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