how to find derivative for: y = x^e * e^x
e^(i*pi) gave you $\displaystyle e^x(ex^{e-1} + x^e)$. Now factor out an additional $\displaystyle x^{e-1}$: $\displaystyle x^e= x^{e-1}(x)$ so $\displaystyle e^xx^{e-1}(e+ x)$.
how do you factor out x^(e-1) when there is only one x^(e-1), the other is x^e ======> $\displaystyle e^x(ex^{e-1} + x^e)
$
wouldn't it have to be $\displaystyle e(e^xx^{e-1}) + x(e^xx^{e-1})
$ in order to write it as $\displaystyle e^xx^{e-1}(e+x)
$
so how do you make $\displaystyle e^x(ex^{e-1} + x^e)$ (which is equal to $\displaystyle e(e^xx^{e-1}) + e^xx^e
$) into $\displaystyle e(e^xx^{e-1}) + x(e^xx^{e-1})
$ in order to write it as the answer given $\displaystyle e^xx^{e-1}(e+x)
$
$\displaystyle e^x(ex^{e-1}\, +\, x^e)\, =\, e^x\left(e x^{e-1}\, +\, x^{e-1+1}\right)$
. . . . .$\displaystyle =\, e^x\left(e x^{e-1}\, +\, x^{e-1}x^1\right)\, =\, e^x \left(e x^{e-1}\, +\, x^{e-1} x\right)$
. . . . .$\displaystyle =\, e^x x^{e-1}\left(e\, +\, x\right)$