Differentiation Question 44
Question: If $ y={{\tan }^{-1}}\sqrt{\frac{x+1}{x-1}} $ . Then $ \frac{dy}{dx} $ is
Options:
A) $ \frac{-1}{2| x |\sqrt{x^{2}-1}} $
B) $ \frac{-1}{2x\sqrt{x^{2}-1}} $
C) $ \frac{1}{2x\sqrt{x^{2}-1}} $
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ x=\sec \theta . $ Then $ y={{\tan }^{-1}}\sqrt{\frac{\sec \theta +1}{\sec \theta -1}} $
$ ={{\tan }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}={{\tan }^{-1}}( \cot \frac{\theta }{2} ) $
$ ={{\tan }^{-1}}{ \tan ( \frac{\pi }{2}-\frac{\theta }{2} ) } $
$ =\frac{\pi }{2}-\frac{1}{2}{{\sec }^{-1}}x $
$ \therefore \frac{dy}{dx}=-\frac{1}{2}.\frac{1}{| x |\sqrt{x^{2}-1}} $
BETA
