Answer
$\frac{1}{tan^{-1}{x}}$ $\frac{1}{{1+x^2}}$
Work Step by Step
Given that $y$ = $\ln tan^{-1}{x}$, on substituting $tan^{-1}{x}$ = $u$ and applying the chain rule derivative, we get:
$\frac{dy}{dx}$ = $\frac{{d\ln{u}}}{du}$ $\times$ $\frac{du}{dx}$
or $\frac{dy}{dx}$ = $\frac{1}{u}$ $\times$ $\frac{du}{dx}$
so $\frac{dy}{dx}$ = $\frac{1}{tan^{-1}{x}}$ $\times$ $\frac{dtan^{-1}{x}}{dx}$
or $\frac{dy}{dx}$ = $\frac{1}{tan^{-1}{x}}$ $\times$ $\frac{1}{{1+x^2}}$
The final answer is: $\frac{1}{tan^{-1}{x}}$ $\frac{1}{{1+x^2}}$
