Question

If $y=\sqrt{\frac{1+e^x}{1-e^x}},\mathrm{show}\mathrm{that}\frac{dy}{dx}=\frac{e^x}{\left(1-e^x\right)\sqrt{1-e^{2x}}}$

Answer 1

$\mathrm{We}\mathrm{have},y=\sqrt{\frac{1+e^x}{1-e^x}}$
Differentiate it with respect to x,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left(\sqrt{\frac{1+e^x}{1-e^x}}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{2\sqrt{\left({\displaystyle \frac{1+e^x}{1-e^x}}\right)}}\times \frac{d}{dx}\left(\frac{1+e^x}{1-e^x}\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{2}\times \sqrt{\frac{1-e^x}{1+e^x}}\left[\frac{\left\{\left(1-e^x\right){\displaystyle \frac{d}{dx}}\left(1+e^x\right)-\left(1+e^x\right){\displaystyle \frac{d}{dx}}\left(1-e^x\right)\right\}}{{\left(1-e^x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{2}\sqrt{\frac{1-e^x}{1+e^x}}\left[\frac{\left(1-e^x\right)e^x+\left(1+e^x\right)e^x}{{\left(1-e^x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{2}\sqrt{\frac{1-e^x}{1+e^x}}\times \left[\frac{2e^x}{{\left(1-e^x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{e^x}{\sqrt{\left(1+e^x\right)\sqrt{\left(1-e^x\right)}}}\frac{1}{1-e^x} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{e^x}{\left(1-e^x\right)\sqrt{1-e^{2x}}} \end{gathered}$$