If $y=x+tan\, x$, then value of $\frac{d^2y}{dx^2}$ at $x=\frac{\pi}{4}$ is :
Answer & explanation
Correct answer: option 3
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 $\left.\frac{d^2y}{dx^2}\right]_{x=\frac{\pi}{4}}=2×{\sqrt{2}}^2×1=4$