Question

The derivative of \( f(\tan x) \) w.r.t. \( g(\sec x) \) at \( x = \frac{\pi}{4} \), where \( f'(1) = 2 \) and \( g'( \sqrt{2} ) = 4 \), is

• A. \( \frac{1}{\sqrt{2}} \)
• B. 2
• C. \( \sqrt{2} \)
• D. \( \frac{1}{2} \)

Hint:

For derivatives involving composite functions, always apply the chain rule and evaluate at the given points carefully.

Answer 1

The Correct Option is A

Step 1: Apply the chain rule.
The derivative of \( f(\tan x) \) w.r.t. \( g(\sec x) \) is given by: \[ \frac{d}{dx} \left[ f(\tan x) \right] = f'( \tan x) \cdot \sec^2 x \] and \[ \frac{d}{dx} \left[ g(\sec x) \right] = g'( \sec x) \cdot \sec x \tan x \]
Step 2: Evaluate at \( x = \frac{\pi}{4} \).
At \( x = \frac{\pi}{4} \), \( \tan x = 1 \) and \( \sec x = \sqrt{2} \). So, substituting the given values: \[ f'( \tan \frac{\pi}{4}) = f'(1) = 2 \quad \text{and} \quad g'( \sec \frac{\pi}{4}) = g'(\sqrt{2}) = 4 \] Thus, \[ \frac{d}{dx} \left[ f(\tan x) \right] \Bigg/ \frac{d}{dx} \left[ g(\sec x) \right] = \frac{2 \cdot \sec^2 \frac{\pi}{4}}{4 \cdot \sec \frac{\pi}{4} \cdot \tan \frac{\pi}{4}} = \frac{2 \cdot 2}{4 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} \]
Step 3: Conclusion.
The correct answer is \( \frac{1}{\sqrt{2}} \).