Jika f(x)=x²(2x+3)³,mata turunan pertama f(x) adalah f'(x), maka f'(x) adalah

f(x) = x^2 (2x + 3)^3 = u . v
u = x^2 => u' = 2x
v = (2x + 3)^3 => v' = 3(2x + 3)^2 . 2 = 6(2x + 3)^2
f'(x) = u' v + v' u
= 2x (2x + 3)^3 + 6(2x + 3)^2 . x^2
= 2x (2x + 3)^3 + 6x^2 (2x + 3)^2

= (2x + 3)^2 (2x(2x + 3) + 6x^2)
= (2x + 3)^2 (4x^2 + 6x + 6x^2)
= (2x + 3)^2 (10x^2 + 6x)
= (10x^2 + 6x)(2x + 3)^2

= 2x(5x + 3)(2x + 3)^2

F(x)= (2x+3).(2x+3)
= 4x^2+6x+6x+9
=(4x^2+13x+9).(2x+3)
=8x^3+12x^2+24x^2+36x+18x+27
= x^2(8x^3+3x^2+36x+45)
=8x^5+3x^4+36x^3+4x^2
F'(x)=40x^4+12x^3+108x^2+8x