Differentiation Question 180
Question: If $ y={{\sec }^{-1}}( \frac{\sqrt{x}+1}{\sqrt{x}-1} )+{{\sin }^{-1}}( \frac{\sqrt{x}-1}{\sqrt{x}+1} ) $ , then $ \frac{dy}{dx}= $
[UPSEAT 1999; AMU 2002; Kerala (Engg.) 2005]
Options:
A) 0
B) $ \frac{1}{\sqrt{x}+1} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\sec }^{-1}}( \frac{\sqrt{x}+1}{\sqrt{x}-1} )+{{\sin }^{-1}}( \frac{\sqrt{x}-1}{\sqrt{x}+1} ) $
$ ={{\cos }^{-1}}( \frac{\sqrt{x}-1}{\sqrt{x}+1} )+{{\sin }^{-1}}( \frac{\sqrt{x}-1}{\sqrt{x}+1} )=\frac{\pi }{2} $
Therefore $ \frac{dy}{dx}=0 $ , $ { \because {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} } $ .
BETA
