Differentiate `log(1+x^2)` with respect to `tan^(-1)x` .

Explanation

`log (1+x^{2})` with respect to `tan ^{-1} x`

Let `u=log (1+x^{2})`

`{V}=tan ^{-1} {x}`

`frac{d u}{d x}=frac{2 x}{1+x^{2}}`

`frac{d v}{d x}=frac{1}{1+x^{2}}`

`frac{frac{d x}{d x}}{frac{d u}{d x}}=frac{frac{2 x}{(1+x^{2})}}{frac{1}{1+x^{2}}}`

`frac{d u}{d v}=frac{2 x}{(1+x^{2})} times(1+x^{2})`

`frac{d u}{d v}=2 x`