Differentiation Question 345
Question: If $ y={{\sin }^{-1}}\frac{2x}{1+x^{2}}+{{\sec }^{-1}}\frac{1+x^{2}}{1-x^{2}} $ , then $ \frac{dy}{dx} $ =
[RPET 1996]
Options:
A) $ \frac{4}{1-x^{2}} $
B) $ \frac{1}{1+x^{2}} $
C) $ \frac{4}{1-x^{2}} $
D) $ \frac{-4}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ x=\tan \theta $
$ y={{\sin }^{-1}}( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } )+{{\sec }^{-1}}( \frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } ) $
$ =2\theta +2\theta =4{{\tan }^{-1}}x $ . $ y={{\sin }^{-1}}( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } )+{{\sec }^{-1}}( \frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } ) $
Therefore $ x=0 $ .
BETA
