Differentiate `sin^(-1)((2x)/(1+2x))` wrt to `tan^(-1)x , -1

Explanation

Let `x-tan theta`

`y=sin^(-1)((2x)/(1+2x))=sin^(-1)((2tan theta)/(1+2tan theta))`

`=sin^(-1)((2tan theta)/(sec^2 theta))`

`sin^(-1)((2frac{sin theta}{cos theta})/(frac{1}{cos^2 theta}))`

`=sin^(-1) 2sin theta cos theta`

`=sin^(-1) sin 2 theta=> 2 theta`

So, `y=2 theta=> 2tan^(-1)x`

`y=2 frac{d}{dx}(tan^(-1)x)=>frac{2}{1+x^2}`