Question

Find \( \frac{d}{dx} \log (\sec x + \tan x) \):

• A. \( \frac{1}{\sec x + \tan x} \)
• B. \( \sec x \)
• C. \( \tan x \)
• D. \( \sec x + \tan x \)

Hint:

When differentiating \( \log(f(x)) \), use the chain rule: \( \frac{d}{dx} \log(f(x)) = \frac{f'(x)}{f(x)} \).

Answer 1

The Correct Option is A

We are given: \[ \frac{d}{dx} \log (\sec x + \tan x) \] Using the chain rule: \[ \frac{d}{dx} \log f(x) = \frac{1}{f(x)} \cdot f'(x) \] Let \( f(x) = \sec x + \tan x \). Then: \[ f'(x) = \frac{d}{dx} (\sec x + \tan x) = \sec x \tan x + \sec^2 x \] So: \[ \frac{d}{dx} \log (\sec x + \tan x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \] But notice: \[ \frac{d}{dx} \log (\sec x + \tan x) = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{d}{dx} (\log (\sec x + \tan x)) \] This simplifies to: \[ \frac{d}{dx} \log (\sec x + \tan x) = \frac{d}{dx} [\log(\sec x + \tan x)] = \boxed{ \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) } \] But none of the options contain this expression — only (A) is a partial derivative rule form. However, since: \[ \frac{d}{dx} \log(\sec x + \tan x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \] So the correct answer isn't just (A) — it should be the full expression. Correction: The correct answer based on derivative computation is: \[ \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \] But none of the given options match this. So either the options are incorrect or incomplete.