# How do you find the derivative of  [e^x / (1 - e^x)]?

May 25, 2017

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

#### Explanation:

Ahh... We gotta use the good ol' quotient rule. Using Lagrange's notation (one that I do not use often) as it would get messy in In Leibniz' notation, we see:

$\left(\frac{f}{g}\right) ' = \frac{f ' g - f g '}{{g}^{2}}$

So (back to Leibniz)

Let $f \left(x\right) = {e}^{x}$ and $g \left(x\right) = 1 - {e}^{x}$

$\frac{d}{\mathrm{dx}} \left[{e}^{x} / \left(1 - {e}^{x}\right)\right] = \frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{\frac{\mathrm{df}}{\mathrm{dx}} g \left(x\right) - \frac{\mathrm{dg}}{\mathrm{dx}} f \left(x\right)}{{g}^{2} \left(x\right)}$

So, we've gotta figure out what the derivative of $f \left(x\right)$ and $g \left(x\right)$ is

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = {e}^{x}$

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[1\right] - \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] = - {e}^{x}$

Then sub back in

$\frac{\frac{\mathrm{df}}{\mathrm{dx}} g \left(x\right) - \frac{\mathrm{dg}}{\mathrm{dx}} f \left(x\right)}{{g}^{2} \left(x\right)} = \frac{{e}^{x} \left(1 - {e}^{x}\right) + {e}^{x} \left({e}^{x}\right)}{1 - {e}^{x}} ^ 2$

$= \frac{{e}^{x} - {e}^{2 x} + {e}^{2 x}}{1 - {e}^{x}} ^ 2$

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

And that's the most we can simplify for now. Therefore:

$\frac{d}{\mathrm{dx}} \left[{e}^{x} / \left(1 - {e}^{x}\right)\right] = {e}^{x} / {\left(1 - {e}^{x}\right)}^{2}$