Question:
Differentiate $\log \left(1+x^{2}\right)$ with respect to $\tan ^{-1} x$
Solution:
Let $u=\log \left(1+x^{2}\right)$ and $v=\tan ^{-1} x$
$\Rightarrow \frac{d u}{d x}=\frac{1}{\left(1+x^{2}\right)} \frac{d}{d x}\left(1+x^{2}\right)=\frac{2 x}{\left(1+x^{2}\right)}$ and $\frac{d v}{d x}=\frac{1}{1+x^{2}}$
$\therefore \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{2 x}{1+x^{2}} \times \frac{1+x^{2}}{1}=2 x$