Question
Differentiate $\sin ^{-1}\left(\frac{2 \cos x+3 \sin x}{\sqrt{13}}\right)$ w.r. to $x$

Answer

$ \text { Let } y =\sin ^{-1}\left(\frac{2 \cos x+3 \sin x}{\sqrt{13}}\right)$
$=\sin ^{-1}\left(\frac{2 \cos x}{\sqrt{13}}+\frac{3 \sin x}{\sqrt{13}}\right) $
Put $\frac{2}{\sqrt{13}}=\sin t$ and $\frac{3}{\sqrt{13}}=\cos t$
Also, $\sin ^2 t +\cos ^2 t =\frac{4}{13}+\frac{9}{13}=1$
and $\tan t =\frac{2}{3}$
$\therefore t =\tan ^{-1}\left(\frac{2}{3}\right)$
$\therefore y=\sin ^{-1}(\sin t \cdot \cos x+\cos t \cdot \sin x)$
$=\sin ^{-1}[\sin (t+x)]$
$= t + x$
$=\tan ^{-1}\left(\frac{2}{3}\right)+x$
Differentiating w. r. t. x, we get
$ \frac{ d y}{ d x}=\frac{ d }{ d x}\left[\tan ^{-1}\left(\frac{2}{3}\right)+x\right]$
$=0+1$
$=1 $

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