If $y=x+tan\, x$, then value of $\frac{d^2y}{dx^2}$ at $x=\frac{\pi}{4}$ is :

1 2 4 $2\sqrt{2}$

4

The correct answer is Option (3) → 4 $y=x+tan\, x$ $\frac{dy}{dx}=1+\sec^2x$ $\frac{d^2y}{dx^2}=2\sec^2x\tan x$ so $\left.\frac{d^2y}{dx^2}\right]_{x=\frac{\pi}{4}}=2×{\sqrt{2}}^2×1=4$