Finding the derivative? Hello everyone how would I find derivative $f'(-2)$ if  
$$f(x)=(x^3-5)(x^2-3)(x+1)$$
What I did is I made $(x^2-3)(x+1)$ the other part in the product rule.
This is what I did.
$(x^3-5)(2x^2+2x-3)+(3x^2)(x^3+x^2-3x-3)$
Then I plug I in $x=-2$ and got $-25$ but I am not sure if I did it correctly.

Your strategy works fine here, since $(x^2 - 3)(x+1)$ is fairly easy to expand:
$$f(x) = (x^3 - 5)[(x^2-3)(x+1)]=(x^3 - 5)(x^3 + x^2 - 3x -3)$$
Using the product rule:
$$f'(x) = (x^3-5)(\color{blue}{\bf{3}}x^2+2x-3)+(3x^2)(x^3+x^2-3x-3)$$
(i.e., your only error was in miscalculating the derivative of $x^3$..
Now, evaluate $f'(-2)$. And if you'd like to check your work, feel free to hover the calculation below.

$$f'(-2) = -77$$

Note: The product rule for two factors can be expanded to a product rule for three factors, which is handy, to use with more complicated polynomial factors, and/or other functions of $x$: If you have a function that is a product of three functions,  e.g.
$$f(x)=g(x)\cdot h(x)\cdot j(x)$$
Then can apply use the following (Product Rule, three factors):
$$f'(x) = \left(g'(x)\cdot h(x)\cdot j(x)\right) + \left(g(x) \cdot h'(x)\cdot j(x)\right) + \left(g(x)\cdot h(x)\cdot j'(x)\right)$$
