jika f(x) = (2x+3) (x²+3x+5)²,maka f'(-2) =

f(x) = (2x + 3)[tex](x^{2} + 3x + 5)^{2}[/tex]
u = 2x + 3
u' = 2
v = [tex](x^{2} + 3x + 5)^{2}[/tex]
v' = 2([tex]x^{2}[/tex] + 3x + 5)(2x + 3)
v' = [tex]4x^{3}[/tex] + [tex]12x^{2}[/tex] + 20x + [tex]6x^{2}[/tex] + 18x + 30
v' = [tex]4x^{3}[/tex] + [tex]18x^{2}[/tex] + 38x + 30
f'(x) = u'v + uv'
f'(x) = 2[tex](x^{2} + 3x + 5)^{2}[/tex] + (2x + 3)([tex]4x^{3}[/tex] + [tex]18x^{2}[/tex] + 38x + 30)
f'(-2) = 2[tex]((-2)^{2} + 3(-2) + 5)^{2}[/tex] + (2(-2) + 3)([tex]4(-2)^{3}[/tex] + [tex]18(-2)^{2}[/tex] + 38(-2) + 30)
f'(-2) = 2[tex](4 - 6 + 5)^{2}[/tex] + (-1)(-32 + 72 - 76 + 30)
f'(-2) = 18 + 6
f'(-2) = 24 ... (E)