If $y =e^{\frac{1}{2}}\log(1+\tan^2x)$, then $\frac{d^2y}{dx^2}$ is equal to:

$\sec x (\sec^2x + \tan^2x)$

The correct answer is Option (3) → $\sec x (\sec^2x + \tan^2x)$ **

Given:

$y = e^{\frac12 \log(1+\tan^2 x)}$

Simplify using $e^{\log A}=A$:

$y = (1+\tan^2 x)^{1/2}$

Use identity $1+\tan^2 x = \sec^2 x$:

$y = (\sec^2 x)^{1/2} = \sec x$

Now compute derivatives.

$\displaystyle y' = \sec x \tan x$

$\displaystyle y'' = \frac{d}{dx}(\sec x \tan x)$

Use product rule:

$y'' = \sec x \cdot \sec^2 x + \tan x \cdot \sec x \tan x$

$y'' = \sec x (\sec^2 x + \tan^2 x)$

Correct answer: $\sec x \,(\sec^2 x + \tan^2 x)$