# How do you use the limit definition of the derivative to find the derivative of f(x)=-4x^3-3x^2+1?

Oct 28, 2016

$f ' \left(x\right) = - 12 {x}^{2} - 6 x$

#### Explanation:

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

So for $f \left(x\right) = - 4 {x}^{3} - 3 {x}^{2} + 1$ we have:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left\{- 4 {\left(x + h\right)}^{3} - 3 {\left(x + h\right)}^{2} + 1\right\} - \left\{- 4 {x}^{3} - 3 {x}^{2} + 1\right\}}{h}$
$\therefore f ' \left(x\right) = {\lim}_{h \to 0} \frac{\left\{- 4 \left({x}^{3} + 3 h {x}^{2} + 3 {h}^{2} x + {h}^{3}\right) - 3 \left({x}^{2} + 2 h x + {h}^{2}\right) + 1\right\} - \left\{- 4 {x}^{3} - 3 {x}^{2} + 1\right\}}{h}$

$\therefore f ' \left(x\right) = {\lim}_{h \to 0} \frac{- 4 {x}^{3} - 12 h {x}^{2} - 12 {h}^{2} x - 4 {h}^{3} - 3 {x}^{2} - 6 h x - 3 {h}^{2} + 1 + 4 {x}^{3} + 3 {x}^{2} - 1}{h}$

$\therefore f ' \left(x\right) = {\lim}_{h \to 0} \frac{- 12 h {x}^{2} - 12 {h}^{2} x - 4 {h}^{3} - 6 h x - 3 {h}^{2}}{h}$

$\therefore f ' \left(x\right) = {\lim}_{h \to 0} \left(- 12 {x}^{2} - 12 h x - 4 {h}^{2} - 6 x - 3 h\right)$
$\therefore f ' \left(x\right) = - 12 {x}^{2} - 6 x$