Question

If $y=({\cot }^{-1}x{)}^2$, show that $(1+x^2{)}^2\frac{d^{2}y}{d{x}^2}+2x(1+x^2)\frac{dy}{dx}=2$.

Answer 1

$y=({\cot }^{-1}x{)}^2$
$\frac{dy}{dx}=2{\cot }^{-1}x\frac{d}{dx}\left({\cot }^{-1}x\right)$
$\Rightarrow \frac{dy}{dx}=2{\cot }^{-1}x\left(\frac{-1}{1+x^2}\right)$
$\Rightarrow (1+x^2)\frac{dy}{dx}=-2{\cot }^{-1}x$

Differentiating once again w.r.t. $x$
$(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}=-2.\frac{(-1)}{1+x^2}$

$(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}=\frac{2}{1+x^2}$

$\Rightarrow (1+x^2{)}^2\frac{d^{2}y}{d{x}^2}+2x(1+x^2)\frac{dy}{dx}=2$