Diffrentiate the following w. r. t. x. ^ -1 ( x+ x 2 ) - Vidyadip

Question

Diffrentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{\cos \sqrt{x}+\sin \sqrt{x}}{\sqrt{2}}\right)$

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Answer

$y=\sin ^{-1}\left(\frac{\cos \sqrt{x}+\sin \sqrt{x}}{\sqrt{2}}\right)$
$=\sin ^{-1}\left(\frac{1}{\sqrt{2}} \cos \sqrt{x}+\frac{1}{\sqrt{2}} \sin \sqrt{x}\right)$
Put,
$\frac{1}{\sqrt{2}}=\sin x$
$\frac{1}{\sqrt{2}}=\cos \alpha$
Also,
$\sin ^2 \alpha+\cos ^2 \alpha=\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2=1$
And,
$\tan \alpha=1$
$\therefore \alpha=\tan ^{-1} 1$
$y=\sin ^{-1}(\sin \alpha \cdot \cos \sqrt{x}+\cos \alpha \cdot \sin (x)$
$=\sin ^{-1}(\sin (\alpha+\sqrt{x}))$
$y=\alpha+\sqrt{x}$
$y=\tan ^{-1}(1)+\sqrt{x}$
Differentiating w.r.t. $x$, we get
$\frac{ dy }{ dx }=\frac{ d }{ dx }\left(\tan ^{-1}+\sqrt{x}\right)$
$=0+\frac{1}{2 \sqrt{x}}$
$=\frac{1}{2 \sqrt{x}} .$

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