Question

f(x)=x^3-3x^2+4, x0=-2

Answer 1

f' ( x) = 3x^2 -6x
f'( -2) = 12 + 12 = 24

Answer 2

f'(x) = (x^3-3x^2+4)' = (x^3)' - (3x^2)' + (4)' = 3*x^(3-1) - 3*2*x^(2-1) + 0 = 3x^2-6x
f'(x0) = f'(-2) = 3*(-2)^2 - 6*(-2) = 3*4 + 12 = 24