If \(y = \log_e(\sec x)\), where \(-\displaystyle\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1 + \left(\displaystyle\frac{dy}{dx}\right)^{2}\)
By the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{1}{\sec x} \cdot \sec x\tan x = \tan x \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \sec^{2}x \)
By the identity \(\sec^{2}x = 1 + \tan^{2}x\):
\( \displaystyle \frac{d^{2}y}{dx^{2}} = 1 + \tan^{2}x = 1 + \left(\frac{dy}{dx}\right)^{2} \)
Why the other options are incorrect
Option (1) \(\left(\frac{dy}{dx}\right)^{2}\) — Incorrect. This is \(\tan^{2}x\), which is 1 less than \(\sec^{2}x\).
Option (3) \(1 - \left(\frac{dy}{dx}\right)^{2}\) — Incorrect. This is \(1 - \tan^{2}x\); for example at \(x = \frac{\pi}{4}\) it is 0, while \(\sec^{2}\frac{\pi}{4} = 2\).
Option (4) \(1 + \frac{dy}{dx}\) — Incorrect. This is \(1 + \tan x\); for example at \(x = -\frac{\pi}{4}\) it is 0, while \(\sec^{2}\left(-\frac{\pi}{4}\right) = 2\).