A) \[\frac{2}{1-{{x}^{2}}}\]
B) \[\frac{1}{1+{{x}^{2}}}\]
C) \[\pm \frac{2}{1+{{x}^{2}}}\]
D) \[-\frac{2}{1+{{x}^{2}}}\]
Correct Answer: C
Solution :
\[y={{\sin }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] Put \[x=\tan \theta \]Þ \[\theta ={{\tan }^{-1}}x\] \[\therefore y={{\sin }^{-1}}\cos 2\theta =\frac{\pi }{2}\pm 2\theta \] \[y=\frac{\pi }{2}\pm 2{{\tan }^{-1}}x\]\[\Rightarrow \]\[\frac{dy}{dx}=\frac{\pm 2}{1+{{x}^{2}}}\].
You need to login to perform this action.
You will be redirected in
3 sec
