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)

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

Answer 2

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