By definition of the derivative,
[tex]y'=\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^x}h[/tex]
We can employ the standard manipulation for proving the product rule:
[tex]\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}+xe^{x+h}-xe^x}h[/tex]
[tex]\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}}h+xe^x\lim_{h\to0}\frac{e^h-1}h[/tex]
[tex]\displaystyle\lim_{h\to0}e^{x+h}\lim_{h\to0}\frac{(x+h)-x}h+xe^x\lim_{h\to0}\frac{e^h-1}h[/tex]
[tex]e^{x+0}\displaystyle\lim_{h\to0}\frac hh+xe^x\lim_{h\to0}\frac{e^h-1}h[/tex]
[tex]e^x+xe^x\displaystyle\lim_{h\to0}\frac{e^h-1}h[/tex]
The remaining limit is pretty well-known and has a value of 1. We can derive it from the definition of [tex]e[/tex],
[tex]e=\displaystyle\lim_{n\to0}(1+n)^{1/n}[/tex]
In the limit above, we substitute [tex]\eta=e^h-1[/tex], so that [tex]h=\ln(\eta+1)[/tex]. As [tex]h\to0[/tex], we have [tex]\eta\to e^0-1=0[/tex]:
[tex]\displaystyle\lim_{h\to0}\frac{e^h-1}h=\lim_{\eta\to0}\frac\eta{\ln(\eta+1)}[/tex]
[tex]\displaystyle=\lim_{\eta\to0}\frac1{\frac1\eta\ln(\eta+1)}[/tex]
[tex]\displaystyle=\frac1{\lim\limits_{\eta\to0}\ln(\eta+1)^{1/\eta}}[/tex]
[tex]\displaystyle=\frac1{\ln\left(\lim\limits_{\eta\to0}(\eta+1)^{1/\eta}\right)}[/tex]
[tex]=\dfrac1{\ln e}=\dfrac11=1[/tex]
After all this, we've shown that
[tex](xe^x)'=e^x+xe^x=e^x(x+1)[/tex]
