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

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- Mar 19th 2009, 12:17 PMalgebra2derivative
how to find derivative for: y = x^e * e^x

- Mar 19th 2009, 12:27 PMe^(i*pi)
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)$ - Mar 19th 2009, 02:51 PMalgebra2
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 - Mar 19th 2009, 03:47 PMHallsofIvy
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 $\displaystyle e^xx^{e-1}(e+ x)$.

- Mar 19th 2009, 05:08 PMalgebra2
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)

$ - Mar 20th 2009, 04:33 AMstapel
$\displaystyle e^x(ex^{e-1}\, +\, x^e)\, =\, e^x\left(e x^{e-1}\, +\, x^{e-1+1}\right)$

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

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

(Wink)