Differentiation Question 88
Question: If $ y={{\sin }^{-1}}\frac{2x}{1+x^{2}}, $ where $ 0<x<1 $ and $ 0<y<\frac{\pi }{2}, $ then $ \frac{dy}{dx}= $
Options:
A) $ \frac{2}{1+x^{2}} $
B) $ \frac{2x}{1+x^{2}} $
C) $ \frac{-2}{1+x^{2}} $
D) $ \frac{-x}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\sin }^{-1}}\frac{2x}{1+x^{2}} $
Put $ x=\tan \theta $
$ x $
$ y=2\theta =2{{\tan }^{-1}}x $
$ (\because \theta ={{\tan }^{-1}}x) $
Differentiating with respect to x, we get $ \frac{dy}{dx}=\frac{2}{1+x^{2}} $
$ ( \text{Since }0<x<1\text{ and }0<y<\frac{\pi }{2} ) $ .
BETA
