Question

If $y=\log \left(\frac{(1+\sqrt{x})}{(1-\sqrt{x})}\right)$, then $\frac{dy}{dx}$

• A. $\frac{\sqrt{x}}{(1-x)}$
• B. $\frac{1}{\sqrt{x}(1-x)}$
• C. $\frac{\sqrt{x}}{(1+x)}$
• D. $\frac{1}{\sqrt{x}(1+x)}$

Answer 1

The correct option is B

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

Find the $\frac{dy}{dx}$:

Given that, $y=\log \left(\frac{(1+\sqrt{x})}{(1-\sqrt{x})}\right)$

Using logarithmic property, $\log \frac{a}{b}=\log a-\log b$ we have:

$y=\log (1+\sqrt{x})-\log (1-\sqrt{x})$

Now differentiate $y$ with respect to $x$we have:

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{1+\sqrt{x}}·\frac{1}{2\sqrt{x}}-\frac{1}{1-\sqrt{x}}·\left(-\frac{1}{2\sqrt{x}}\right)\\ & =& \frac{1}{1+\sqrt{x}}·\frac{1}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}·\left(\frac{1}{2\sqrt{x}}\right)\\ & =& \frac{1}{2\sqrt{x}}\left(\frac{1}{1+\sqrt{x}}+\frac{1}{1-\sqrt{x}}\right)\\ & =& \frac{1}{2\sqrt{x}}\left(\frac{1-\sqrt{x}+1+\sqrt{x}}{1^2-{\left(\sqrt{x}\right)}^2}\right)\\ & =& \frac{1}{2\sqrt{x}}\left(\frac{2}{1^2-{\left(\sqrt{x}\right)}^2}\right)\\ & =& \frac{1}{\left(1-x\right)\sqrt{x}}\end{array}$

Hence, option B is correct.