# How do you differentiate f(x)= ( 4- x )/ (e^x + 2)  using the quotient rule?

Mar 30, 2018

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{{e}^{x} \left(x - 5\right) - 2}{{e}^{x} + 2} ^ 2$.

#### Explanation:

The quotient rule states

$\frac{\mathrm{dg} \frac{x}{h \left(x\right)}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \cdot h \left(x\right) - g \left(x\right) \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}}{h \left(x\right)} ^ 2$.

In this case $g \left(x\right) = 4 - x$, $\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = - 1$, $h \left(x\right) = {e}^{x} + 2$, and $\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = {e}^{x}$, so

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left(- 1\right) \left({e}^{x} + 2\right) - \left(4 - x\right) {e}^{x}}{{e}^{x} + 2} ^ 2 = \frac{{e}^{x} \left(x - 5\right) - 2}{{e}^{x} + 2} ^ 2$.