The derivative of $\cot ^{-1} x$ w.r.t $\log \left(1+x^{2}\right)$ is

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Aicharya · Accepted Answer

Let $u=\cot ^{-1} x$ and $v=\log \left(1+x^{2}ight)$ $\begin{array}{l}\frac{d u}{d x}=\frac{-1}{1+x^{2}} 	ext { and } \frac{d v}{d x}=\frac{2 x}{1+x^{2}} \ 	herefore \frac{d u}{d v}=\frac{\left(\frac{d u}{d x}ight)}{\left(\frac{d v}{d x}ight)}=\frac{\left(\frac{-1}{1+x^{2}}ight)}{\left(\frac{2 x}{1+x^{2}}ight)}=\frac{-1}{2 x}\end{array}$

The derivative of $\cot ^{-1} x$ w.r.t $\log \left(1+x^{2}\right)$ is

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