# Using the limit definition, how do you find the derivative of f(x) = x/(x+4)?

Jul 3, 2016

$= \frac{4}{{\left(x + 4\right)}^{2}}$

#### Explanation:

$f \left(x\right) = \frac{x}{x + 4}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{1}{h} \left(\frac{x + h}{x + h + 4} - \frac{x}{x + 4}\right)$

$= {\lim}_{h \to 0} \frac{1}{h} \frac{\left(x + h\right) \left(x + 4\right) - x \left(x + h + 4\right)}{\left(x + h + 4\right) \left(x + 4\right)}$

$= {\lim}_{h \to 0} \frac{1}{h} \frac{\left({x}^{2} + 4 x + h x + 4 h - {x}^{2} - x h - 4 x\right)}{\left(x + h + 4\right) \left(x + 4\right)}$

$= {\lim}_{h \to 0} \frac{1}{h} \frac{4 h}{\left(x + h + 4\right) \left(x + 4\right)}$

$= {\lim}_{h \to 0} \frac{4}{\left(x + h + 4\right) \left(x + 4\right)}$

$= \frac{4}{\left(x + 4\right) \left(x + 4\right)}$

$= \frac{4}{{\left(x + 4\right)}^{2}}$