Question

If \( f(x) = \log(\sec x + \tan x) \), then \[ f'\left( \frac{\pi}{4} \right) = \]

• A. 1
• B. \( \frac{2}{\sqrt{3}} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \sqrt{2} \)

Hint:

When differentiating logarithmic functions, always apply the chain rule and simplify the trigonometric expressions carefully.

Answer 1

The Correct Option is D

Step 1: Differentiating the function.
The given function is \( f(x) = \log(\sec x + \tan x) \). We differentiate it using the chain rule: \[ f'(x) = \frac{d}{dx} \log(\sec x + \tan x) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} (\sec x + \tan x) \] We know that \( \frac{d}{dx} \sec x = \sec x \tan x \) and \( \frac{d}{dx} \tan x = \sec^2 x \), so: \[ f'(x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \]
Step 2: Evaluating the derivative at \( x = \frac{\pi}{4} \).
At \( x = \frac{\pi}{4} \), \( \sec \frac{\pi}{4} = \sqrt{2} \) and \( \tan \frac{\pi}{4} = 1 \). Substituting these values into the derivative: \[ f'\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2} + 1} \cdot \left( \sqrt{2} \times 1 + 2 \right) \] After simplifying, we find that \( f'\left( \frac{\pi}{4} \right) = \sqrt{2} \).

Step 3: Conclusion.
Thus, \( f'\left( \frac{\pi}{4} \right) = \sqrt{2} \), which makes option (D) the correct answer.