Diketahui f(x)=x³. nilai f'(3) adalah​

Nilai f'(3) adalah 27

༺ Penyelesaian ༻

f(x) = x³ ==> f'(x) = 3 . 1 . x^(3-1)

= 3x²

f'(3) = 3(3)²

= 3 . 9

= 27

༺ Pembahasan ༻

Sifat-sifat turunan fungsi:

1. f(x) = k ==> f'(x) = 0

k = konstanta

Contoh soal:

f(x) = 6 ==> f'(x) = 0

2. f(x) = x ==> f'(x) = 1

Contoh soal:

f(x) = 2x ==> f'(x) = 2 . 1 = 2

3. f(x) = a . xⁿ ==> f'(x) = n . a . x^(n – 1)

Contoh soal:

f(x) = 4x² ==> f'(x) = 2 . 4 . x^(2-1)

= 8x

4. h(x) ± g(x) ==> h'(x) ± g'(x)

Contoh soal:

h(x) = 3x

g(x) = 5x

h(x) + g(x) = 3x + 5x

==> h'(x) + g'(x) = 3 . 1 + 5 . 1

= 3 + 5 = 8

5. f(x) = u(x) . v(x)

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

Contoh soal:

f(x) = 2x . 4x ==> f'(x) = 2x . 4x + 2x . 4x

= 8x² + 8x²

= 16x²

6. f(x) = u(x)/v(x)

==> f'(x) = [u'(x) . v(x) + u(x) . v'(x)]/[v²(x)]

Contoh soal:

f(x) = 3x – 2/x

f'(x) = (3 (x – 4) + (x ) -3)/(x )²

= 3x – 12 – 3x/x²

= 12/x²

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__ Detail Jawaban:

• Kelas : XI
• Mapel : Matematika
• Materi : Turunan Fungsi Aljabar
• Kode soal : 2

Jawab:

27

Penjelasan dengan langkah-langkah:

f'(x)=3x²

f'(3)=3.3²=3.9=27