Differentiation Question 41
Question: If $ y={{\tan }^{-1}}( \frac{x}{1+\sqrt{1-x^{2}}} ) $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{1}{2\sqrt{1-x^{2}}} $
B) $ 1-\sqrt{1-x^{2}} $
C) $ \frac{1}{2} $
D) $ \frac{1}{\sqrt{1-x^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\tan }^{-1}}( \frac{x}{1+\sqrt{1-x^{2}}} ) $
Put $ x=\sin \theta $
$ \therefore y={{\tan }^{-1}}( \frac{\sin \theta }{1+\sqrt{1-{{\sin }^{2}}\theta }} )={{\tan }^{-1}}( \frac{\sin \theta }{1+\cos \theta } ) $
$ ={{\tan }^{-1}}\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2} $
So, $ y=\frac{{{\sin }^{-1}}x}{2}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}} $ .
BETA
