Q. math~—bagi yg mau sj.f(x) = 10x^20 + 20x^10f”‘(x) = … (turunan ke 3)​

Jawab:
f'''(x) = 68.400x¹⁷ + 14.400x⁷

Penjelasan dengan langkah-langkah:
Diketahui :
∵ f(x) = 10x²⁰ + 20x¹⁰
∴ f'''(x) = (10x²⁰)''' + (20x¹⁰)'''
∵
Turunan ketiga
(axⁿ)''' = a(n)(n-1)(n-2)xⁿ⁻³
∴
f'''(x) = 10(20)(20-1)(20-2)x²⁰⁻³ + 20(10)(10-1)(10-2)x¹⁰⁻³
f'''(x) = 200(19)(18)x¹⁷ + 200(9)(8)x⁷
f'''(x) = 200(342)x¹⁷ + 200(72)x⁷
f'''(x) = 68.400x¹⁷ + 14.400x⁷

[[ KLF ]]

Jawaban:

–

Penjelasan dengan langkah-langkah:

f(x) = f"'(x)

Turunan ke–1.

• = (20 . 10x^(20 – 1)) + (10 . 20x^(10 – 1))
• = (200x^(20 – 1)) + (200x^(10 – 1))
• = 200x¹⁹ + 200x⁹

Turunan ke–2

• = (19 . 200x^(19 – 1)) + (9 . 200x^(9 – 1))
• = (3.800x^(19 – 1)) + (1.800x^(9 – 1))
• = 3.800x¹⁸ + 1.800x⁸

Turunan ke–3

• = (18 . 3.800x^(18 – 1)) + (8 . 1.800^(9 – 1))
• = (68.400x^(18 – 1)) + (14.400x^(9 – 1))
• = 68.400x¹⁷ + 14.400x⁷