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

Mar 28, 2017

Call the function $f \left(x\right)$. By laws of logarithms and the definition $\ln \left(e\right) = 1$, we have

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

We now use the quotient rule to find the derivative. If $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then $f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{h \left(x\right)} ^ 2$.

Here we have $g \left(x\right) = x$ and $h \left(x\right) = {e}^{x} - 1$. Then $g ' \left(x\right) = 1$ and $h ' \left(x\right) = {e}^{x}$.

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

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

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

Hopefully this helps!