derivative

**algebra2** · March 19th 2009, 12:17 PM

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

**e^(i*pi)** · March 19th 2009, 12:27 PM

Originally Posted by algebra2

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

Use the product rule:

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

$v = e^x$ and $\frac{dv}{dx} = e^x$
$ \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)$

**algebra2** · March 19th 2009, 02:51 PM

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

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

is there a way to get this

**HallsofIvy** · March 19th 2009, 03:47 PM

e^(i*pi) gave you $e^x(ex^{e-1} + x^e)$ . Now factor out an additional $x^{e-1}$ : $x^e= x^{e-1}(x)$ so $e^xx^{e-1}(e+ x)$ .

**algebra2** · March 19th 2009, 05:08 PM

how do you factor out x^(e-1) when there is only one x^(e-1), the other is x^e ======> $e^x(ex^{e-1} + x^e) $

wouldn't it have to be $e(e^xx^{e-1}) + x(e^xx^{e-1}) $ in order to write it as $e^xx^{e-1}(e+x) $

so how do you make $e^x(ex^{e-1} + x^e)$ (which is equal to $e(e^xx^{e-1}) + e^xx^e $ ) into $e(e^xx^{e-1}) + x(e^xx^{e-1}) $ in order to write it as the answer given $e^xx^{e-1}(e+x) $

**stapel** · March 20th 2009, 04:33 AM

Originally Posted by algebra2

l used the product rule and got the same answer as you, but the answer is suppose to be $e^xx^{e-1}(x+e)$

is there a way to get this

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

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

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

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