Question

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

• A. $\frac{3}{\sqrt{1-x^2}}$
• B. $\frac{12}{\sqrt{1-x^2}}$
• C. $\frac{-1}{\sqrt{1-x^2}}$
• D. $\frac{1}{\sqrt{1-x^2}}$

Answer 1

Answer: (D)

$y={\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$
Put $x=\sin \theta \Rightarrow \theta ={\sin }^{-1}x$
$y={\sin }^{-1}\left(\frac{5\sin \theta +12\cos \theta }{13}\right)$
Now, put $\frac{5}{13}=\cos t$ and $\frac{12}{13}=\sin t$
$\Rightarrow \tan t=\frac{12}{5}\Rightarrow t={\tan }^{-1}\frac{12}{5}$
$\therefore y={\sin }^{-1}\left(\cos t\sin \theta +\sin t\cos \theta \right)$
$={\sin }^{-1}\sin \left(\theta +t\right)$
$y=\theta +t\Rightarrow y={\sin }^{-1}x+{\tan }^{-1}\left(\frac{12}{5}\right)$
Now, $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$