Diffrentiate the following w. r. t. x. ^ -1 (4 x+5 x 41 ) - Vidyadip

Question

Diffrentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{4 \sin x+5 \cos x}{\sqrt{41}}\right)$

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Answer

Let $y=\sin ^{-1}\left(\frac{4 \sin x+5 \cos x}{\sqrt{41}}\right)$$=\sin ^{-1}\left[(\sin x)\left(\frac{4}{\sqrt{41}}\right)+(\cos x)\left(\frac{5}{\sqrt{41}}\right)\right]$
Since, $\left(\frac{4}{\sqrt{41}}\right)^2+\left(\frac{5}{\sqrt{41}}\right)^2=\frac{16}{41}+\frac{25}{41}=1$,
we can write, $\frac{4}{\sqrt{41}}=\cos \alpha$ and $\frac{5}{\sqrt{41}}=\sin \alpha$.
$ \therefore y & =\sin ^{-1}(\sin x \cos \alpha+\cos x \sin \alpha)$
$=\sin ^{-1}[\sin (x+\alpha)]$
$=x+\alpha, \quad \text { where } \alpha \text { is a constant }$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}(x+\alpha)$
$=\frac{d}{d x}(x)+\frac{d}{d x}(\alpha)$
$=1+0=1$

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