Thread: [SOLVED] The definition of derivative

1. [SOLVED] The definition of derivative

Anyone know how to find the derivative using
5x^3+2

lim
x->c

$\displaystyle \frac {f(x)-f(c)}{x-c}$

2. We are given: $\displaystyle f(x) = 5x^3 + 2$

$\displaystyle \lim_{x \to c} \frac{{\color{red}f(x)} - {\color{blue}f(c)}}{x - c}$

$\displaystyle = \lim_{x \to c} \frac{{\color{red}(5x^3 + 2)} - {\color{blue}(5c^3 + 2)}}{x - c}$

$\displaystyle = \lim_{x \to c} \frac{5(x^3 - c^3)}{x - c}$

Recall the difference of cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Apply that to the numerator and something should cancel out Then you can plug in x = c without worrying about dividing by 0!

3. Does it look something like this?

$\displaystyle \frac {5(x-c)(x^2+xc+c^2)}{x-c}$

$\displaystyle 5(x^2+xc+c^2)$

$\displaystyle 5(c^2+(c)(c)+c^2)$

$\displaystyle 5(2c^2+c^2)$

4. Don't forget the limit in front of each step!!! :

$\displaystyle = {\color{red}\lim_{x \to c}} \ \frac {5(x-c)(x^2+xc+c^2)}{x-c}$

$\displaystyle = {\color{red}\lim_{x \to c}} \ 5(x^2+xc+c^2)$

$\displaystyle = 5(c^2 + c^2 +c^2)$ (You don't need the limit sign here since you already took the limit by replacing x with c)

And you can simplify the inside by collecting like terms:

$\displaystyle = 5(3c^2) = 15c^2$

So given any point c, the derivative at that point will be 15 times its square

5. i guess i didn't simplify all the way

i see now

thanks for the help