Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

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

Solution:

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

Let $x=\cos 2 \theta$

Now

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

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

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

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,

$0

$0<\cos 2 \theta<1$

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

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

Now,

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

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

$y=\frac{1}{2} \cos ^{-1} x+\frac{\pi}{4}$

Differentiating w.r.t $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2} \cos ^{-1} \mathrm{x}+\frac{\pi}{4}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \times \frac{-1}{\sqrt{1-\mathrm{x}^{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}}$

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