tolong dibantu dengan cara kak

f(x) = (2x^3 + 3x+ 3)^2/3
f'(x) = 2/3 × (6x^2 + 3) × (2x^3 +3x +3)^2/3-3/3
f'(x) = 4x^2 + 2 ×(2x^3 + 3x + 3)^-1/3
f'(x) = 4x^2 + 2 / akar 3 (2x^3 + 3x + 3)
f'(1) = 4(1)^2 + 2 / akar 3 (2(1)^3 + 3(1) + 3)
f'(1) = 4 + 2 / akar 3 (2+3+3)
f'(1) = 6 / akar 3 (8)
f'(1) = 6 / 2
f'(1) = 3 (c)