If \(y =e^{\frac{1}{2}\log(1+\tan^2x)}\), then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sec x (\sec^2x + \tan^2x)\)
Using the identity \(1 + \tan^2 x = \sec^2 x\) and \(e^{k\log A} = A^k\), the function simplifies first:
\( \displaystyle y = e^{\frac{1}{2}\log(\sec^2 x)} = (\sec^2 x)^{1/2} = \sec x \)
(taking \(\sec x \gt 0\), as the options do). Differentiating:
\( \displaystyle \frac{dy}{dx} = \sec x \tan x \)
By the product rule:
\( \displaystyle \frac{d^2y}{dx^2} = (\sec x \tan x)\tan x + \sec x \cdot \sec^2 x \)
\(= \sec x \tan^2 x + \sec^3 x\)
\(= \sec x(\sec^2 x + \tan^2 x)\)
Why the other options are incorrect
Option (1) \(\sec x\) — Incorrect. This is y itself, not its second derivative.
Option (2) \(\displaystyle\frac{1}{2}\left(\frac{1}{1+\tan^2x}\right)e^{\log(1+\tan^2x)}\) — Incorrect. Since \(e^{\log(1+\tan^2x)} = 1 + \tan^2 x\), this is just the constant \(\frac{1}{2}\).
Option (4) \(\sec x\tan x\) — Incorrect. This is the first derivative \(\frac{dy}{dx}\); it must be differentiated once more.