Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sin ^{-1}\left\{\sqrt{\frac{1-x}{2}}\right\}, 0

Solution:

$y=\sin ^{-1}\left\{\sqrt{\frac{1-x}{2}}\right\}$

let $x=\cos 2 \theta$

Now

$y=\sin ^{-1}\left\{\sqrt{\frac{1-\cos 2 \theta}{2}}\right\}$

$y=\sin ^{-1}\left\{\sqrt{\frac{2 \sin ^{2} \theta}{2}}\right\}$

Using $\cos 2 \theta=1-2 \sin ^{2} \theta$

$y=\sin ^{-1}(\sin \theta)$

Considering the limits,

$0

$0<\cos 2 \theta<1$

$0<2 \theta<\frac{\pi}{2}$

$0<\theta<\frac{\pi}{4}$

Now, $y=\sin ^{-1}(\sin \theta)$

$y=\theta$

$y=\frac{1}{2} \cos ^{-1} x$

Differentiating w.r.t $\mathrm{x}$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now