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:
