Question

Differentiating Exponential functions In Exercise,find the derivative of the function. See Example 2 and 3.

Answer 1

Answer:

$\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=-\frac{2e^x}{\left(e^x-1\right)^2}$.

Step-by-step explanation:

To find the derivative of the function $f(x)=\frac{e^x+1}{e^x-1}$ you must:

Step 1. Apply the Quotient Rule $\left(\frac{f}{g}\right)'=\frac{f\:'\cdot g-g'\cdot f}{g^2}$

$\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=\frac{\frac{d}{dx}\left(e^x+1\right)\left(e^x-1\right)-\frac{d}{dx}\left(e^x-1\right)\left(e^x+1\right)}{\left(e^x-1\right)^2}$

$\frac{d}{dx}\left(e^x+1\right)=e^x\\\\\frac{d}{dx}\left(e^x-1\right)=e^x$

$=\frac{e^x\left(e^x-1\right)-e^x\left(e^x+1\right)}{\left(e^x-1\right)^2}$

Step 2. Simplify

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

Therefore,

$\frac{d}{dx}\left(\frac{e^x+1}{e^x-1}\right)=-\frac{2e^x}{\left(e^x-1\right)^2}$