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

$$\begin{aligned} \mathrm{y}=\left(\tan ^{-1} x\right)^{2} & \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\frac{2 \tan ^{-1} x}{1+x^{2}} \\ & \Rightarrow\left(1+x^{2}\right) \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=2 \tan ^{-1} x, \text { differentiating with respect to ' } x \\ & \Rightarrow\left(1+x^{2}\right)^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2 x\left(x^{2}+1\right) \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=2 \end{aligned} $$