14. Find dy/dx  in the following: 

y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} ) , -\frac{1}{\sqrt2} < x \frac{1}{\sqrt 2 }

Answers (1)
G Gautam harsolia

Given function is
y = \sin ^ { -1 } ( 2x \sqrt {1- x^2} )
Lets take x = \sin t
Then,
\frac{d(x)}{dx} = \frac{(\sin t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)
1 =\cos t.\frac{dt}{dx}
\frac{dt}{dx} = \frac{1}{\cos t } = \frac{1}{\sqrt{1-\sin ^2t}} = \frac{1}{\sqrt{1-x^2}}
                                                                                       (\because \cos x = \sqrt{1-\sin^2x} \ and \ x = \sin t )
And
2x\sqrt{1-x^2} = 2\sin t \sqrt{1-\sin^2t} = 2\sin t \sqrt{\cos^2 t} = 2\sin t\cos t =\sin 2t
                                                                                                                                              (\because \cos x = \sqrt{1-\sin^2x} \ and \ 2\sin x\cos x = \sin2x )
Now, our equation reduces to
y = \sin ^ { -1 } ( \sin 2t )
y = 2t
Now, differentiation w.r.t. x 
\frac{d(y)}{dx} = \frac{d(2t)}{dt}.\frac{dt}{dx}
\frac{dy}{dx} = 2.\frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}
Therefore, the answer is  \frac{2}{\sqrt{1-x^2}}

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