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

Nov 6, 2016

$\frac{d}{\mathrm{dx}} \left[\ln \left(\frac{{e}^{x}}{{e}^{x} + 1}\right)\right] = \frac{1}{{e}^{x} + 1}$

#### Explanation:

Use the derivative of natural logs rule:
$\frac{d}{\mathrm{dx}} \left[\ln u\right] = \frac{u '}{u}$

We want to simplify the following:
$\frac{d}{\mathrm{dx}} \left[\ln \left(\frac{{e}^{x}}{{e}^{x} + 1}\right)\right]$

Dividing by the inner term is the same as multiplying by the reciprocal:
$= \left(\frac{d}{\mathrm{dx}} \left[\frac{{e}^{x}}{{e}^{x} + 1}\right]\right) \cdot \left(\frac{{e}^{x} + 1}{{e}^{x}}\right)$

Use quotient rule to differentiate first term:
$= \left(\frac{\left({e}^{x} + 1\right) \left({e}^{x}\right) - \left({e}^{x}\right) \left({e}^{x}\right)}{{\left({e}^{x} + 1\right)}^{2}}\right) \cdot \left(\frac{{e}^{x} + 1}{{e}^{x}}\right)$

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