Question

Find the derivative of ln(1+ x^2) w.r.t. tan inverse x

Answer 1

Let us define $y=\tan^{-1} x, x=\tan y.$

So we may rewrite the derivative as

$\dfrac{d\ln(1+x^2) } {d \tan^{-1}x } = \dfrac{d\ln(1+\tan^2y) } {d y} =\dfrac{1} {1+\tan^2y} \cdot 2\tan y \cdot \dfrac{1} {\cos^2y} = \dfrac{1} {1+x^2} \cdot2x\cdot(1+\tan^2y) = \dfrac{1} {1+x^2} \cdot2x\cdot(1+x^2)=2x$