Differentiate the following functions with respect to x :
Question:

Differentiate the following functions with respect to $x$ :

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

Solution:

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

Let $x=\sin \theta$

Now

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

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

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

Now

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

$y=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\pi}{4}\right)+\cos \theta \sin \left(\frac{\pi}{4}\right)\right\}$

Using $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$y=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

Considering the limits,

$-1<x<1$

$-1<\sin \theta<1$

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

$-\frac{\pi}{2}+\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}+\frac{\pi}{4}$

$-\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3 \pi}{4}$

Now,

$y=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

$y=\theta+\frac{\pi}{4}$

$y=\sin ^{-1} x+\frac{\pi}{4}$

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

$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x+\frac{\pi}{4}\right)$

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

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.