The derivative of \(\rm sin ^{-1}\dfrac{2x}{1+x^2}\) w.r.t \(\rm tan ^{-1}\dfrac{2x}{1-x^2}\) is equal to:

Concept:

\(\dfrac{\partial u}{\partial v} = \dfrac{\partial u}{\partial θ} \times \dfrac{\partial θ}{\partial v}\)

Given:

u = \(\rm sin ^{-1}\dfrac{2x}{1+x^2}\)  And v = \(\rm tan ^{-1}\dfrac{2x}{1-x^2}\)

Calculation:

Let,

x = tan θ 

Then, 

u = \(\rm sin ^{-1}\dfrac{2 \tanθ}{1+\tan^2θ}\) = \(\rm \sin ^{-1}{(\sin2θ)}\)

u = 2θ 

\(\dfrac{\partial u}{\partial θ} = \) 2

And,

v = \(\rm tan ^{-1}\dfrac{2x}{1-x^2}\) = \(\rm \tan ^{-1}\dfrac{2\tanθ}{1-\tan^2θ}\) = \(\rm tan ^{-1}{(\tan2θ)}\)

v = 2θ 

\(\dfrac{\partial v}{\partial θ} = \) 2

After that,

\(\dfrac{\partial u}{\partial v} = \dfrac{\partial u}{\partial θ} \times \dfrac{\partial θ}{\partial v}\)

\(\dfrac{\partial u}{\partial v} = 2 \times \dfrac{1}{2}\)

= 1