Question

Tentukan turunan pertama fungsi f(x) = (3x - 2)⁵ (x² - 3x + 2)³

Answer 1

Jawaban:

f'(x) = 2673x¹⁰-29970x⁹+147501x⁸-419832x⁷+765996x⁶-937200x⁵+779760x⁴-436224x³+157248x²-33024x+3072

PENJELASAN :

ada di foto

Semoga membantu

Answer 2

Verified answer

Turunan pertama dari fungsi f(x) = (3x - 2)⁵ (x² - 3x + 2)³ adalah:

[tex]\begin{aligned}&\textsf{Alternatif 1:}\\&\boxed{f'(x)=3\left(3x-2\right)^4\left(x^2-3x+2\right)^2\left(11x^2-28x+16\right)}\end{aligned}[/tex]

[tex]\begin{aligned}&\textsf{Alternatif 2:}\\&\boxed{f'(x)=3\left(3x^2-5x+2\right)^2\left(3x^2-8x+4\right)^2\left(11x^2-28x+16\right)}\end{aligned}[/tex]

[tex]\begin{aligned}&\textsf{Alternatif 3:}\\&\boxed{f'(x)=3\left[\left(3x^2-5x+2\right)\left(3x^2-8x+4\right)\right]^2\left(11x^2-28x+16\right)}\end{aligned}[/tex]

Ketiga alternatif tersebut saling ekuivalen.
Alternatif lain: dengan menjabarkan seperti jawaban pertama.
 

Penjelasan

Turunan

Diberikan fungsi:
[tex]f(x)=(3x-2)^5(x^2-3x+2)^3[/tex]

Turunan pertamanya adalah:

[tex]\begin{aligned}f'(x)&=\left[\left(3x-2\right)^5\right]'\left(x^2-3x+2\right)^3\\&\quad{}+\left(3x-2\right)^5\left[\left(x^2-3x+2\right)^3\right]'\\&\qquad{\sf Ingat:\ }\left(u^n\right)'=nu^{n-1}u'\\&=5\left(3x-2\right)^4\left(3x-2\right)'\left(x^2-3x+2\right)^3\\&\quad{}+\left(3x-2\right)^5\cdot3\left(x^2-3x+2\right)^2\left(x^2-3x+2\right)'\\&=15\left(3x-2\right)^4\left(x^2-3x+2\right)^3\\&\quad{}+3\left(3x-2\right)^5\left(2x-3\right)\left(x^2-3x+2\right)^2\end{aligned}[/tex]
[tex]\begin{aligned}&=3\left(3x-2\right)^4\left(x^2-3x+2\right)^2\left[5\left(x^2-3x+2\right)+\left(3x-2\right)\left(2x-3\right)\right]\\&=3\left(3x-2\right)^4\left(x^2-3x+2\right)^2\left(5x^2-15x+10+6x^2-13x+6\right)\\f'(x)&=\boxed{3\left(3x-2\right)^4\left(x^2-3x+2\right)^2\left(11x^2-28x+16\right)}\\&=3\left(3x-2\right)^4\left(x-1\right)^2\left(x-2\right)^2\left(11x^2-28x+16\right)\end{aligned}[/tex]
[tex]\begin{aligned}&=3\left[\left(3x-2\right)\left(x-1\right)\right]^2\left[\left(3x-2\right)\left(x-2\right)\right]^2\left(11x^2-28x+16\right)\\f'(x)&=\boxed{3\left(3x^2-5x+2\right)^2\left(3x^2-8x+4\right)^2\left(11x^2-28x+16\right)}\\f'(x)&=\boxed{3\left[\left(3x^2-5x+2\right)\left(3x^2-8x+4\right)\right]^2\left(11x^2-28x+16\right)}\end{aligned}[/tex]
 
 
[tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]