Correct Answer - Option 2 : 1
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
