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Math Help - Differentiate a mixture of e's

  1. #1
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    Differentiate a mixture of e's

    Hi i'm just looking to see if i've got this right:

    differentiate:

    \frac{e^x}{e^x+e^{-x}}

    I got this:

    \frac{2}{e^{2x}+e^{-2x}}

    Something at the back of my head says "that looks wrong" so i'm seeking clarifcation and/or guidance

    Thank you
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  2. #2
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    Quote Originally Posted by aceband View Post
    Hi i'm just looking to see if i've got this right:

    differentiate:

    \frac{e^x}{e^x+e^{-x}}

    I got this:

    \frac{2}{e^{2x}+e^{-2x}}

    Something at the back of my head says "that looks wrong" so i'm seeking clarifcation and/or guidance

    Thank you
    Not sure how you got the denominator. I got

     <br />
\frac{2}{\left(e^x + e^{-x}\right)^2}<br />
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    \frac{d}{dx}\frac{e^x}{e^x+e^{-x}}= \frac{e^x(e^x+e^{-x})-e^x(e^x-e^{-x})}{(e^x+e^{-x})^2}=\frac{2}{e^{2x}+e^{-2x}+2}
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  4. #4
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    Thanks guys,

    How did you get that +2 in the bottom though?
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  5. #5
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    2e^xe^{-x} = 2e^{x-x} = 2e^0 = 2
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  6. #6
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    Quote Originally Posted by aceband View Post
    Thanks guys,

    How did you get that +2 in the bottom though?
    \left(e^x + e^{-x}\right)^2 = \left(e^{x}\right)^2 + 2e^x \cdot e^{-x} + \left(e^{-x}\right)^2 and the middle term when simplified give 2.
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  7. #7
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    AHH!!! Fantastic now i get it - i stupidly thought 2 * e^0 was 0

    Thank you so much for the help.
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