Respuesta :
f(x) = e^2x (x^3 + 1)
It can also be written as it follows:
[tex]f(x)=e^{2x}\cdot(x^3+1)[/tex]first we need to determine f'(x).
we can see this function as a product of two funcions:
[tex]\begin{gathered} f(x)=g(x)\cdot h(x) \\ g(x)=e^{2x};h(x)=x^3+1 \end{gathered}[/tex]so, we can apply the Product rule
[tex]\frac{d}{dx}\lbrack g(x)\cdot h(x)\rbrack=g(x)\cdot h^{\prime}(x)+g^{\prime}(x)\cdot h(x)[/tex]Let's determine the derivate of those two function
[tex]\begin{gathered} g(x)=e^{2x} \\ g^{\prime}(x)=2\cdot e^{2x} \end{gathered}[/tex][tex]\begin{gathered} h(x)=x^3+1 \\ h^{\prime}(x)=3x^2 \end{gathered}[/tex]Now, we can use the functions g(x) and h(x) and its derivates to determine f'(x) using the Product rule
[tex]\begin{gathered} f^{\prime}(x)=g(x)\cdot h^{\prime}(x)+g^{\prime}(x)\cdot h(x) \\ f^{\prime}(x)=e^{2x}\cdot3x^2+2e^{2x}\cdot(x^3+1) \end{gathered}[/tex]we can simplify this expression as it follows:
[tex]\begin{gathered} f^{\prime}(x)=3e^{2x}x^2+2e^{2x}x^3+2e^{2x} \\ f^{\prime}(x)=2e^{\mleft\{2x\mright\}}x^3+3e^{\mleft\{2x\mright\}}x^2+2e^{\mleft\{2x\mright\}} \end{gathered}[/tex]Finally, we need to replace x = 2 to find f'(2)
[tex]f^{\prime}(2)=2e^{\{2\cdot2\}}\cdot2^3+3e^{\{2\cdot2\}}\cdot2^2+2e^{\{2\cdot2\}}[/tex]simplify:
[tex]\begin{gathered} f^{\prime}(2)=2e^4\cdot8+3e^4\cdot4+2e^4 \\ f^{\prime}(2)=16e^4+12e^4+2e^4^{} \\ f^{\prime}(2)=(16+12+2)\cdot e^4 \\ f^{\prime}(2)=30e^4 \end{gathered}[/tex]