# How do you find f'(x) and f''(x) given f(x)=x/(3+e^x)?

Dec 13, 2017

#### Explanation:

$f \left(x\right) = \frac{x}{3 + {e}^{x}}$

Using $\frac{d}{\mathrm{dx}} \left(\frac{f}{g}\right) = \frac{f ' g - f g '}{g} ^ 2$ with $f = x$ and $g = 3 + {e}^{x}$

$f ' \left(x\right) = \frac{\left(1\right) \left(3 + {e}^{x}\right) - \left(x\right) \left({e}^{x}\right)}{3 + {e}^{x}} ^ 2$

$= \frac{3 + {e}^{x} - x {e}^{x}}{3 + {e}^{x}} ^ 2$

Using $\frac{d}{\mathrm{dx}} \left(\frac{f}{g}\right) = \frac{f ' g - f g '}{g} ^ 2$ again, but this time with $f = 3 + {e}^{x} - x {e}^{x}$ and $g = {\left(3 + {e}^{x}\right)}^{2}$

$f ' ' \left(x\right) = \frac{\left({e}^{x} - \left(1 {e}^{x} + x {e}^{x}\right)\right) {\left(3 + {e}^{x}\right)}^{2} - \left(3 + {e}^{x} - x {e}^{x}\right) \left(2 \left(3 + {e}^{x}\right) {e}^{x}\right)}{3 + {e}^{x}} ^ 4$

$= \frac{\left(- x {e}^{x}\right) \textcolor{red}{{\left(3 + {e}^{x}\right)}^{2}} - 2 {e}^{x} \left(3 + {e}^{x} - x {e}^{x}\right) \left(\textcolor{red}{\left(3 + {e}^{x}\right)}\right)}{3 + {e}^{x}} ^ 4$

$= \frac{\textcolor{red}{\left(3 + {e}^{x}\right)} \left[\left(- x {e}^{x}\right) \textcolor{red}{\left(3 + {e}^{x}\right)} - 2 {e}^{x} \left(3 + {e}^{x} - x {e}^{x}\right]\right)}{3 + {e}^{x}} ^ 4$

$= \frac{\left(- x {e}^{x}\right) \left(3 + {e}^{x}\right) - 2 {e}^{x} \left(3 + {e}^{x} - x {e}^{x}\right)}{3 + {e}^{x}} ^ 3$

Rewrite the numerator to taste.