A) \[-\frac{2}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]
C) \[cos\,2x\]
D) none of these
Correct Answer: B
Solution :
\[\frac{d}{dx}({{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}})\] Put, \[x=\sin \theta \] \[\therefore \] \[\frac{d}{dx}({{\sin }^{-1}}2\sin \theta \sqrt{1-{{\sin }^{2}}\theta })\] \[=\frac{d}{dx}(si{{n}^{-1}}2sin\theta cos\theta )\] \[=\frac{d}{dx}(si{{n}^{-1}}sin2\theta )=\frac{d}{d\theta }=\frac{d}{dx}(2\theta )\] \[=\frac{d}{dx}(2{{\sin }^{-1}}x)=\frac{2}{\sqrt{1-{{x}^{2}}}}\]
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