Tentukan Turunan Dari
f(x) = ( 2x + 1 )⁵ ( x² + 1 )³
Jawaban:
[tex]\large\text{$\begin{aligned}f'(x) = \bf10\left(2x+1\right)^4\left(x^2+1\right)^3+6x\left(2x+1\right)^5\left(x^2+1\right)^2\end{aligned}$}[/tex]
Pembahasan
Turunan dan Aturan Rantai
[tex]\large\text{$\begin{aligned}f(x)&=\left( 2x+1 \right)^5 \left( x^2+1 \right )^3\\&=\left(g(x)\right)^5\left(h(x)\right)^3\\&\textsf{dengan $ g(x)=2x+1 \, ,\ h(x)=x^2+1 $.}\\\\f'(x) & = \left [\left( g(x) \right )^5 \right ]' \left( h(x) \right )^3 + \left(g(x) \right)^5 \left[\left(h(x)\right)^3\right]'\\&=5\left(g(x)\right)^4g'(x)\left(h(x)\right)^3 + \left(g(x)\right)^5\cdot3\left(h(x)\right)^2 h'(x)\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&\quad....\textsf{ substitusi $g(x)$ dan $h(x)$}\\f'(x)&=5\left(2x+1\right)^4(2x+1)'\left(x^2+1\right)^3+\left(2x+1\right)^5\cdot3\left(x^2+1\right)^2(x^2+1)'\\&=5\left(2x+1\right)^4\cdot2\cdot\left(x^2+1\right)^3+\left(2x+1\right)^5\cdot3\left(x^2+1\right)^2\cdot2x\\&=\bf10\left(2x+1\right)^4\left(x^2+1\right)^3+6x\left(2x+1\right)^5\left(x^2+1\right)^2\end{aligned}$}[/tex]
[answer.2.content]