Jawaban:

[1] f(x) = 8x³

bentuk fungsinya masuk dalam suku biasa, jadi turunan nya menggunakan rumus

• f(x) = xⁿ
• f'(x) = nxⁿ⁻¹

f(x) = 8x³

f'(x) = 8(3)x³⁻¹

f'(x) = 24x²

[2] f(x) = (x² - 4)(x² + 2x - 1)

bentuk fungsi nya merupakan perkalian aljabar, jadi turunannya ditentukan dari

• f(x) = u . v
• f'(x) = u'v + uv'

f(x) = (x² - 4)(x² + 2x - 1)

• u = (x² - 4)
• u' = 2x
• v = (x² + 2x - 1)
• v' = 2x + 2

f'(x) = 2x(x² + 2x - 1) + (x² - 4)(2x + 2)

f'(x) = 2x³ + 4x² - 2x + 2x³ + 2x² - 8x - 8

f'(x) = 2x³ + 2x³ + 4x² + 2x² - 2x - 8x - 8

f'(x) = 4x³ + 6x² - 10x - 8

[3] f(x) = x² / (x - 3)

bentuk fungsi pecahan, turunan nya

• f(x) = u/v
• f'(x) = (u'v - uv') / v²

f(x) = x² / (x - 3)

• u = x²
• u' = 2x
• v = (x - 3)
• v' = 1

f'(x) = [2x(x - 3) - x²(1)] / (x - 3)²

f'(x) = [2x² - 6x - x²] / [x² - 6x + 9]

f'(x) = [x² - 6x] / [x² - 6x + 9]

[4] f(x) = x² + 6x + 8

Ini seperti nomor [1], yang mana fungsinya berbentuk suku biasa. Rumus nya sama

• f(x) = xⁿ
• f'(x) = nxⁿ⁻¹

f(x) = x² + 6x + 8

f'(x) = 2x²⁻¹ + 6x¹⁻¹

f'(x) = 2x¹ + 6x⁰

f'(x) = 2x + 6

[5] f(x) = [x² - 2x + 1] / [2x + 3]

Sama seperti nomor [3] karena bentuk fungsi nya sama² pecahan

• f(x) = u/v
• f'(x) = (u'v - uv') / v²

f(x) = [x² - 2x + 1] / [2x + 3]

• u = [x² - 2x + 1]
• u' = [2x - 2]
• v = [2x + 3]
• v' = 2

f'(x) = [(2x - 2)(2x + 3) - (x² - 2x + 1).2] / [2x + 3]²

f'(x) = [4x² + 6x - 4x - 6 - 2x² + 4x - 2] / [4x² + 12x + 9]

f'(x) = [4x² - 2x² + 6x - 4x + 4x - 6 - 2] / [4x² + 12x + 9]

f'(x) = [2x² + 6x - 8] / [4x² + 12x + 9]

Detail jawaban

♬ Mapel : Matematika

♬ Kelas : XI

♬ Materi : Bab 9 – Turunan fungsi aljabar

♬ Kode mapel : 2

♬ Kode kategorisasi : 11.2.9

♬ Kata kunci : Turunan fungsi, perkalian, biasa, pecahan

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