Find dy/dx  in the following: 

y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )

Answers (1)

Given function is
y = \sin ^{-1} \left ( \frac{2x}{1+ x^2 } \right )
Lets consider x = \tan t
Then,
\frac{d(x)}{dx} = \frac{d(\tan t)}{dt}.\frac{dt}{dx} \ \ \ \ \ \ \ \ \ (by \ chain \ rule)
1 = \sec^2 t . \frac{dt}{dx}\\ \frac{dt}{dx} = \frac{1}{\sec^2t} = \frac{1}{1+\tan ^2t} = \frac{1}{1+x^2}\ \ \ \ \ \ \ (\because \sec^2t=1+\tan^2t \ and \ x = \tan t)
Now,
\frac{2x}{1+x^2} = \frac{2\tan t }{1+\tan^2t} = \sin 2t \ \ \ \ \ \ (\because \sin 2x = \frac{2\tan x }{1+\tan^2 x})
Our equation reduces to 
y = \sin^{-1}(\sin 2t)
y = 2t
Now, differentiation w.r.t. x is
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{1}{1+x^2} = \frac{2}{1+x^2}
Therefore, the answer is  \frac{2}{1+x^2}

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