Let y = tan^–1\(\left[\left(\frac{\sqrt{1+\mathrm x^2}}{\mathrm x}\right)^{-1}\right]\)

let x = tan z

⇒ \(\frac{d\mathrm x}{dz}=\sec^2z\)   (By differentiating w.r.t z)

⇒ \(\frac{dz}{d\mathrm x}\) \(=\frac{1}{\sec^2z}\) 

\(=\frac{1}{1+\tan^2z}=\frac{1}{1+\mathrm x^2}\) .................(i)

Then y = tan^–1\(\left[\left(\frac{\sqrt{1+\tan^2z}}{\tan z}\right)^{-1}\right]\)

\(=\tan^{-1}\left[\left(\frac{\sec z}{\tan z}\right)^{-1}\right]\)

\(=\tan^{-1}\left[\frac{\tan z}{\sec z}\right]\) \(=\tan^{-1}\left(\frac{\sin z}{\cos z}\times \cos z\right)\)

\(=\tan^{-1}(\sin z)\) \(=\tan^{-1}\left(\tan \frac{z}{\sqrt{1-z^2}}\right)\)

⇒ y = \(\frac{z}{\sqrt{1-z^2}}\) .........(ii)

Differentiate y w.r.t z, we get

\(\frac{dy}{dz}=\frac{\sqrt{1-z^2}\times 1-z\times \frac{-2z}{2\sqrt{1-z^2}}}{(1-z^2)}\)

\(=\frac{1-z^2+z^2}{(1-z^2)^{\frac{3}{2}}}\)

⇒ \(\frac{dy}{dz}=\frac{1}{(1-z^2)^{\frac{3}{2}}}\) ...........(iii)

Now, \(\frac{dy}{d\mathrm x}\) \(=\frac{dy}{dz}.\frac{dz}{d\mathrm x}\)

\(=\frac{1}{(1-z^2)^{\frac{3}{2}}}\times \frac{1}{1+\mathrm x^2}\)  (From equations (i) and (iii))

\(=\frac{1}{(1+\mathrm x^2)(1-(\tan^{-1}\mathrm x)^2)^{\frac{3}{2}}}\) (\(\because\) x = tan z ⇒ z = tan^-1x)