A) \[\frac{7}{1+{{x}^{2}}}\]
B) \[\frac{4}{1+{{x}^{2}}}\]
C) \[\frac{1}{x}\]
D) None of these
Correct Answer: A
Solution :
We have, \[y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)+{{\sec }^{-1}}\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\] Let\[x=\tan \theta \] Then,\[y={{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+{{\sec }^{-1}}\left( \frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } \right)\] \[=2\theta +2\theta =4{{\tan }^{-1}}x\] \[\therefore \] \[\frac{dy}{dx}=\frac{4}{1+{{x}^{2}}}\]
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