# Using the limit definition, how do you differentiate f(x) =1/(x+2)?

Oct 31, 2016

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

#### Explanation:

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

So, with $f \left(x\right) = \frac{1}{x + 2}$ we have:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\left(\frac{1}{\left(x + h\right) + 2}\right) - \left(\frac{1}{x + 2}\right)}{h}\right)$

$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\left(\frac{1}{x + h + 2}\right) - \left(\frac{1}{x + 2}\right)}{h}\right)$

$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\frac{\left(x + 2\right) - \left(x + h + 2\right)}{\left(x + h + 2\right) \left(x + 2\right)}}{h}\right)$

$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{x + 2 - x - h - 2}{h \left(x + h + 2\right) \left(x + 2\right)}\right)$

$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{- h}{h \left(x + h + 2\right) \left(x + 2\right)}\right)$

$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{- 1}{\left(x + h + 2\right) \left(x + 2\right)}\right)$

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

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