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