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

Aug 6, 2015

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

#### Explanation:

You can differentiate this function by using the quotient rule, which tells you that you can differentiate functions that take the form

$f \left(x\right) = g \frac{x}{h \left(x\right)}$, with $g \left(x\right) \ne 0$

by using the formula

color(blue)(d/dx(f(x)) = ([d/dx(g(x))] * h(x) - g(x) * d/dx(h(x)))/[(g(x)]^2)

In your case, you have $g \left(x\right) = {e}^{x}$ and $h \left(x\right) = {e}^{x} + 1$, which means that you can write

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

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

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}} + {e}^{x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}}}{{e}^{x} + 1} ^ 2 = \textcolor{g r e e n}{{e}^{x} / {\left({e}^{x} + 1\right)}^{2}}$

Alternatively, you can play around with the function a bit and write it as

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

In this case, you can the product rule and the chain rule to differentiate the function

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

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

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

Once again, you have

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

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}} + {e}^{x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2 x}}}}}{{e}^{x} + 1} ^ 2 = \textcolor{g r e e n}{{e}^{x} / {\left({e}^{x} + 1\right)}^{2}}$