$\text{If }y={{\tan }^{-1}}x\text{ prove that }(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\frac{dy}{dx}=0$
$\begin{align} & \text{Let }y={{\tan }^{-1}}x \\ & \tan y=x \\ & \text{Differentiate with respect to }x \\ & {{\sec }^{2}}y\frac{dy}{dx}=1 \\ & \frac{dy}{dx}=\frac{1}{{{\sec }^{2}}y}=\frac{1}{1+{{\tan }^{2}}y}=\frac{1}{1+{{x}^{2}}} \\ & \frac{d}{dx}\left( {{\tan }^{-1}}x \right)=\frac{1}{1+{{x}^{2}}} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-2x}{{{(1+{{x}^{2}})}^{2}}} \\ & \therefore \text{Mulitply boths sides by }(1+{{x}^{2}}) \\ & (1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-2x}{(1+{{x}^{2}})} \\ & (1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\left( \frac{1}{1+{{x}^{2}}} \right)=0 \\ & (1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\frac{dy}{dx}=0\text{ (proved)} \\\end{align}$