Solve this following
Question:

Mark $(\sqrt{ })$ against the correct answer in the following:

If $y=\log \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)$ then $\frac{d y}{d x}=?$

A. $\frac{1}{\sqrt{x}(1-x)}$

B. $\frac{-1}{x(1-\sqrt{x})^{2}}$

C. $\frac{-\sqrt{x}}{2(1-\sqrt{x})}$

D. none of these

Solution:

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