A) \[\frac{1}{2\sqrt{x}}\]
B) \[\frac{\sqrt{x}}{\sqrt{1-x}}\]
C) \[1\]
D) none of these
Correct Answer: D
Solution :
Let\[u={{\sin }^{-1}}\frac{1-x}{1+x}\]and\[v=\sqrt{x}\] On differentiating w.r.t.\[x,\] respectively \[\therefore \] \[\frac{du}{dx}=\frac{1}{\sqrt{1-{{\left( \frac{1-x}{1+x} \right)}^{2}}}}\] \[\times \frac{(1+x)(-1)-(1-x)(1)}{{{(1+x)}^{2}}}\] \[=\frac{1}{\sqrt{4x}}\times \frac{(-2)}{1+x}\] and \[\frac{dv}{dx}=\frac{1}{2\sqrt{x}}\] \[\therefore \] \[\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{-2}{\sqrt{4x}(1+x)}\times 2\sqrt{x}\] \[=\frac{-2}{1+x}\]
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