# How do you use the limit definition to find the derivative of f(x)=(4-3x)/(2+x)?

Nov 1, 2016

$f \left(x\right) = - \frac{10}{x + 2} ^ 2$

#### 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 $f \left(x\right) = \frac{4 - 3 x}{2 + x}$ we have:
$f \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\left(\frac{4 - 3 \left(x + h\right)}{2 + \left(x + h\right)}\right) - \left(\frac{4 - 3 x}{2 + x}\right)}{h}\right)$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\left(\frac{4 - 3 x - 3 h}{2 + x + h}\right) - \left(\frac{4 - 3 x}{2 + x}\right)}{h}\right)$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{\frac{\left(4 - 3 x - 3 h\right) \left(2 + x\right) - \left(4 - 3 x\right) \left(2 + x + h\right)}{\left(2 + x + h\right) \left(2 + x\right)}}{h}\right)$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left(8 + 4 x - 6 x - 3 {x}^{2} - 6 h - 3 h x\right) - \left(8 + 4 x + 4 h - 6 x - 3 {x}^{2} - 3 x h\right)}{h \left(2 + x + h\right) \left(2 + x\right)}$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \frac{8 + 4 x - 6 x - 3 {x}^{2} - 6 h - 3 h x - 8 - 4 x - 4 h + 6 x + 3 {x}^{2} + 3 x h}{h \left(2 + x + h\right) \left(2 + x\right)}$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{- 10 h}{h \left(2 + x + h\right) \left(2 + x\right)}\right)$

$\therefore f \left(x\right) = {\lim}_{h \rightarrow 0} \left(\frac{- 10}{\left(2 + x + h\right) \left(2 + x\right)}\right)$

$\therefore f \left(x\right) = \frac{- 10}{\left(2 + x\right) \left(2 + x\right)}$

$H e n c e , f \left(x\right) = - \frac{10}{x + 2} ^ 2$