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

Berikut ini adalah pertanyaan dari peesbedrf pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

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

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Turunan pertamadari fungsif(x) = (3x - 2)⁵ (x² - 3x + 2)³ adalah:

\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}

\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}

\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}

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

Penjelasan

Turunan

Diberikan fungsi:
f(x)=(3x-2)^5(x^2-3x+2)^3

Turunan pertamanya adalah:

\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}
\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}
\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}
 
 
\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}

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. PenjelasanTurunanDiberikan 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]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. PenjelasanTurunanDiberikan 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]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. PenjelasanTurunanDiberikan 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]

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Last Update: Mon, 22 May 23
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