1. ## derivative

how to find derivative for: y = x^e * e^x

2. Originally Posted by algebra2
how to find derivative for: y = x^e * e^x
Use the product rule:

$\displaystyle u = x^e$ and $\displaystyle \frac{du}{dx} = ex^{e-1}$

$\displaystyle v = e^x$ and $\displaystyle \frac{dv}{dx} = e^x$
$\displaystyle \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} = ee^xx^{e-1} + x^ee^x = e^x(ex^{e-1} + x^e)$

3. l used the product rule and got the same answer as you, but the answer is suppose to be

$\displaystyle e^xx^{e-1}(x+e)$

is there a way to get this

4. 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 $\displaystyle e^xx^{e-1}(e+ x)$.

5. 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 $\displaystyle e^xx^{e-1}(e+x)$

6. Originally Posted by algebra2
l used the product rule and got the same answer as you, but the answer is suppose to be $\displaystyle e^xx^{e-1}(x+e)$

is there a way to get this
$\displaystyle e^x(ex^{e-1}\, +\, x^e)\, =\, e^x\left(e x^{e-1}\, +\, x^{e-1+1}\right)$

. . . . .$\displaystyle =\, e^x\left(e x^{e-1}\, +\, x^{e-1}x^1\right)\, =\, e^x \left(e x^{e-1}\, +\, x^{e-1} x\right)$

. . . . .$\displaystyle =\, e^x x^{e-1}\left(e\, +\, x\right)$