Find dy/dx in the following: y equals sec inverse 1 upon 2 x square minus 1

15. Find dy/dx in the following:

$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right ) , 0 < x < 1/ \sqrt 2$

Answers (1)

Given function is
$y = \sec ^{-1} \left ( \frac{1}{2x ^2 -1} \right )$
Let's take $x = \cos t$
Then,
$\frac{d(x)}{dx} = \frac{(\cos t)}{dt}.\frac{dt}{dx} \ \ \ \ \ (by \ chain \ rule)$
$1 =-\sin t.\frac{dt}{dx}$
$\frac{dt}{dx} = \frac{-1}{\sin t } = \frac{-1}{\sqrt{1-\cos ^2t}} = \frac{-1}{\sqrt{1-x^2}}$
$(\because \sin x = \sqrt{1-\cos^2x} \ and \ x = \cos t )$
And
$\frac{1}{2x^2-1} =\frac{1}{2\cos^2 t - 1} = \frac{1}{\cos2t} = \sec2t$
$(\because \cos 2x = \sqrt{2\cos^2x-1} \ and \ \cos x = \frac{1}{\sec x} )$

Now, our equation reduces to
$y = \sec ^{-1} \left ( \sec 2t \right )$
$y = 2t$
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 $\frac{-2}{\sqrt{1-x^2}}$