Differentiate $\log(1+x^{2})$ with respect $\tan^{-1}x$.

Let $y=\log(1+x^{2})$

$y=\tan^{-1}(x)$

To find $\dfrac{dy}{dz}$

first we find $\dfrac{dy}{dx}=\dfrac{d}{dx}(\log(1+x^{2})=\dfrac{2x}{1+x^{2}}$

Next we find

$\dfrac{dz}{dx}=\dfrac{d}{dx}(\tan^{-1}(x))=\dfrac{1}{1+x^{2}}$

Now $\dfrac{dy}{dz}=\dfrac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}}=\dfrac{\dfrac{2x}{1+x^{2}}}{\dfrac{1}{1+x^{2}}}=2x$