A) \[\frac{1}{2\sqrt{x}}\]
B) \[\frac{\sqrt{x}}{\sqrt{1-x}}\]
C) 1
D) None of these
Correct Answer: D
Solution :
Let\[y={{\sin }^{-1}}\frac{1-x}{1+x}\]Þ\[\frac{dy}{dx}=\frac{-1}{\sqrt{x}(1+x)}\] ....(i) and\[z=\sqrt{x}\]Þ\[\frac{dz}{dx}=\frac{1}{2\sqrt{x}}\] .....(ii) Therefore by (i) and (ii) \[\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{-2}{1+x}\].
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