Answer:
[tex]\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=-\frac{2e^x}{\left(e^x-1\right)^2}[/tex].
Step-by-step explanation:
To find the derivative of the function [tex]f(x)=\frac{e^x+1}{e^x-1}[/tex] you must:
Step 1. Apply the Quotient Rule [tex]\left(\frac{f}{g}\right)'=\frac{f\:'\cdot g-g'\cdot f}{g^2}[/tex]
[tex]\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=\frac{\frac{d}{dx}\left(e^x+1\right)\left(e^x-1\right)-\frac{d}{dx}\left(e^x-1\right)\left(e^x+1\right)}{\left(e^x-1\right)^2}[/tex]
[tex]\frac{d}{dx}\left(e^x+1\right)=e^x\\\\\frac{d}{dx}\left(e^x-1\right)=e^x[/tex]
[tex]=\frac{e^x\left(e^x-1\right)-e^x\left(e^x+1\right)}{\left(e^x-1\right)^2}[/tex]
Step 2. Simplify
[tex]\frac{{e^{2x}-e^x-e^{2x}-e^x}}{\left(e^x-1\right)^2} \\\\\frac{-2e^x}{\left(e^x-1\right)^2}[/tex]
Therefore,
[tex]\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=-\frac{2e^x}{\left(e^x-1\right)^2}[/tex]
