Question

Differentiate log (1 + x²) with respect to tan⁻¹ x

Answer 1

$$\begin{gathered} \mathrm{Let}u=\log \left(1+x^2\right)\mathrm{and}v={\tan }^{-1}x \\ \Rightarrow \frac{du}{dx}=\frac{1}{\left(1+x^2\right)}\frac{d}{dx}\left(1+x^2\right)=\frac{2x}{\left(1+x^2\right)}\mathrm{and}\frac{dv}{dx}=\frac{1}{1+x^2} \\ \therefore \frac{du}{dv}=\frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{2x}{1+x^2}\times \frac{1+x^2}{1}=2x \end{gathered}$$