JEE Main & Advanced
Mathematics
Applications of Derivatives
Question Bank
Critical Thinking
question_answer
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]
A)1
B)\[\frac{x-1}{x+1}\]
C)Does not exist
D)None of these
Correct Answer:
C
Solution :
Let \[y={{\sec }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\]
\[\because \] \[-1\le \frac{2x}{1+{{x}^{2}}}\le 1\]; So, \[{{\sec }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] is defined only at \[x=-1\] and 1. So, \[y\] is not differentiable.