# How do you find the instantaneous rate of change of the function f(x)=x/(x+2) when x=2?

Feb 7, 2016

$f ' \left(2\right) = \frac{1}{8}$, so the instantaneous rate of change at $x = 2$ is $\frac{1}{8}$

#### Explanation:

The instantaneous rate of change at $x = 2$ is equal to the value of the derivative at $x = 2$.

To find the derivative, use the quotient rule, which states that

$\frac{d}{\mathrm{dx}} \left[g \frac{x}{h \left(x\right)}\right] = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

Applying this to the function at hand, we see that

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

Note that both of these derivatives are equal to $1$.

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

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

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

The instantaneous rate of change at $x = 2$ is

$f ' \left(2\right) = \frac{2}{2 + 2} ^ 2 = \frac{2}{4} ^ 2 = \frac{2}{16} = \frac{1}{8}$