Question

If $y=\sqrt{\frac{1+e^x}{1-e^x}}$ , then $\frac{dy}{dx}$ is equal to?

• A. $\frac{e^x}{(1-e^x)\sqrt{(1-e^{2x})}}$
• B. $\frac{e^x}{(1-e^x)\sqrt{(1-e^x)}}$
• C. $\frac{e^x}{(1-e^x)\sqrt{(1+e^{2x})}}$
• D. $\frac{e^x}{(1-e^x)\sqrt{(1+e^x)}}$

Answer 1

The correct option is A

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

Explanation for the correct option:

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

Given,

$y=\sqrt{\frac{1+e^x}{1-e^x}}$

Now, differentiate the given function with respect to $x$,

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{2\sqrt{{\displaystyle \frac{1+e^x}{1-e^x}}}}\times \frac{\left(1-e^x\right)e^x-\left(1+e^x\right)\left(-e^x\right)}{{\left(1-e^x\right)}^2}\left[\because \frac{d\left(e^x\right)}{dx}=e^x,\frac{d\left(y^m\right)}{dx}=m{y}^{m-1}\times \frac{dy}{dx}\right]\\ & =& \frac{1}{2\sqrt{{\displaystyle \frac{1+e^x}{1-e^x}}}}\times \frac{2e^x}{{\left(1-e^x\right)}^2}\\ & =& \frac{e^x}{\left(1-e^x\right)\sqrt{1-e^{2x}}}\end{array}$

Hence, the correct option is A.