Respuesta :

dalendrk
[tex]f(x)=xe^{2x}\\\\f'(x)=x'e^{2x}+x(e^{2x})'=e^{2x}+xe^{2x}\cdot2=2xe^{2x}+e^{2x}=e^{2x}(2x+1)\\\\f''(x)=(e^{2x})'(2x+1)+e^{2x}(2x+1)'=e^{2x}\cdot2(2x+1)+e^{2x}\cdot2\\\\=e^{2x}(4x+2)+2e^{2x}=(4x+2+2)e^{2x}=(4x+4)e^{2x}\\\\Answer:\\\boxed{f'(x)=(2x+1)e^{2x}\ and\ f''(x)=(4x+4)e^{2x}}[/tex]
meerkat18
The derivative of a function xe^(2x) can be determined by applying differential calculus techniques to the equation. In this case, the applicable law is the law of products

y = xe^(2x)
y' = 2x  e^2x + e^2x
y'' = 4x e^2x + 2e^2x + 2 e^2x
Simplifying
y'' = (4x+4) e^2x 
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