Question

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

Answer 1

$log(1+x^2)$ with respect to $ta{n}^{-1}x$

Let $u=log(1+x^2)$

$v=ta{n}^{-1}x$

$\frac{du}{dx}=\frac{2x}{1+x^2}$

$\frac{dv}{dx}=\frac{1}{1+x^2}$

$\frac{\frac{du}{dx}}{\frac{du}{dx}}=\frac{\frac{2x}{(1+x^2)}}{\frac{1}{1+x^2}}$

$\frac{du}{dv}=\frac{2x}{(1+x^2)}\times (1+x^2)$

$\frac{du}{dv}=2x$