$\begin{align}  & y={{\tan }^{-1}}\left( \frac{1+x}{1-x} \right) \\ & \text{Let }u=\frac{1+x}{1-x},\text{   }\frac{du}{dx}=\frac{(1-x)-(1+x)(-1)}{{{(1-x)}^{2}}} \\ & \frac{du}{dx}=\frac{2}{{{(1-x)}^{2}}} \\ & y={{\tan }^{-1}}u,\text{   }\frac{dy}{du}=\frac{1}{1+{{u}^{2}}} \\ & \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=\frac{1}{1+{{u}^{2}}}\times \frac{2}{{{(1-x)}^{2}}} \\ & \frac{dy}{dx}=\frac{1}{1+{{\left( \frac{1+x}{1-x} \right)}^{2}}}\times \frac{2}{{{(1-x)}^{2}}} \\ & \frac{dy}{dx}=\frac{{{(1-x)}^{2}}}{{{(1-x)}^{2}}+{{(1+x)}^{2}}}\times \frac{2}{{{(1-x)}^{2}}} \\ & \frac{dy}{dx}=\frac{2}{(1-2x+{{x}^{2}}+1+2x+{{x}^{2}})} \\ & \frac{dy}{dx}=\frac{2}{2+2{{x}^{2}}}=\frac{1}{1+{{x}^{2}}} \\\end{align}$