A) \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C) \[-\frac{1}{\sqrt{1+{{x}^{2}}}}\]
D) None of these
Correct Answer: D
Solution :
Let\[u={{\sin }^{-1}}x\]and \[v={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\] \[\therefore \] \[\frac{du}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}\] and \[\frac{dv}{dx}=-\frac{1}{\sqrt{1-{{(\sqrt{1-{{x}^{2}}})}^{2}}}}\] \[\times \frac{1}{2}{{(1-{{x}^{2}})}^{\frac{1}{2}-1}}.(-2x)\] \[=\frac{x{{(1-{{x}^{2}})}^{-\frac{1}{2}}}}{\sqrt{1-1+{{x}^{2}}}}\] \[=\frac{x}{\sqrt{1-{{x}^{2}}}.x}\] \[=\frac{1}{\sqrt{1-{{x}^{2}}}}\] Now, \[\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{\frac{1}{\sqrt{1-{{x}^{2}}}}}{\frac{1}{\sqrt{1-{{x}^{2}}}}}=1\]
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