y'=6x^2+6x-4; when x=2, y'=24+12-4=32.
If you need to know how to differentiate ax^n, it's nax^(n-1).
The reason is if h is very small, the differential of ax^n is given by a(x+h)^n-ax^n=a(x^n+nhx^(n-1)+... -ax^n where ... involves terms with higher powers of h that can be ignored because they are very small. So the result is nhax^(n-1).
We write y=ax^n and y+k=a(x+h)^n, so the small change in y occurs as a result of a small change in x. k=nhax^(n-1).
k/h=nax^(n-1) and k/h becomes dy/dx=y'=nax^(n-1). We apply this differentiation to each term in turn. A constant doesn't change so its differential is zero.