A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
D) \[\frac{2}{\sqrt{1+{{x}^{2}}}}\]
Correct Answer: A
Solution :
Given, \[y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right)\] Put, \[x=\sin \theta \] \[\therefore \] \[y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{\sin }^{2}}\theta }} \right)\] \[\Rightarrow \] \[y={{\sec }^{-1}}(\sec \theta )=\theta \] \[\Rightarrow \] \[y={{\sin }^{-1}}x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}\]
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