In order to find the equation of a tanget line, you need the slope.

the problem I was given is

f(x) = $\displaystyle ((x^3)+4x-1)(x-2)$ and the point they want me to find it at is (1,-4)

my work so far has been

f'(x) =$\displaystyle (x-2)((3x^2)+4) + ((x^3) + 4x-1)$

f'(x) =$\displaystyle ((3x^3)+4x-(6x^2)-8 + (x^3) +4x -1$

and finally

f'(x) =$\displaystyle (4x^3)+8x-(6x^2)-9$

Would I be correct in plugging two points into my final f'(x) in order to get the slope?

I thought I was done but, something is off in my logic and I'm not sure what.