Question

If $ y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)+{{\sec }^{-1}}\left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right) $ , then $ \frac{dy}{dx} $ is equal to

• A. $ \frac{7}{1+{{x}^{2}}} $
• B. $ \frac{4}{1+{{x}^{2}}} $
• C. $ \frac{1}{x} $
• D. None of these

Answer 1

Answer: (A)

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}}} $