Find dy/dx in the following: y= sin ^-1 2 x 1

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)

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 $\frac{2}{\sqrt{1-x^2}}$