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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:

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

    v = e^x and \frac{dv}{dx} = e^x
    <br />
\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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  3. #3
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    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)<br />

    is there a way to get this
    Last edited by algebra2; March 19th 2009 at 03:20 PM.
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  4. #4
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    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).
    Last edited by mr fantastic; March 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 ======> e^x(ex^{e-1} + x^e)<br />

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

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