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Math Help - finding y'

  1. #1
    Junior Member
    Joined
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    finding y'

    <br /> <br />
y = \frac{e^x - e^{-x}}{e^x + e^{-x}}<br />

    use the quotient rule and get:

    <br /> <br />
y' = \frac{(e^x + e^{-x}) \cdot (e^x - e^{-x}) \cdot (-1) - (e^x - e^{-x}) \cdot (e^x + e^{-x}) \cdot (-1)}{(e^x + e^{-x})^2}<br />

    but when i simplify this, i get 0
    what am i doing wrong...
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    what am i doing wrong...
    you're not taking the derivative correctly ...

    the derivative of e^x - e^{-x} is e^x - e^{-x}\cdot -1 = e^x + e^{-x}

    the derivative of e^x + e^{-x} is e^x + e^{-x}\cdot -1 = e^x - e^{-x}


    <br />
y = \frac{e^x - e^{-x}}{e^x + e^{-x}}<br />

    use the quotient rule

    <br />
y' = \frac{(e^x + e^{-x}) \cdot (e^x + e^{-x}) - (e^x - e^{-x}) \cdot (e^x - e^{-x})}{(e^x + e^{-x})^2}<br />
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