Respuesta :
Answer:
[tex]\frac{d^2}{dx^2}\left(\left(2x+1\right)e^{4x}\right)=32e^{4x}x+32e^{4x}[/tex].
Step-by-step explanation:
To find the second derivative of the function [tex]f(x)=\left(2 x + 1\right) e^{4 x}\right)[/tex] you must:
Step 1. Find the first derivative [tex]\frac{d}{dx}\left(\left(1+2x\right)e^{4x}\right)[/tex]
[tex]\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\\\\f=1+2x,\:g=e^{4x}[/tex]
[tex]\frac{d}{dx}\left(\left(1+2x\right)e^{4x}\right)=\frac{d}{dx}\left(1+2x\right)e^{4x}+\frac{d}{dx}\left(e^{4x}\right)\left(1+2x\right)[/tex]
[tex]\frac{d}{dx}\left(1+2x\right)=2\\\\\frac{d}{dx}\left(e^{4x}\right)=e^{4x}\cdot \:4[/tex]
[tex]\frac{d}{dx}\left(\left(1+2x\right)e^{4x}\right)=2e^{4x}+e^{4x}\cdot \:4\left(1+2x\right)\\\\\frac{d}{dx}\left(\left(1+2x\right)e^{4x}\right)=6e^{4x}+8e^{4x}x[/tex]
Step 2. Find the second derivative [tex]\frac{d}{dx} (6e^{4x}+8e^{4x}x)[/tex]
[tex]\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}\left(6e^{4x}+8e^{4x}x\right)=\frac{d}{dx}\left(6e^{4x}\right)+\frac{d}{dx}\left(8e^{4x}x\right)\\\\=24e^{4x}+8\left(4e^{4x}x+e^{4x}\right)\\\\32e^{4x}x+32e^{4x}[/tex]
[tex]\frac{d^2}{dx^2}\left(\left(2x+1\right)e^{4x}\right)=32e^{4x}x+32e^{4x}[/tex].
