If $y=\tan ^{-1} x$, show that $\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=0$
Formula: -
(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$
(ii) $\frac{\mathrm{d}\left(\tan ^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}$
(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$
(iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$
Given: -
$Y=\tan ^{-1} x$
Differentiating w.r.t x
$\frac{d y}{d x}=\frac{d\left(\tan ^{-1} x\right)}{d x}$
Using formula(ii)
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+\mathrm{x}^{2}}$
$\Rightarrow\left(1+\mathrm{x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}=1$
Again Differentiating w.r.t $\mathrm{x}$
Using formula(iii)
$\left(1+x^{2}\right) \frac{d y}{d x}+2 x \frac{d y}{d x}=0$
Hence proved.