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}}}$