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  1. #1
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    derivative

    how to find derivative for: y = x^e * e^x
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  2. #2
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    Quote Originally Posted by algebra2 View Post
    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 $\displaystyle \frac{dv}{dx} = e^x$
    $\displaystyle
    \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx} = ee^xx^{e-1} + x^ee^x = e^x(ex^{e-1} + x^e)$
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    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
    Last edited by algebra2; Mar 19th 2009 at 03:20 PM.
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  4. #4
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    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 $\displaystyle e^xx^{e-1}(e+ x)$.
    Last edited by mr fantastic; Mar 19th 2009 at 05:50 PM. Reason: Added a missing latex tag
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  5. #5
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    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)
    $
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  6. #6
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    Quote Originally Posted by algebra2 View Post
    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
    $\displaystyle e^x(ex^{e-1}\, +\, x^e)\, =\, e^x\left(e x^{e-1}\, +\, x^{e-1+1}\right)$

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

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

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