Question

Find the second order derivatives of the function
$tan^{-1}x$

Answer 1

The correct answer is $=\frac{-2x}{(1+x^2)^2}$
Let $y=tan^{-1}x$
Then,
$\frac{dy}{dx}=\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^2}$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{1}{1+x^2}]$
$=\frac{d}{dx}(1+x^2)^{-1}$
$=(-1).(1+x^2)^{-2}.\frac{d}{dx}(1+x^2)$
$=\frac{-1}{(1+x^2)^2}\times 2x$
$=\frac{-2x}{(1+x^2)^2}$