If \(y = \log_e(\sec x + \tan x)\), \(0 \lt x \lt \displaystyle\frac{\pi}{2}\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sec x\)
By the chain rule, \(\frac{d}{dx}\log_e u = \frac{1}{u}\frac{du}{dx}\) with \(u = \sec x + \tan x\):
\( \displaystyle \frac{dy}{dx} = \frac{\sec x\tan x + \sec^{2}x}{\sec x + \tan x} \)
Taking \(\sec x\) common in the numerator:
\( \displaystyle \frac{dy}{dx} = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} \)
\( = \sec x \)