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

I think that this was not the function that you intended to write (probably you omitted parentheses, which are essential when one writes in-line fractions). In fact, you can rewrite that function as follows:
$c \left(x\right) = \ln \left(1 + {e}^{x} / 1 - {e}^{x}\right) = \ln \left(1 + {e}^{x} - {e}^{x}\right) = \ln \left(1\right) = 0$
The derivative of a constant function is $0$, so
$c ' \left(x\right) = 0$

I'll try to infer the function that you wanted to write:
$f \left(x\right) = \ln \left(\frac{1 + {e}^{x}}{1 - {e}^{x}}\right)$
The code to render this properly is f(x)=ln((1+e^x)/(1-e^x)) (notice the usage of parentheses).

This is a function composition. Given two functions $y = g \left(x\right)$ and $z = h \left(y\right)$ you can compute the derivative of the composition $f \left(x\right) = h \left(g \left(x\right)\right)$ as follows:
$f ' \left(x\right) = h ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
In this particular case $g \left(x\right) = \frac{1 + {e}^{x}}{1 - {e}^{x}}$ and $h \left(y\right) = \ln \left(y\right)$.

The derivative of $g$ we is the derivative of a quotient
$g ' \left(x\right) = \frac{{e}^{x} \left(1 - {e}^{x}\right) - \left(1 + {e}^{x}\right) \left(- {e}^{x}\right)}{1 - {e}^{x}} ^ 2 = \frac{2 {e}^{x}}{1 - {e}^{x}} ^ 2$
and the derivative of $h$ is the derivative of the logarithmic function
$h ' \left(y\right) = \frac{1}{y}$
So
$f ' \left(x\right) = h ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \frac{1}{\frac{1 + {e}^{x}}{1 - {e}^{x}}} \cdot \frac{2 {e}^{x}}{1 - {e}^{x}} ^ 2 = \frac{2 {e}^{x}}{\left(1 + {e}^{x}\right) \left(1 - {e}^{x}\right)} = \frac{2 {e}^{x}}{1 - {e}^{2 x}}$

Apr 26, 2015

I assume you want to differentiate:

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

You could use $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$ with $u = \frac{1 + {e}^{x}}{1 - {e}^{x}}$ and use the quotient rule to find $\frac{\mathrm{du}}{\mathrm{dx}}$, but it's a little quicker to use $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$ to rewrite the function as:

$y = \ln \left(1 + {e}^{x}\right) - \ln \left(1 - {e}^{x}\right)$. before differentiating:

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

Or, simplifying:

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