# derivative

• Mar 19th 2009, 12:17 PM
algebra2
derivative
how to find derivative for: y = x^e * e^x
• Mar 19th 2009, 12:27 PM
e^(i*pi)
Quote:

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)$
• Mar 19th 2009, 02:51 PM
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
• Mar 19th 2009, 03:47 PM
HallsofIvy
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 $e^xx^{e-1}(e+ x)$.
• Mar 19th 2009, 05:08 PM
algebra2
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)
$
• Mar 20th 2009, 04:33 AM
stapel
Quote:

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

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

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

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

(Wink)