# Thread: Differentiate the following function

1. ## Differentiate the following function

f(x) = 5x^3 - 6x + 8

f ' (x) =

2. f'(x) = limit as h approaches 0 of $\displaystyle \frac {f(x+h) - f(x)}{h}$

Can you solve it now?

3. ## Is this right?

5(x+h)^3-6(x+h)+8-(5x^3-6x+8)/h = 5x^3+5h^3-6x+6h+8-5x^3+6x-8/h =

5h^3+6h/h = h(5^3+6)/h = 131

4. Looks good to me.

5. I put this in and its coming up wrong, can u help me find where i went wrong

6. 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:

$\displaystyle \lim_{h \to 0} \frac {5(x + h)^3 - 6(x + h) + 8 - (5x^3 - 6x + 8)}{h}$

$\displaystyle \lim_{h \to 0} \frac {5(x+h)(x^2 + 2xh + h^2) - 6x - 6h + 8 - 5x^3 + 6x - 8}{h}$

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:

$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h} = 15x^2 - 6$