A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
C) \[-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
D) \[-\frac{1}{\sqrt{{{x}^{2}}-1}}\]
Correct Answer: C
Solution :
\[y={{\sin }^{-1}}(\sqrt{1-{{x}^{2}}})\] Let \[\sqrt{1-{{x}^{2}}}=\sin \theta \Rightarrow 1-{{x}^{2}}={{\sin }^{2}}\theta \] Þ \[{{x}^{2}}=1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] \[\therefore x=\cos \theta \] or \[\theta ={{\cos }^{-1}}x\] \[\Rightarrow y={{\cos }^{-1}}x\] Differentiating w.r.t. x of y, we get \[\frac{dy}{dx}=-\frac{1}{\sqrt{1-{{x}^{2}}}}\].
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