Question

If $y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$, then $\frac{dy}{dx}$ is equal to

• A. $-\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{3}{\sqrt{1-x^2}}$
• D. $-\frac{x}{\sqrt{1-x^2}}$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{1-x^2}}$

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$
Let $x=\sin \theta$
$\therefore y={\sin }^{-1}\left(\frac{5\sin \theta +12\cos \theta }{13}\right)={\sin }^{-1}(\frac{5}{13}\sin \theta +\frac{12}{13}\cos \theta )$
$\therefore y={\sin }^{-1}(\sin \theta \cos \alpha +\sin \alpha \cos \theta )$
where $\alpha ={\tan }^{-1}\left(\frac{12}{5}\right)$
$={\sin }^{-1}\sin (\theta +\alpha )=\theta +\alpha$
$\therefore y={\sin }^{-1}x+\alpha$
$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$