Question

If y = tan⁻¹ x, show that $\left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}=0$.

Answer 1

Here,
$$\begin{gathered} y={\tan }^{-1}x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{1+x^2} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=\frac{-2x}{{\left(1+x^2\right)}^2} \\ \Rightarrow \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=\frac{-2x}{1+x^2}\times \frac{1}{1+x^2} \\ \Rightarrow \frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=\frac{-2x{\displaystyle \frac{dy}{dx}}}{1+x^2} \\ \Rightarrow \left(1+x^2\right)\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}=-2x\frac{dy}{dx} \\ \Rightarrow \left(1+x^2\right)\frac{{\mathit{d}}^{2}y}{\mathit{d}{x}^2}+2x\frac{dy}{dx}=0 \end{gathered}$$

Hence proved.