# What is the derivative of ln((e^x)/(1+e^x))?

Sep 27, 2015

$\frac{1}{1 + {e}^{x}}$

#### Explanation:

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

$= \frac{1}{e} ^ x \cdot {e}^{x} / \left(1 + {e}^{x}\right) = \frac{1}{1 + {e}^{x}}$

Sep 28, 2015

$\frac{1}{1 + {e}^{x}}$

#### Explanation:

You can bypass using the quoitient rule for derivatives by using the quotient rule for logarithms

$\ln \left(\frac{a}{b}\right) = \ln a - \ln b$

In your case, you will have

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

Now all you have to do is use the chain rule twice, once for $\ln u$, with $u = {e}^{x}$, and once for $\ln v$, with $v = 1 + {e}^{x}$.

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

You have

$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{d}{\mathrm{du}} \ln u \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \cdot \frac{d}{\mathrm{dx}} {e}^{x} = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{x}}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{x}}}} = 1$

and

$\frac{d}{\mathrm{dx}} \left(\ln v\right) = \frac{d}{\mathrm{dv}} \ln v \cdot \frac{d}{\mathrm{dx}} \left(v\right)$

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

The target derivative will thus be

d/dxln(e^x/(1+e^x)) = 1 - e^x/(1+e^x) = (1 + color(red)(cancel(color(black)(e^x))) - color(red)(cancel(color(black)(e^x))))/(1+e^x) = color(green)(1/(1+e^x)