Question

Use first principle to find y'all for y=xe^x

Answer 1

By definition of the derivative,

$y'=\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^x}h$

We can employ the standard manipulation for proving the product rule:

$\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}+xe^{x+h}-xe^x}h$

$\displaystyle\lim_{h\to0}\frac{(x+h)e^{x+h}-xe^{x+h}}h+xe^x\lim_{h\to0}\frac{e^h-1}h$

$\displaystyle\lim_{h\to0}e^{x+h}\lim_{h\to0}\frac{(x+h)-x}h+xe^x\lim_{h\to0}\frac{e^h-1}h$

$e^{x+0}\displaystyle\lim_{h\to0}\frac hh+xe^x\lim_{h\to0}\frac{e^h-1}h$

$e^x+xe^x\displaystyle\lim_{h\to0}\frac{e^h-1}h$

The remaining limit is pretty well-known and has a value of 1. We can derive it from the definition of $e$,

$e=\displaystyle\lim_{n\to0}(1+n)^{1/n}$

In the limit above, we substitute $\eta=e^h-1$, so that $h=\ln(\eta+1)$. As $h\to0$, we have $\eta\to e^0-1=0$:

$\displaystyle\lim_{h\to0}\frac{e^h-1}h=\lim_{\eta\to0}\frac\eta{\ln(\eta+1)}$

$\displaystyle=\lim_{\eta\to0}\frac1{\frac1\eta\ln(\eta+1)}$

$\displaystyle=\frac1{\lim\limits_{\eta\to0}\ln(\eta+1)^{1/\eta}}$

$\displaystyle=\frac1{\ln\left(\lim\limits_{\eta\to0}(\eta+1)^{1/\eta}\right)}$

$=\dfrac1{\ln e}=\dfrac11=1$

After all this, we've shown that

$(xe^x)'=e^x+xe^x=e^x(x+1)$