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}}

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