Question

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

Answer 1

Given $y=({\tan }^{-1}x{)}^2$----(1)
Differentiating w.r.t to x, we get
$\frac{dy}{dx}=2ta{n}^{-1}x.\frac{1}{1+x^2}$----(2)
or $(1+x^2)y^{\prime }=2{\tan }^{-1}x$
Again differentiating with w.r.t to x, we get
$(1+x^2)\frac{d{y}^{\prime }}{dx}+y^{\prime }\frac{d(1+x^2)}{dx}=2.\frac{1}{1+x^2}$
$\Rightarrow (1+x^2).y^{\prime \prime }+y^{\prime }.2x=\frac{2}{1+x^2}$
$\Rightarrow (1+x^2){.}^2{y}^{\prime \prime }+y^{\prime }.2x(1+x^2)=2$
Therefore, $\Rightarrow (1+x^2){.}^2\frac{d^{2}y}{dx}+2x(1+x^2)\frac{dy}{dx}=2$