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Solution:

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.