Question

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

Answer 1

Let $y={\tan }^{-1}x$
$\Rightarrow \frac{d}{dx}=\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$
$\therefore$ $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{1+x^2}\right)=\frac{d}{dx}(1+x^2{)}^{-1}=(-1).(1+x^2).\frac{d}{dx}(1+x^2)$
$=\frac{1}{(1+x^2{)}^2}\times 2x=\frac{-2x}{(1+x^2{)}^2}$