Carilah turunan pertama dari

Contoh Soal 1

Carilah turunan pertama dari :
a. y = 3x⁵ – 12x³ + 5x
b. y = 2x – 5x² + 7x⁵
c. y = [latex]\frac {1} {3}[/latex] x² – [latex]\frac {2} {3}[/latex] x² + 3x

Pembahasan

Jawaban (a) :

y = 3x⁵ – 12x³ + 5x
y’ = 5 . 3x^{5 – 1} – 3 . 12x^{3 – 1} + 1 . 5x^{1 – 1}
y’ = 15x⁴ – 36x² + 5

Jawaban (b) :

y = 2x – 5x² + 7x⁵
y’ = 1 . 2x^{1 – 1} – 2 . 5x^{2 – 1} + 5 . 7 x^{5 – 1}
y’ = 2 – 10x + 35x⁴

Jawaban (c) :

y = [latex]\frac {1} {3}[/latex] x² – [latex]\frac {2} {3}[/latex] x² + 3x
y’ = 2 . [latex]\frac {1} {3}[/latex] x^{2 – 1} – 2 . [latex]\frac {2} {3}[/latex] x^{2 – 1} + 1 . 3x^{1 – 1}
y’ = [latex]\frac {2} {3}[/latex]x – [latex]\frac {4} {3}[/latex] x + 3

Contoh Soal 2

Carilah turunan pertama dari:
a. y = (x + 2) (2x – 7)
b. y = (3x + 4) (5x – 2)
c. y = (5x + 2) (x² – 3)

Pembahasan

Jawaban (a) :

y = (x + 2) (2x – 7)
U = x + 2 maka U’ = 1
V = 2x – 7 maka V’ = 2
y = U . V
y’ = U’ . V + U . V’
y’ = 1 . (2x – 7) + (x + 2) . 2
y’ = (2x – 7) + (2x + 4)
y’ = 2x + 2x – 7 + 4 = 4x – 3

Jawaban (b) :

y = (3x + 4) (5x – 2)
U = 3x + 4 maka U’ = 3
V = 5x – 2 maka V’ = 5
y = U . V
y’ = U’ . V + U . V’
y’ = 3 (5x – 2) + (3x + 4) . 5
y’ = 15x – 6 + 15x + 20
y’ = 30x + 24

Jawaban (c):

y = (5x + 2) (x² – 3)
U = 5x + 2 maka U’ = 5
V = x² – 3 maka V’ = 2x
y = U . V
y’ = U’ . V + U . V’
y’ = 5 (x² – 3) + (5x + 2) . 2x
y’ = 5x² – 15 + 10x² + 4x
y’ = 15x² + 4x – 15

Contoh soal 3

Carilah turunan pertama dari:
a. y = [latex]\frac {x – 5} {4x + 2}[/latex]
b. y = [latex]\frac {2 – 5x} {x + 2}[/latex]
c. y = [latex]\frac {x^{2} + 1} {1 – x}[/latex]

Pembahasan

Jawaban (a) :

y = [latex]\frac {x – 5} {4x + 2}[/latex]
U = x – 5 maka U’ = 1
V = 4x + 2 maka V’ = 4
y = [latex]\frac {U} {V}[/latex]
y’ = [latex]\frac {U’ . V – U . V’} {V^2}[/latex]
y’ = [latex]\frac {1 (4x + 2) – (x – 5) . 4} {(4x + 2)^2}[/latex]
y’ = [latex]\frac {4x + 2 – 4x + 20} {(4x + 2)^2}[/latex]
y’ = [latex]\frac {22} {(4x + 2)^2}[/latex]

Jawaban (b) :

y = [latex]\frac {2 – 5x} {x + 2}[/latex]
U = 2 – 5x maka U’ = -5
V = x + 2 maka V’ = 1
y = [latex]\frac {U} {V}[/latex]
y’ = [latex]\frac {U’ . V – U . V’} {V^2}[/latex]
y’ = [latex]\frac {-5 (x + 2) – (2 – 5x) . 1} {(x + 2)^2}[/latex]
y’ = [latex]\frac {-5x – 10 – 2 + 10x} {(x + 2)^2}[/latex]
y’ = [latex]\frac {5x – 12} {(x + 2)^2}[/latex]

Jawaban (c) :

y = [latex]\frac {x^{2} + 1} {1 – x}[/latex]
U = x² + 1 maka U’ = 2x
V = 1 – x maka V’ = -1
y = [latex]\frac {U} {V}[/latex]
y’ = [latex]\frac {U’ . V – U . V’} {V^2}[/latex]
y’ = [latex]\frac {2x (1 – x) – (x^{2} + 1) . -1} {(1 – x)^2}[/latex]
y’ = [latex]\frac {2x – 2x^{2} + x^{2} + 1} {(1 – x)^2}[/latex]
y’ = [latex]\frac {-x^{2} + 2x + 1} {(1 – x)^2}[/latex]

Contoh soal 4

Carilah turunan pertama dari :
a. y = (2x + 3)³
b. y = (2 – x)⁵
c. y = [latex]\sqrt {x^{2} + 5}[/latex]

Pembahasan

Jawaban (a) :

y = (2x + 3)³
U = 2x + 3 maka U’ = 2
y(U) = U³ maka y'(U) = 3U²
y’ = U’ . y'(U)
y’ = 2 . 3U²
y’ = 6 (2x + 3)²

Jawaban (b) :

y = (2 – x)⁵
U = 2 – x maka U’ = -1
y(U) = U⁵ maka y'(U) = 5U⁴
y’ = U’ . y'(U)
y’ = -1 . 5U⁴
y’ = -5 (2 – x)⁴

Jawaban (c) :

y = [latex]\sqrt {x^{2} + 5}[/latex]
y = (x² + 5)^1/2
U = x² + 5 maka U’ = 2x
y(U) = U^1/2 maka y'(U) = 1/2U^-1\2
y’ = U’ . y'(U)
y’ = 2x . 1/2U^-1/2
y’ = x (x² + 5)^-1/2
y’ = [latex]\frac {x} {\sqrt {x^{2} + 5}}[/latex]