Jawaban:
f'(x) =
[tex] - \frac{4x}{( {2x}^{2} + 1) \: \sqrt[3]{ {2x}^{2} + 1 } } [/tex]
semoga benar...
[tex]\large\text{$\begin{aligned}f'(x)&=\boxed{\ \bf{-}\frac{4x}{\left(2x^2+1\right)^{{}^{\bf4}\!/_{\bf3}}}\ }\\&\quad\text{atau}\\f'(x)&=\boxed{\ \bf{-}\frac{4x}{\sqrt[\bf3]{\bf\left(2x^2+1\right)^4}}\ }\end{aligned}$}[/tex]
Pembahasan
Turunan
Diberikan fungsi:
[tex]\large\text{$\begin{aligned}f(x)&=\frac{3}{\sqrt[3]{2x^2+1}}\\\end{aligned}$}[/tex]
Turunan pertamanya adalah:
[tex]\begin{aligned}f'(x)&=\tfrac{d}{dx}\left[\frac{3}{\sqrt[3]{2x^2+1}}\right]\\&\quad...\text{keluarkan konstanta}\\f'(x)&=3\cdot\tfrac{d}{dx}\left[\frac{1}{\sqrt[3]{2x^2+1}}\right]\\&\quad...\ \text{ubah bentuk}\\f'(x)&=3\cdot\tfrac{d}{dx}\left[\left(2x^2+1\right)^{-{}^{\bf1}\!/_{\bf3}}\right]\\&\quad...\ \text{aturan rantai}\end{aligned}[/tex]
[tex]\begin{aligned}f'(x)&=3\cdot\left(-\tfrac{1}{3}\right)\left(2x^2+1\right)^{\left(-{}^{\bf1}\!/_{\bf3}\:-\:1\right)}\cdot\tfrac{d}{dx}\left(2x^2+1\right)\\&=(-1)\left(2x^2+1\right)^{-{}^{\bf4}\!/_{\bf3}}\cdot4x\\&=-4x\left(2x^2+1\right)^{-{}^{\bf4}\!/_{\bf3}}\end{aligned}[/tex]
[tex]\begin{aligned}f'(x)&=\boxed{\ \bf{-}\frac{4x}{\left(2x^2+1\right)^{{}^{\bf4}\!/_{\bf3}}}\ }\\&\quad\text{atau}\\f'(x)&=\boxed{\ \bf{-}\frac{4x}{\sqrt[\bf3]{\bf\left(2x^2+1\right)^4}}\ }\end{aligned}[/tex]
[tex]\blacksquare[/tex]
[answer.2.content]
