Differentiation Question 62
Question: If $ y={{\sin }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) $ , then $ \frac{dy}{dx} $ equals
[EAMCET 1991; RPET 1996]
Options:
A) $ \frac{2}{1-x^{2}} $
B) $ \frac{1}{1+x^{2}} $
C) $ \pm \frac{2}{1+x^{2}} $
D) $ -\frac{2}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{\sin }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) $
Put $ x=\tan \theta $
Therefore $ \theta ={{\tan }^{-1}}x $
$ \therefore y={{\sin }^{-1}}\cos 2\theta =\frac{\pi }{2}\pm 2\theta $
$ y=\frac{\pi }{2}\pm 2{{\tan }^{-1}}x $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{\pm 2}{1+x^{2}} $ .
BETA
