# How do you use the limit definition to find the derivative of y=-4x-5?

Oct 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4$

#### 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 $y = - 4 x - 5$ we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \left(\frac{\left(- 4 \left(x + h\right) - 5\right) - \left(- 4 x - 5\right)}{h}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \left(\frac{\left(- 4 x - 4 h - 5\right) - \left(- 4 x - 5\right)}{h}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \left(\frac{- 4 x - 4 h - 5 + 4 x + 5}{h}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \left(- \frac{4 h}{h}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\lim}_{h \rightarrow 0} \left(- 4\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - 4$