f(x) = 5x^3 - 6x + 8
f ' (x) =
Ok let me walk you through it.
$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$
So plug in $\displaystyle 5x^3 - 6x + 8$ to get:
$\displaystyle \lim_{h \to 0} \frac {5(x + h)^3 - 6(x + h) + 8 - (5x^3 - 6x + 8)}{h}$
Now use your factoring skills:
$\displaystyle \lim_{h \to 0} \frac {5(x+h)(x^2 + 2xh + h^2) - 6x - 6h + 8 - 5x^3 + 6x - 8}{h}$
Becomes:
$\displaystyle \lim_{h \to 0} \frac {5x^3 + 10x^2h + 5xh^2 + 5x^2h + 10xh^2 + 5h^3 - 6x - 6h + 8 - 5x^3 + 6x - 8}{h}$
Combine like terms:
$\displaystyle \lim_{h \to 0} \frac {15x^2h + 15xh^2 + 5h^3 - 6h}{h}$
Factor out an $\displaystyle h$:
$\displaystyle \lim_{h \to 0}15x^2 + 15xh + 5h^2 - 6$
As $\displaystyle h$ approaches 0, so do $\displaystyle 15xh$ and $\displaystyle 5h^2$, so we have that:
$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h} = 15x^2 - 6$