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