Question

If $y={\sin }^{-1}\left(\frac{4\cos x+3\sin x}{5}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{4}{5}\sin x$
• B. $\frac{3}{5}\cos x$
• C. $1$
• D. $\frac{x^2}{2}$

Answer 1

The correct option is C $1$

$y={\sin }^{-1}(\frac{4}{5}\cos x+\frac{3}{5}\sin x)$
Let $\sin \theta =\frac{4}{5}$
$\therefore \cos \theta =\frac{\sqrt{5^2-4^2}}{5}=\frac{3}{5}$
$\therefore y={\sin }^{-1}(\sin \theta \cos x+\cos \theta \sin x)\Rightarrow y={\sin }^{-1}\sin (\theta +x)$
$\Rightarrow y=\theta +x\Rightarrow \frac{dy}{dx}=1$