A) \[\frac{-1}{\sqrt{x(1-x)}}\]
B) \[\frac{1}{\sqrt{x(1+x)}}\]
C) \[\frac{1}{\sqrt{x(1-x)}}\]
D) None of these
Correct Answer: A
Solution :
\[y={{\sin }^{-1}}(\sqrt{1-x})+{{\cos }^{-1}}\sqrt{x}\] \[\therefore \] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-{{(\sqrt{1-x})}^{2}}}}.\frac{-1}{2\sqrt{1-x}}\] \[+\left( \frac{-1}{\sqrt{1-{{(\sqrt{x})}^{2}}}} \right).\frac{1}{2\sqrt{x}}\] \[=\frac{1}{2\sqrt{1-1+x}\sqrt{1-x}}-\frac{1}{2\sqrt{1-x}\sqrt{x}}\] \[=\frac{-1}{2\sqrt{x}\sqrt{1-x}}-\frac{1}{2\sqrt{1-x}\sqrt{x}}\] \[=-\frac{2}{2\sqrt{x}\sqrt{1-x}}=-\frac{1}{\sqrt{x(1-x)}}\]
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