1st derivative of 3x3+12 is 9x2.
PROOF
Let y=3x3+12. When x=x+h where h is small then y becomes y+k where k is small.
So y+k=3(x+h)3+12=3(x3+3x2h+3xh2+h3)+12.
If h is small, 3xh2+h3 will be even smaller, and will be negligible as h approaches zero.
So we can write y+k=3(x3+3x2h)=3x3+9x2h+12.
If we subtract y=3x3+12 we get: k=9x2h, k/h=9x2. As h→0, k/h→dy/dx (first derivative)=9x2.