RBSE Solutions Class 12 Maths Chapter 7 Differentiation Exercise 7.1

Find the derivative of following functions w.r.t. x :

 

Question 1. \( \text{sin} (x^2) \)
Answer: Let the function be \( y = \text{sin}(x^2) \). We need to differentiate this with respect to \( x \). Using the chain rule, we first differentiate \( \text{sin}(u) \) to get \( \text{cos}(u) \), where \( u = x^2 \). Then we multiply by the derivative of \( u \) with respect to \( x \), which is \( \frac{d}{dx}(x^2) = 2x \). So, \( \frac{dy}{dx} = \text{cos}(x^2) \cdot (2x) \). This gives us \( 2x \text{cos}(x^2) \). The chain rule helps break down complex derivatives into simpler parts.
In simple words: To find the derivative of \( \text{sin}(x^2) \), we first get \( \text{cos}(x^2) \). Then, we multiply this by the derivative of \( x^2 \), which is \( 2x \). So the final answer is \( 2x \text{cos}(x^2) \).

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating composite functions like \( \text{sin}(x^2) \). Differentiate the outer function first, then multiply by the derivative of the inner function.

 

Question 2. \( \text{tan} (2x + 3) \)
Answer: Let the given function be \( y = \text{tan}(2x + 3) \). To differentiate this, we use the chain rule. First, we differentiate \( \text{tan}(u) \) to get \( \text{sec}^2(u) \), where \( u = 2x + 3 \). Next, we find the derivative of \( u \) with respect to \( x \), which is \( \frac{d}{dx}(2x + 3) = 2 \). Therefore, the derivative \( \frac{dy}{dx} \) is \( \text{sec}^2(2x + 3) \cdot 2 \). This simplifies to \( 2 \text{sec}^2(2x + 3) \). The chain rule is very useful for functions within functions.
In simple words: To find the derivative of \( \text{tan}(2x + 3) \), we first write \( \text{sec}^2(2x + 3) \). Then we multiply this by the derivative of \( 2x + 3 \), which is \( 2 \). So, the final answer is \( 2 \text{sec}^2(2x + 3) \).

🎯 Exam Tip: Always identify the inner and outer functions clearly when using the chain rule. For \( \text{tan}(f(x)) \), the derivative is \( \text{sec}^2(f(x)) \cdot f'(x) \).

 

Question 3. \( \text{sin} \{\text{cos} (x^2)\} \)
Answer: Let \( y = \text{sin} \{\text{cos}(x^2)\} \). This function involves multiple layers, so we apply the chain rule repeatedly. First, differentiate \( \text{sin}(u) \) with respect to \( u \), where \( u = \text{cos}(x^2) \). This gives \( \text{cos}(u) \), or \( \text{cos}(\text{cos}(x^2)) \). Next, differentiate \( u = \text{cos}(x^2) \) with respect to \( x \). This involves another chain rule step. Differentiate \( \text{cos}(v) \) with respect to \( v \), where \( v = x^2 \), which gives \( -\text{sin}(v) \), or \( -\text{sin}(x^2) \). Finally, differentiate \( v = x^2 \) with respect to \( x \), which gives \( 2x \). Combining these steps: \( \frac{dy}{dx} = \text{cos}(\text{cos}(x^2)) \cdot (-\text{sin}(x^2)) \cdot (2x) \) \( \implies \frac{dy}{dx} = -2x \text{sin}(x^2) \text{cos}(\text{cos}(x^2)) \). Layered functions require careful application of the chain rule from outside to inside.
In simple words: To find the derivative of \( \text{sin}\{\text{cos}(x^2)\} \), we differentiate it step-by-step. First, \( \text{sin} \) becomes \( \text{cos} \), keeping \( \text{cos}(x^2) \) inside. Then, \( \text{cos}(x^2) \) becomes \( -\text{sin}(x^2) \), and finally \( x^2 \) becomes \( 2x \). Multiply all these parts together to get \( -2x \text{sin}(x^2) \text{cos}(\text{cos}(x^2)) \).

🎯 Exam Tip: For nested functions like \( f(g(h(x))) \), apply the chain rule from the outermost function inwards: \( f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \).

 

Question 4. \( \frac{\text{sec}x - 1}{\text{sec}x + 1} \)
Answer: Let \( y = \frac{\text{sec}x - 1}{\text{sec}x + 1} \). We can simplify this expression first using the identity \( \text{sec}x = \frac{1}{\text{cos}x} \): \( y = \frac{\frac{1}{\text{cos}x} - 1}{\frac{1}{\text{cos}x} + 1} = \frac{\frac{1 - \text{cos}x}{\text{cos}x}}{\frac{1 + \text{cos}x}{\text{cos}x}} = \frac{1 - \text{cos}x}{1 + \text{cos}x} \) Now, we differentiate \( y = \frac{1 - \text{cos}x}{1 + \text{cos}x} \) using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = 1 - \text{cos}x \) and \( v = 1 + \text{cos}x \). \( u' = \frac{d}{dx}(1 - \text{cos}x) = 0 - (-\text{sin}x) = \text{sin}x \) \( v' = \frac{d}{dx}(1 + \text{cos}x) = 0 + (-\text{sin}x) = -\text{sin}x \) Applying the quotient rule: \( \frac{dy}{dx} = \frac{(\text{sin}x)(1 + \text{cos}x) - (1 - \text{cos}x)(-\text{sin}x)}{(1 + \text{cos}x)^2} \) \( \implies \frac{dy}{dx} = \frac{\text{sin}x + \text{sin}x \text{cos}x + \text{sin}x - \text{sin}x \text{cos}x}{(1 + \text{cos}x)^2} \) \( \implies \frac{dy}{dx} = \frac{2 \text{sin}x}{(1 + \text{cos}x)^2} \) This derivative is much simpler after the initial trigonometric simplification. We can also use the half-angle formulas: \( 1 - \text{cos}x = 2\text{sin}^2 \left( \frac{x}{2} \right) \) and \( 1 + \text{cos}x = 2\text{cos}^2 \left( \frac{x}{2} \right) \). So, \( y = \frac{2\text{sin}^2 \left( \frac{x}{2} \right)}{2\text{cos}^2 \left( \frac{x}{2} \right)} = \text{tan}^2 \left( \frac{x}{2} \right) \). Differentiating \( \text{tan}^2 \left( \frac{x}{2} \right) \) using the chain rule: \( \frac{dy}{dx} = 2 \text{tan} \left( \frac{x}{2} \right) \cdot \text{sec}^2 \left( \frac{x}{2} \right) \cdot \frac{d}{dx} \left( \frac{x}{2} \right) \) \( \implies \frac{dy}{dx} = 2 \text{tan} \left( \frac{x}{2} \right) \cdot \text{sec}^2 \left( \frac{x}{2} \right) \cdot \frac{1}{2} \) \( \implies \frac{dy}{dx} = \text{tan} \left( \frac{x}{2} \right) \text{sec}^2 \left( \frac{x}{2} \right) \) We know \( \text{tan} \left( \frac{x}{2} \right) = \frac{\text{sin} \left( \frac{x}{2} \right)}{\text{cos} \left( \frac{x}{2} \right)} \) and \( \text{sec}^2 \left( \frac{x}{2} \right) = \frac{1}{\text{cos}^2 \left( \frac{x}{2} \right)} \). \( \implies \frac{dy}{dx} = \frac{\text{sin} \left( \frac{x}{2} \right)}{\text{cos}^3 \left( \frac{x}{2} \right)} \). This looks different. Let's convert \( \frac{2 \text{sin}x}{(1 + \text{cos}x)^2} \) to the half-angle form. \( 2 \text{sin}x = 2 (2 \text{sin} \left( \frac{x}{2} \right) \text{cos} \left( \frac{x}{2} \right)) = 4 \text{sin} \left( \frac{x}{2} \right) \text{cos} \left( \frac{x}{2} \right) \) \( (1 + \text{cos}x)^2 = (2\text{cos}^2 \left( \frac{x}{2} \right))^2 = 4\text{cos}^4 \left( \frac{x}{2} \right) \) So, \( \frac{dy}{dx} = \frac{4 \text{sin} \left( \frac{x}{2} \right) \text{cos} \left( \frac{x}{2} \right)}{4\text{cos}^4 \left( \frac{x}{2} \right)} = \frac{\text{sin} \left( \frac{x}{2} \right)}{\text{cos}^3 \left( \frac{x}{2} \right)} \), which matches. Both methods yield the same result. Pre-simplifying the expression often makes differentiation easier.
In simple words: First, rewrite the expression using \( \text{sec}x = \frac{1}{\text{cos}x} \), which simplifies it to \( \frac{1 - \text{cos}x}{1 + \text{cos}x} \). Then, use the quotient rule for differentiation. The derivative of \( 1 - \text{cos}x \) is \( \text{sin}x \), and the derivative of \( 1 + \text{cos}x \) is \( -\text{sin}x \). Putting these into the quotient rule formula gives \( \frac{2 \text{sin}x}{(1 + \text{cos}x)^2} \).

🎯 Exam Tip: Always simplify trigonometric expressions before differentiating, if possible. This can significantly reduce the complexity of the calculation and potential errors.

 

Question 5. \( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \)
Answer: Let \( y = \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \). It's helpful to simplify this expression first by multiplying the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{1+x} - \sqrt{1-x} \). \( y = \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})}{(\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})} \) Using \( (a-b)^2 = a^2 - 2ab + b^2 \) for the numerator and \( (a+b)(a-b) = a^2 - b^2 \) for the denominator: Numerator: \( (\sqrt{1+x})^2 - 2\sqrt{1+x}\sqrt{1-x} + (\sqrt{1-x})^2 = (1+x) - 2\sqrt{1-x^2} + (1-x) = 2 - 2\sqrt{1-x^2} \) Denominator: \( (\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 1+x-1+x = 2x \) So, \( y = \frac{2 - 2\sqrt{1-x^2}}{2x} = \frac{1 - \sqrt{1-x^2}}{x} \). Now we differentiate \( y = \frac{1 - \sqrt{1-x^2}}{x} \) using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = 1 - \sqrt{1-x^2} \) and \( v = x \). \( u' = \frac{d}{dx}(1 - \sqrt{1-x^2}) = 0 - \frac{1}{2\sqrt{1-x^2}} \cdot \frac{d}{dx}(1-x^2) = - \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{x}{\sqrt{1-x^2}} \) \( v' = \frac{d}{dx}(x) = 1 \) Applying the quotient rule: \( \frac{dy}{dx} = \frac{\left( \frac{x}{\sqrt{1-x^2}} \right) (x) - (1 - \sqrt{1-x^2})(1)}{x^2} \) \( \implies \frac{dy}{dx} = \frac{\frac{x^2}{\sqrt{1-x^2}} - (1 - \sqrt{1-x^2})}{x^2} \) To combine the terms in the numerator, find a common denominator: \( \implies \frac{dy}{dx} = \frac{\frac{x^2 - \sqrt{1-x^2}(1 - \sqrt{1-x^2})}{\sqrt{1-x^2}}}{x^2} \) \( \implies \frac{dy}{dx} = \frac{x^2 - \sqrt{1-x^2} + (1-x^2)}{x^2 \sqrt{1-x^2}} \) \( \implies \frac{dy}{dx} = \frac{x^2 - \sqrt{1-x^2} + 1 - x^2}{x^2 \sqrt{1-x^2}} \) \( \implies \frac{dy}{dx} = \frac{1 - \sqrt{1-x^2}}{x^2 \sqrt{1-x^2}} \) The process of rationalizing the expression simplifies the derivative steps significantly.
In simple words: First, simplify the given expression by multiplying the top and bottom by the conjugate of the bottom part. This turns the expression into \( \frac{1 - \sqrt{1-x^2}}{x} \). Then, use the quotient rule to find the derivative. The derivative of the top part is \( \frac{x}{\sqrt{1-x^2}} \), and the derivative of the bottom part \( x \) is \( 1 \). Substitute these into the quotient rule formula and simplify to get \( \frac{1 - \sqrt{1-x^2}}{x^2 \sqrt{1-x^2}} \).

🎯 Exam Tip: For expressions involving square roots in the numerator and denominator, rationalize the denominator (or the whole expression) first. This pre-computation step often simplifies the differentiation process significantly.

 

Question 6. \( \text{sin}(x^\circ) \)
Answer: Let \( y = \text{sin}(x^\circ) \). Before differentiating, we must convert the angle from degrees to radians, as trigonometric derivative formulas typically use radians. We know that \( 1^\circ = \frac{\pi}{180} \) radians. Therefore, \( x^\circ = x \cdot \frac{\pi}{180} \) radians. So, \( y = \text{sin} \left( \frac{\pi x}{180} \right) \). Now, we differentiate with respect to \( x \) using the chain rule. \( \frac{dy}{dx} = \text{cos} \left( \frac{\pi x}{180} \right) \cdot \frac{d}{dx} \left( \frac{\pi x}{180} \right) \) \( \implies \frac{dy}{dx} = \text{cos} \left( \frac{\pi x}{180} \right) \cdot \frac{\pi}{180} \) We can write this back in terms of degrees: \( \implies \frac{dy}{dx} = \frac{\pi}{180} \text{cos}(x^\circ) \). This shows that differentiating an angle in degrees introduces a factor of \( \frac{\pi}{180} \).
In simple words: First, change \( x^\circ \) to radians using the rule \( x^\circ = \frac{\pi x}{180} \) radians. So, the function becomes \( y = \text{sin} \left( \frac{\pi x}{180} \right) \). Then, differentiate this using the chain rule. The derivative will be \( \text{cos} \left( \frac{\pi x}{180} \right) \) multiplied by \( \frac{\pi}{180} \). This simplifies to \( \frac{\pi}{180} \text{cos}(x^\circ) \).

🎯 Exam Tip: When dealing with trigonometric functions involving angles in degrees, always convert the angle to radians first before applying differentiation formulas. This is a common point of error.

 

Question 7. \( \text{log}_e \sqrt{\frac{1-\text{cos}x}{1+\text{cos}x}} \)
Answer: Let \( y = \text{log}_e \sqrt{\frac{1-\text{cos}x}{1+\text{cos}x}} \). We should simplify the expression inside the logarithm first. We know that \( \frac{1-\text{cos}x}{1+\text{cos}x} = \text{tan}^2 \left( \frac{x}{2} \right) \). So, \( y = \text{log}_e \sqrt{\text{tan}^2 \left( \frac{x}{2} \right)} = \text{log}_e \left| \text{tan} \left( \frac{x}{2} \right) \right| \). For most differentiation problems, we assume the function is defined where \( \text{tan} \left( \frac{x}{2} \right) \) is positive. So, \( y = \text{log}_e \left( \text{tan} \left( \frac{x}{2} \right) \right) \). Now, we differentiate using the chain rule. Recall that \( \frac{d}{dx}(\text{log}_e u) = \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \text{tan} \left( \frac{x}{2} \right) \). \( \frac{du}{dx} = \frac{d}{dx} \left( \text{tan} \left( \frac{x}{2} \right) \right) = \text{sec}^2 \left( \frac{x}{2} \right) \cdot \frac{d}{dx} \left( \frac{x}{2} \right) = \text{sec}^2 \left( \frac{x}{2} \right) \cdot \frac{1}{2} \). So, \( \frac{dy}{dx} = \frac{1}{\text{tan} \left( \frac{x}{2} \right)} \cdot \frac{1}{2} \text{sec}^2 \left( \frac{x}{2} \right) \). We know \( \text{tan} \left( \frac{x}{2} \right) = \frac{\text{sin} \left( \frac{x}{2} \right)}{\text{cos} \left( \frac{x}{2} \right)} \) and \( \text{sec}^2 \left( \frac{x}{2} \right) = \frac{1}{\text{cos}^2 \left( \frac{x}{2} \right)} \). \( \implies \frac{dy}{dx} = \frac{\text{cos} \left( \frac{x}{2} \right)}{\text{sin} \left( \frac{x}{2} \right)} \cdot \frac{1}{2} \cdot \frac{1}{\text{cos}^2 \left( \frac{x}{2} \right)} \) \( \implies \frac{dy}{dx} = \frac{1}{2 \text{sin} \left( \frac{x}{2} \right) \text{cos} \left( \frac{x}{2} \right)} \) Using the identity \( \text{sin}x = 2 \text{sin} \left( \frac{x}{2} \right) \text{cos} \left( \frac{x}{2} \right) \): \( \implies \frac{dy}{dx} = \frac{1}{\text{sin}x} = \text{cosec}x \). Simplifying trigonometric expressions before differentiation is a powerful technique.
In simple words: First, simplify the inside of the logarithm using the identity \( \frac{1-\text{cos}x}{1+\text{cos}x} = \text{tan}^2 \left( \frac{x}{2} \right) \). This makes the function \( y = \text{log}_e \left( \text{tan} \left( \frac{x}{2} \right) \right) \). Then, differentiate this using the chain rule. \( \text{log}_e u \) becomes \( \frac{1}{u} \cdot u' \). The derivative of \( \text{tan} \left( \frac{x}{2} \right) \) is \( \frac{1}{2} \text{sec}^2 \left( \frac{x}{2} \right) \). Combine and simplify using trigonometric identities to get \( \text{cosec}x \).

🎯 Exam Tip: For logarithmic functions, always simplify the argument of the logarithm first using log properties or trigonometric identities. This often turns a complex derivative into a much simpler one.

 

Question 8. \( \text{sec}(x^\circ) \)
Answer: Let \( y = \text{sec}(x^\circ) \). Similar to Question 6, we must convert the angle from degrees to radians. \( x^\circ = x \cdot \frac{\pi}{180} \) radians. So, \( y = \text{sec} \left( \frac{\pi x}{180} \right) \). Now, we differentiate using the chain rule. Recall that \( \frac{d}{dx}(\text{sec} u) = \text{sec} u \text{tan} u \cdot \frac{du}{dx} \). Here, \( u = \frac{\pi x}{180} \). \( \frac{du}{dx} = \frac{d}{dx} \left( \frac{\pi x}{180} \right) = \frac{\pi}{180} \). So, \( \frac{dy}{dx} = \text{sec} \left( \frac{\pi x}{180} \right) \text{tan} \left( \frac{\pi x}{180} \right) \cdot \frac{\pi}{180} \). Converting back to degrees notation: \( \implies \frac{dy}{dx} = \frac{\pi}{180} \text{sec}(x^\circ) \text{tan}(x^\circ) \). This emphasizes the importance of consistent units in calculus.
In simple words: First, change \( x^\circ \) to radians using \( x^\circ = \frac{\pi x}{180} \) radians, so the function is \( y = \text{sec} \left( \frac{\pi x}{180} \right) \). Then, use the chain rule to differentiate. The derivative of \( \text{sec}(u) \) is \( \text{sec}(u)\text{tan}(u) \) multiplied by the derivative of \( u \). The derivative of \( \frac{\pi x}{180} \) is \( \frac{\pi}{180} \). So, the final answer is \( \frac{\pi}{180} \text{sec}(x^\circ) \text{tan}(x^\circ) \).

🎯 Exam Tip: Always convert angles in degrees to radians (by multiplying by \( \frac{\pi}{180} \)) before differentiating trigonometric functions to ensure correct application of derivative rules.

 

Question 9. \( \text{log} \sqrt{\frac{1+\text{sin}x}{1-\text{sin}x}} \)
Answer: Let \( y = \text{log} \sqrt{\frac{1+\text{sin}x}{1-\text{sin}x}} \). We need to simplify the expression inside the logarithm using trigonometric identities. We know \( 1+\text{sin}x = \left( \text{cos} \frac{x}{2} + \text{sin} \frac{x}{2} \right)^2 \) and \( 1-\text{sin}x = \left( \text{cos} \frac{x}{2} - \text{sin} \frac{x}{2} \right)^2 \). So, \( \frac{1+\text{sin}x}{1-\text{sin}x} = \frac{\left( \text{cos} \frac{x}{2} + \text{sin} \frac{x}{2} \right)^2}{\left( \text{cos} \frac{x}{2} - \text{sin} \frac{x}{2} \right)^2} = \left( \frac{\text{cos} \frac{x}{2} + \text{sin} \frac{x}{2}}{\text{cos} \frac{x}{2} - \text{sin} \frac{x}{2}} \right)^2 \). Dividing numerator and denominator by \( \text{cos} \frac{x}{2} \): \( \frac{1+\text{tan} \frac{x}{2}}{1-\text{tan} \frac{x}{2}} = \text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right) \). So, \( y = \text{log} \sqrt{\left( \text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right) \right)^2} = \text{log} \left( \text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) \). Now, we differentiate using the chain rule: \( \frac{dy}{dx} = \frac{1}{\text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \frac{d}{dx} \left( \text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) \) \( \implies \frac{dy}{dx} = \text{cot} \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \text{sec}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} \right) \) \( \implies \frac{dy}{dx} = \text{cot} \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \text{sec}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \frac{1}{2} \) Rewrite in terms of \( \text{sin} \) and \( \text{cos} \): \( \implies \frac{dy}{dx} = \frac{\text{cos} \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\text{sin} \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \frac{1}{2 \text{cos}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)} \) \( \implies \frac{dy}{dx} = \frac{1}{2 \text{sin} \left( \frac{\pi}{4} + \frac{x}{2} \right) \text{cos} \left( \frac{\pi}{4} + \frac{x}{2} \right)} \) Using the identity \( 2 \text{sin} A \text{cos} A = \text{sin}(2A) \): \( \implies \frac{dy}{dx} = \frac{1}{\text{sin} \left( 2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \right)} = \frac{1}{\text{sin} \left( \frac{\pi}{2} + x \right)} \) We know \( \text{sin} \left( \frac{\pi}{2} + x \right) = \text{cos}x \). \( \implies \frac{dy}{dx} = \frac{1}{\text{cos}x} = \text{sec}x \). This extensive simplification greatly aids in finding the derivative.
In simple words: First, simplify the expression inside the logarithm using trigonometric identities. The term \( \frac{1+\text{sin}x}{1-\text{sin}x} \) becomes \( \text{tan}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \). So, the function is \( y = \text{log} \left( \text{tan} \left( \frac{\pi}{4} + \frac{x}{2} \right) \right) \). Differentiate using the chain rule, which involves the derivative of \( \text{log}(u) \), then \( \text{tan}(v) \), then \( \frac{\pi}{4} + \frac{x}{2} \). After simplifying all the terms, the final derivative is \( \text{sec}x \).

🎯 Exam Tip: When dealing with complex logarithmic and trigonometric expressions, remember to utilize identity transformations to simplify the function as much as possible before differentiation. This strategy often leads to a much simpler derivative.

 

Question 10. \( \text{log}_e \frac{x + \sqrt{x^2 + a^2}}{a} \)
Answer: Let \( y = \text{log}_e \frac{x + \sqrt{x^2 + a^2}}{a} \). We can simplify this using logarithm properties: \( \text{log}_e \left( \frac{A}{B} \right) = \text{log}_e A - \text{log}_e B \). So, \( y = \text{log}_e (x + \sqrt{x^2 + a^2}) - \text{log}_e a \). Now, differentiate with respect to \( x \). The derivative of \( \text{log}_e a \) (where \( a \) is a constant) is \( 0 \). For the first term, let \( u = x + \sqrt{x^2 + a^2} \). Then \( \frac{d}{dx}(\text{log}_e u) = \frac{1}{u} \cdot \frac{du}{dx} \). First, find \( \frac{du}{dx} \): \( \frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{x^2 + a^2}) \) \( \frac{d}{dx}(x) = 1 \) \( \frac{d}{dx}(\sqrt{x^2 + a^2}) = \frac{d}{dx}((x^2 + a^2)^{1/2}) = \frac{1}{2}(x^2 + a^2)^{-1/2} \cdot \frac{d}{dx}(x^2 + a^2) \) \( \implies = \frac{1}{2\sqrt{x^2 + a^2}} \cdot (2x) = \frac{x}{\sqrt{x^2 + a^2}} \) So, \( \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \). Now, substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of the first term: \( \frac{d}{dx}(\text{log}_e (x + \sqrt{x^2 + a^2})) = \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{x + \sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2}} \) \( \implies = \frac{1}{\sqrt{x^2 + a^2}} \). Therefore, \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 + a^2}} - 0 = \frac{1}{\sqrt{x^2 + a^2}} \). Logarithm properties greatly simplify the differentiation process for such expressions.
In simple words: First, use log properties to split the function into \( \text{log}_e (x + \sqrt{x^2 + a^2}) - \text{log}_e a \). The derivative of \( \text{log}_e a \) is zero. For the first part, differentiate \( \text{log}_e(\text{something}) \) as \( \frac{1}{\text{something}} \) times the derivative of \( \text{something} \). The derivative of \( x + \sqrt{x^2 + a^2} \) is \( 1 + \frac{x}{\sqrt{x^2 + a^2}} \). After combining and simplifying, the final derivative is \( \frac{1}{\sqrt{x^2 + a^2}} \).

🎯 Exam Tip: When differentiating a logarithm with a complex argument, use logarithm properties to expand and simplify the expression first. This often separates constant terms and simplifies the argument, making differentiation much easier.

 

Question 11. \( \text{log}_e \frac{x^2+x+1}{x^2-x+1} \)
Answer: Let \( y = \text{log}_e \frac{x^2+x+1}{x^2-x+1} \). Using logarithm properties, we can rewrite this as: \( y = \text{log}_e (x^2+x+1) - \text{log}_e (x^2-x+1) \). Now, differentiate each term with respect to \( x \). For the first term: \( \frac{d}{dx}(\text{log}_e (x^2+x+1)) = \frac{1}{x^2+x+1} \cdot \frac{d}{dx}(x^2+x+1) = \frac{2x+1}{x^2+x+1} \). For the second term: \( \frac{d}{dx}(\text{log}_e (x^2-x+1)) = \frac{1}{x^2-x+1} \cdot \frac{d}{dx}(x^2-x+1) = \frac{2x-1}{x^2-x+1} \). So, \( \frac{dy}{dx} = \frac{2x+1}{x^2+x+1} - \frac{2x-1}{x^2-x+1} \). To combine these, find a common denominator: \( \frac{dy}{dx} = \frac{(2x+1)(x^2-x+1) - (2x-1)(x^2+x+1)}{(x^2+x+1)(x^2-x+1)} \) Expand the terms in the numerator: Numerator: \( (2x+1)(x^2-x+1) = 2x^3 - 2x^2 + 2x + x^2 - x + 1 = 2x^3 - x^2 + x + 1 \) \( (2x-1)(x^2+x+1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \) Subtract the second expansion from the first: \( (2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1) = 2x^3 - x^2 + x + 1 - 2x^3 - x^2 - x + 1 = -2x^2 + 2 \) So, \( \frac{dy}{dx} = \frac{-2x^2 + 2}{(x^2+x+1)(x^2-x+1)} \). The denominator can be simplified as \( (x^2+1)^2 - x^2 = x^4 + 2x^2 + 1 - x^2 = x^4 + x^2 + 1 \). Therefore, \( \frac{dy}{dx} = \frac{2(1-x^2)}{x^4+x^2+1} \). Using log properties simplifies this problem significantly compared to differentiating the original quotient.
In simple words: First, rewrite the logarithm as the difference of two logarithms: \( \text{log}_e (x^2+x+1) - \text{log}_e (x^2-x+1) \). Then, differentiate each part separately. The derivative of \( \text{log}_e u \) is \( \frac{1}{u} \) times the derivative of \( u \). Combine the two resulting fractions by finding a common denominator and simplify the numerator. This results in \( \frac{2(1-x^2)}{x^4+x^2+1} \).

🎯 Exam Tip: Always use the property \( \text{log}_e \left( \frac{f(x)}{g(x)} \right) = \text{log}_e f(x) - \text{log}_e g(x) \) for differentiation problems. It converts a complex quotient rule problem into two simpler chain rule problems.

 

Question 12. \( \text{tan} \{\text{log}_e \sqrt{1 + x^2} \} \)
Answer: Let \( y = \text{tan} \{\text{log}_e \sqrt{1 + x^2} \} \). We can simplify the argument of the logarithm first. \( \sqrt{1 + x^2} = (1 + x^2)^{1/2} \). So, \( \text{log}_e \sqrt{1 + x^2} = \text{log}_e (1 + x^2)^{1/2} = \frac{1}{2} \text{log}_e (1 + x^2) \). Thus, the function becomes \( y = \text{tan} \left\{ \frac{1}{2} \text{log}_e (1 + x^2) \right\} \). Now, differentiate using the chain rule. First, differentiate \( \text{tan}(u) \) to get \( \text{sec}^2(u) \), where \( u = \frac{1}{2} \text{log}_e (1 + x^2) \). So, \( \frac{dy}{dx} = \text{sec}^2 \left\{ \frac{1}{2} \text{log}_e (1 + x^2) \right\} \cdot \frac{d}{dx} \left\{ \frac{1}{2} \text{log}_e (1 + x^2) \right\} \). Next, differentiate \( \frac{1}{2} \text{log}_e (1 + x^2) \): \( \frac{d}{dx} \left\{ \frac{1}{2} \text{log}_e (1 + x^2) \right\} = \frac{1}{2} \cdot \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) \) \( \implies = \frac{1}{2} \cdot \frac{1}{1 + x^2} \cdot (2x) = \frac{x}{1 + x^2} \). Substitute this back: \( \frac{dy}{dx} = \text{sec}^2 \left\{ \text{log}_e \sqrt{1 + x^2} \right\} \cdot \frac{x}{1 + x^2} \). This demonstrates how simplifying the function before differentiation can streamline the process.
In simple words: First, simplify the part inside the \( \text{tan} \) function. \( \text{log}_e \sqrt{1 + x^2} \) becomes \( \frac{1}{2} \text{log}_e (1 + x^2) \). So, \( y = \text{tan} \left( \frac{1}{2} \text{log}_e (1 + x^2) \right) \). Now, use the chain rule. The derivative of \( \text{tan}(u) \) is \( \text{sec}^2(u) \) times the derivative of \( u \). The derivative of \( \frac{1}{2} \text{log}_e (1 + x^2) \) is \( \frac{x}{1+x^2} \). Combine these to get \( \text{sec}^2 \left\{ \text{log}_e \sqrt{1 + x^2} \right\} \cdot \frac{x}{1 + x^2} \).

🎯 Exam Tip: When faced with functions involving nested operations (like \( \text{tan} \) of a \( \text{log} \) of a square root), simplify the innermost parts using properties of logarithms or powers first. This step often makes the chain rule application much clearer.

 

Question 13. \( a^{\text{tan}3x} \)
Answer: Let \( y = a^{\text{tan}3x} \). This is an exponential function where the base is a constant \( a \) and the exponent is a function of \( x \). Recall the derivative formula \( \frac{d}{dx}(a^u) = a^u \text{log}_e a \cdot \frac{du}{dx} \). Here, \( u = \text{tan}3x \). First, find \( \frac{du}{dx} \): \( \frac{d}{dx}(\text{tan}3x) = \text{sec}^2(3x) \cdot \frac{d}{dx}(3x) = \text{sec}^2(3x) \cdot 3 \). So, \( \frac{du}{dx} = 3\text{sec}^2(3x) \). Now, apply the formula for \( \frac{d}{dx}(a^u) \): \( \frac{dy}{dx} = a^{\text{tan}3x} \text{log}_e a \cdot (3\text{sec}^2(3x)) \). Rearranging the terms for clarity: \( \implies \frac{dy}{dx} = 3 a^{\text{tan}3x} \text{log}_e a \text{sec}^2(3x) \). This differentiation highlights the chain rule's role in exponential functions.
In simple words: To find the derivative of \( a^{\text{tan}3x} \), use the formula for differentiating \( a^u \), which is \( a^u \text{log}_e a \) times the derivative of \( u \). Here, \( u \) is \( \text{tan}3x \). The derivative of \( \text{tan}3x \) is \( \text{sec}^2(3x) \) multiplied by the derivative of \( 3x \), which is \( 3 \). Combine these parts to get \( 3 a^{\text{tan}3x} \text{log}_e a \text{sec}^2(3x) \).

🎯 Exam Tip: For derivatives of the form \( a^{f(x)} \), remember the chain rule: \( a^{f(x)} \text{log}_e a \cdot f'(x) \). Do not forget the \( \text{log}_e a \) term and the derivative of the exponent.

 

Question 14. \( \text{log}_e (\text{sec} x + \text{tan} x) \)
Answer: Let \( y = \text{log}_e (\text{sec} x + \text{tan} x) \). We use the chain rule for differentiating \( \text{log}_e u \), which is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \text{sec} x + \text{tan} x \). First, find \( \frac{du}{dx} \): \( \frac{d}{dx}(\text{sec} x + \text{tan} x) = \frac{d}{dx}(\text{sec} x) + \frac{d}{dx}(\text{tan} x) \) We know \( \frac{d}{dx}(\text{sec} x) = \text{sec} x \text{tan} x \) and \( \frac{d}{dx}(\text{tan} x) = \text{sec}^2 x \). So, \( \frac{du}{dx} = \text{sec} x \text{tan} x + \text{sec}^2 x \). We can factor out \( \text{sec} x \) from this expression: \( \frac{du}{dx} = \text{sec} x (\text{tan} x + \text{sec} x) \). Now, substitute \( u \) and \( \frac{du}{dx} \) back into the formula for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{1}{\text{sec} x + \text{tan} x} \cdot \text{sec} x (\text{tan} x + \text{sec} x) \) Notice that \( (\text{sec} x + \text{tan} x) \) in the denominator cancels with the \( (\text{tan} x + \text{sec} x) \) in the numerator. \( \implies \frac{dy}{dx} = \text{sec} x \). This derivative is remarkably simple, showcasing how some complex-looking functions simplify beautifully.
In simple words: To differentiate \( \text{log}_e (\text{sec} x + \text{tan} x) \), use the chain rule. This means \( \frac{1}{\text{sec} x + \text{tan} x} \) multiplied by the derivative of \( (\text{sec} x + \text{tan} x) \). The derivative of \( \text{sec} x \) is \( \text{sec} x \text{tan} x \), and the derivative of \( \text{tan} x \) is \( \text{sec}^2 x \). So, the derivative of the inner part is \( \text{sec} x \text{tan} x + \text{sec}^2 x \). Factor out \( \text{sec} x \), and you'll see a cancellation, leaving the final answer as \( \text{sec} x \).

🎯 Exam Tip: Be familiar with the derivatives of common trigonometric functions like \( \text{sec}x \) and \( \text{tan}x \). For logarithmic functions with trigonometric arguments, the result often simplifies after factoring terms in the numerator.

 

Question 15. \( \text{sin}^3 x \cdot \text{sin} 3x \)
Answer: Let \( y = \text{sin}^3 x \cdot \text{sin} 3x \). We will use the product rule \( \frac{d}{dx}(uv) = u'v + uv' \). First, use the triple angle identity for \( \text{sin}3x \): \( \text{sin}3x = 3\text{sin}x - 4\text{sin}^3 x \). From this, \( \text{sin}^3 x = \frac{3\text{sin}x - \text{sin}3x}{4} \). Substitute this back into the original expression for \( y \): \( y = \left( \frac{3\text{sin}x - \text{sin}3x}{4} \right) \cdot \text{sin} 3x \) \( \implies y = \frac{1}{4} (3\text{sin}x \text{sin}3x - \text{sin}^2 3x) \) This looks complicated to differentiate directly. Let's try differentiating the original \( y = \text{sin}^3 x \cdot \text{sin} 3x \) directly using the product rule. Here, \( u = \text{sin}^3 x \) and \( v = \text{sin} 3x \). First, find \( u' = \frac{d}{dx}(\text{sin}^3 x) \): Using the chain rule, \( \frac{d}{dx}((\text{sin}x)^3) = 3(\text{sin}x)^2 \cdot \frac{d}{dx}(\text{sin}x) = 3\text{sin}^2 x \text{cos}x \). Next, find \( v' = \frac{d}{dx}(\text{sin} 3x) \): Using the chain rule, \( \frac{d}{dx}(\text{sin} 3x) = \text{cos} 3x \cdot \frac{d}{dx}(3x) = 3\text{cos} 3x \). Now, apply the product rule: \( \frac{dy}{dx} = u'v + uv' \). \( \frac{dy}{dx} = (3\text{sin}^2 x \text{cos}x)(\text{sin}3x) + (\text{sin}^3 x)(3\text{cos}3x) \) Factor out \( 3\text{sin}^2 x \): \( \implies \frac{dy}{dx} = 3\text{sin}^2 x (\text{cos}x \text{sin}3x + \text{sin}x \text{cos}3x) \) Using the sum identity \( \text{sin}(A+B) = \text{sin}A \text{cos}B + \text{cos}A \text{sin}B \), with \( A=x \) and \( B=3x \): The term \( (\text{cos}x \text{sin}3x + \text{sin}x \text{cos}3x) \) is equivalent to \( \text{sin}(x+3x) = \text{sin}(4x) \). Therefore, \( \frac{dy}{dx} = 3\text{sin}^2 x \text{sin}(4x) \). This method is much more direct and yields a clean result. While identities can be used to transform expressions, direct differentiation with product/chain rules is often simpler.
In simple words: Treat \( \text{sin}^3 x \) as the first part and \( \text{sin} 3x \) as the second part. Use the product rule for differentiation. First, find the derivative of \( \text{sin}^3 x \), which is \( 3\text{sin}^2 x \text{cos}x \). Then, find the derivative of \( \text{sin} 3x \), which is \( 3\text{cos} 3x \). Put these into the product rule formula: \( (3\text{sin}^2 x \text{cos}x)(\text{sin}3x) + (\text{sin}^3 x)(3\text{cos}3x) \). Factor out \( 3\text{sin}^2 x \). The remaining part matches the identity for \( \text{sin}(A+B) \), which gives \( \text{sin}(4x) \). So, the final answer is \( 3\text{sin}^2 x \text{sin}(4x) \).

🎯 Exam Tip: For products of trigonometric functions, carefully apply the product rule and chain rule. Look for opportunities to simplify the resulting expression using sum-product identities to get the most concise form.

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