x} \)
Combine the terms inside the parenthesis on the left side:
\( \frac{dy}{dx} \left( \frac{1 - y \log x \log y}{y \log y} \right) = \frac{y}{x} \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \) \( \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y \log y}{1 - y \log x \log y} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y^2 \log y}{x(1 - y \log x \log y)} \).
In simple words: For a very complex nested power, you first simplify it by calling the repeating part \( y \). Then, you take logarithms twice to get rid of all the powers. After that, you differentiate carefully step by step to find the derivative.

🎯 Exam Tip: For problems with multiple nested exponents, consider taking logarithms multiple times until all variables in the exponents are brought down to the base level. Remember that \( \log(\log a) \) is a constant.

Question 24. If \( y = x^{x^{x \ldots \ldots \infty}} \), prove that \( x\frac{dy}{dx} = \frac{y^2}{1-y \log x} \).
Answer: Given the function is \( y = x^{x^{x \ldots \ldots \infty}} \).
Similar to earlier problems, the infinitely repeating exponent allows us to write this as:
\( y = x^y \)
To differentiate this, take the natural logarithm on both sides:
\( \log y = \log(x^y) \)
Using the logarithm property \( \log(a^b) = b \log a \):
\( \log y = y \log x \)
Now, differentiate both sides with respect to \( x \). Remember to use the product rule for \( y \log x \) and the chain rule for \( \log y \):
\( \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
Group the terms containing \( \frac{dy}{dx} \):
\( \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log x = \frac{y}{x} \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx} \left( \frac{1}{y} - \log x \right) = \frac{y}{x} \)
Combine the terms inside the parenthesis on the left side:
\( \frac{dy}{dx} \left( \frac{1 - y \log x}{y} \right) = \frac{y}{x} \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{y}{x} \cdot \frac{y}{1 - y \log x} \)
\( \frac{dy}{dx} = \frac{y^2}{x(1 - y \log x)} \)
Now, we need to prove that \( x\frac{dy}{dx} = \frac{y^2}{1-y \log x} \). Multiply both sides of the last equation by \( x \):
\( x \frac{dy}{dx} = x \cdot \frac{y^2}{x(1 - y \log x)} \)
\( \implies \) \( x\frac{dy}{dx} = \frac{y^2}{1 - y \log x} \).
In simple words: This problem shows how to find the derivative of a never-ending power expression. First, you rewrite it, then use logarithms to simplify, and finally, differentiate. The last step is to multiply by \( x \) to get the desired result.

🎯 Exam Tip: Pay attention to what the question asks you to prove. Sometimes, you need to perform an extra step (like multiplying by \( x \)) after finding the standard derivative to match the required form.

Question 25. Find \( \frac{dy}{dx} \) when \( x^y + y^x = c \).
Answer: Given the equation is \( x^y + y^x = c \), where \( c \) is a constant.
To differentiate this implicitly, let's break it into two parts: \( u = x^y \) and \( v = y^x \). So, \( u + v = c \).
Differentiating \( u + v = c \) with respect to \( x \), we get: \( \frac{du}{dx} + \frac{dv}{dx} = 0 \).

First, find \( \frac{du}{dx} \) for \( u = x^y \):
Take the natural logarithm on both sides: \( \log u = \log(x^y) \)
\( \log u = y \log x \)
Differentiate both sides with respect to \( x \):
\( \frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \)
\( \implies \) \( \frac{du}{dx} = u \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) \)
\( \implies \) \( \frac{du}{dx} = x^y \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) \)
\( \implies \) \( \frac{du}{dx} = y x^{y-1} + x^y \log x \frac{dy}{dx} \).

Next, find \( \frac{dv}{dx} \) for \( v = y^x \):
Take the natural logarithm on both sides: \( \log v = \log(y^x) \)
\( \log v = x \log y \)
Differentiate both sides with respect to \( x \):
\( \frac{1}{v} \frac{dv}{dx} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx} \)
\( \implies \) \( \frac{dv}{dx} = v \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) \)
\( \implies \) \( \frac{dv}{dx} = y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) \)
\( \implies \) \( \frac{dv}{dx} = y^x \log y + x y^{x-1} \frac{dy}{dx} \).

Now, substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into \( \frac{du}{dx} + \frac{dv}{dx} = 0 \):
\( \left( y x^{y-1} + x^y \log x \frac{dy}{dx} \right) + \left( y^x \log y + x y^{x-1} \frac{dy}{dx} \right) = 0 \)
Group the terms containing \( \frac{dy}{dx} \):
\( \left( x^y \log x + x y^{x-1} \right) \frac{dy}{dx} = - \left( y x^{y-1} + y^x \log y \right) \)
Finally, solve for \( \frac{dy}{dx} \):
\( \implies \) \( \frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \).
In simple words: When you have an equation with two complex parts added together, you can find the derivative of each part separately using logarithms. Then, you combine these derivatives and solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: For equations like \( u(x,y) + v(x,y) = c \), always differentiate each term separately and then sum them up. Remember to apply implicit differentiation rules correctly for terms involving \( y \).

Question 26. If \( x^y = e^{y-x} \), find \( \frac{dy}{dx} \).
Answer: Given the equation is \( x^y = e^{y-x} \).
To simplify and find \( \frac{dy}{dx} \), take the natural logarithm on both sides:
\( \log(x^y) = \log(e^{y-x}) \)
Using logarithm properties \( \log(A^B) = B \log A \) and \( \log e = 1 \):
\( y \log x = (y - x) \cdot 1 \)
\( y \log x = y - x \)
Now, rearrange the terms to isolate \( y \):
\( x = y - y \log x \)
Factor out \( y \) from the right side:
\( x = y(1 - \log x) \)
Now, express \( y \) explicitly in terms of \( x \):
\( y = \frac{x}{1 - \log x} \)
Next, differentiate \( y \) with respect to \( x \) using the quotient rule, \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).
Here, \( u = x \) and \( v = 1 - \log x \).
So, \( u' = \frac{d}{dx}(x) = 1 \).
And \( v' = \frac{d}{dx}(1 - \log x) = 0 - \frac{1}{x} = - \frac{1}{x} \).
Substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{1 \cdot (1 - \log x) - x \cdot \left(-\frac{1}{x}\right)}{(1 - \log x)^2} \)
Simplify the numerator:
\( \frac{dy}{dx} = \frac{1 - \log x + 1}{(1 - \log x)^2} \)
\( \implies \) \( \frac{dy}{dx} = \frac{2 - \log x}{(1 - \log x)^2} \).
In simple words: First, you use logarithms to get rid of the powers. Then, you rearrange the equation to make \( y \) stand alone. After that, you use the division rule for derivatives to find the final answer.

🎯 Exam Tip: When dealing with equations where variables are in exponents, always consider taking logarithms to linearize the exponents. The quotient rule is essential for rational functions.

Question 27. Differentiate \( (\sin x)^x \) with respect to \( x^2 \).
Answer: We need to differentiate one function with respect to another. Let \( y = (\sin x)^x \) and \( z = x^2 \). We want to find \( \frac{dy}{dz} \). This can be found using the chain rule: \( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \).

Step 1: Find \( \frac{dy}{dx} \) for \( y = (\sin x)^x \).
Take the natural logarithm on both sides:
\( \log y = \log((\sin x)^x) \)
Using logarithm property \( \log(A^B) = B \log A \):
\( \log y = x \log(\sin x) \)
Differentiate both sides with respect to \( x \). Use the product rule on the right side and the chain rule on the left side:
\( \frac{1}{y} \frac{dy}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x \)
\( \frac{1}{y} \frac{dy}{dx} = \log(\sin x) + x \cot x \)
Solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = y (\log(\sin x) + x \cot x) \)
\( \implies \) \( \frac{dy}{dx} = (\sin x)^x (\log(\sin x) + x \cot x) \).

Step 2: Find \( \frac{dz}{dx} \) for \( z = x^2 \).
Differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(x^2) \)
\( \implies \) \( \frac{dz}{dx} = 2x \).

Step 3: Calculate \( \frac{dy}{dz} \).
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \)
Substitute the expressions for \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \):
\( \implies \) \( \frac{dy}{dz} = \frac{(\sin x)^x (x \cot x + \log(\sin x))}{2x} \).
In simple words: When you need to differentiate one function with respect to another, you first find the derivative of each function separately concerning \( x \). Then, you divide the first derivative by the second derivative to get your final answer.

🎯 Exam Tip: Remember the formula for differentiation of a function with respect to another: \( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \). This involves finding two separate derivatives with respect to \( x \) and then dividing them.