Turunan fungsi f(x)=(x²+2x+3)(x²+3x-4)​

Jawaban:

turunan fungsi

f(x)=(x²+2x+3) (x²+3x-4)

f'(x)=(2x+2) (2x+3)

Penyelesaian:

f (x) = (x² + 2x + 3) (x² + 3x – 4)

misal

u = x^2 + 2x + 3

u' = 2x + 2

v = x^2 + 3x – 4

v' = 2x + 3

f'(x) = u' v + v' u

f'(x) = (2x + 2)(x^2 + 3x – 4) + (2x + 3)(x^2 + 2x + 3)

f'(x) = 2x^3 + 8x^2 – 2x – 8 + 2x^3 + 7x^2 + 12x + 9

f'(x) = 4x^3 + 15x^2 + 10x + 1

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Detil Jawaban

Kelas: 11

Mapel: Matematika

Bab: Turunan Fungsi Aljabar

Kode: 11.2.9

Kata Kunci: Turunan pertama