Applications Of Derivatives Question 204
Question: If $ y={{\sec }^{-1}}\frac{2x}{1+x^{2}}+{{\sin }^{-1}}\frac{x-1}{x+1} $ ,then $ \frac{dy}{dx} $ is equal to
[Pb. CET 2000]
Options:
A) 1
B) $ \frac{x-1}{x+1} $
C) Does not exist
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y={{\sec }^{-1}}\frac{2x}{1+x^{2}}+{{\sin }^{-1}}( \frac{x-1}{x+1} ) $
$ \because $ $ -1\le \frac{2x}{1+x^{2}}\le 1 $ ; So, $ {{\sec }^{-1}}( \frac{2x}{1+x^{2}} ) $ is defined only at $ x=-1 $ and 1. So, $ y $ is not differentiable.
$ \therefore $ $ \frac{dy}{dx} $ does not exist.
BETA
