Diketahui:
f'(x) = x²-3
g(x) = f(2x+3)
Ditanyakan:
g'(x)
Jawab:
f(x) = ∫ f'(x) dx
= ∫ x²-3 dx
= (1/(2+1))x^(2+1)-(1/(0+1))3x^(0+1)
= (1/3)x³ - (1/1)3x¹
= (1/3)x³ - 3x + C
= (1/3)(2x+3)³ - 3(2x+3) + C
= (1/3)(2x+3)³ - 6x - 9 + C
g'(x) = (1/3)(3)(2)(2x+3)² - 6
= 2(2x+3)²-6
= 2[(2x+3)²-3] (A)
Mapel: Matematika
Kelas: 11
Materi: Turunan
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Diketahui:
f'(x) = x²-3
g(x) = f(2x+3)
Ditanyakan:
g'(x)
Jawab:
f(x) = ∫ f'(x) dx
= ∫ x²-3 dx
= (1/(2+1))x^(2+1)-(1/(0+1))3x^(0+1)
= (1/3)x³ - (1/1)3x¹
= (1/3)x³ - 3x + C
g(x) = f(2x+3)
= (1/3)(2x+3)³ - 3(2x+3) + C
= (1/3)(2x+3)³ - 6x - 9 + C
g'(x) = (1/3)(3)(2)(2x+3)² - 6
= 2(2x+3)²-6
= 2[(2x+3)²-3] (A)
Mapel: Matematika
Kelas: 11
Materi: Turunan