Anyone know how to find the derivative using
5x^3+2
lim
x->c
$\displaystyle
\frac {f(x)-f(c)}{x-c}
$
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!
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