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Math Help - Constant Multiple Rule

  1. #1
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    Constant Multiple Rule

    Finding the derivative of this rule;

    y = (e^x - e^-x) / 2

    y = 1/2 (e^x - e^-x)

    y' = 1/2 (e^x + e^-x)

    The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    Finding the derivative of this rule;

    y = (e^x - e^-x) / 2

    y = 1/2 (e^x - e^-x)

    y' = 1/2 (e^x + e^-x)

    The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?
    if v = -e^u , then v' = -e^u \cdot u'

    v = -e^{-x} ... v' = -e^{-x} \cdot (-1) = +e^{-x}
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  3. #3
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    Quote Originally Posted by Archduke01 View Post
    Finding the derivative of this rule;

    y = (e^x - e^-x) / 2

    y = 1/2 (e^x - e^-x)

    y' = 1/2 (e^x + e^-x)

    The constant multiple rule was applied to that last line to flip the minus sign to a plus ... but I don't understand the logic. How does that rule work?
    \frac{d}{dx}y = \frac{d}{dx} \frac{1}{2} (e^x - e^{-x})

    \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (e^x - e^{-x})

    \frac{dy}{dx} = \frac{1}{2} \left(\frac{d}{dx} e^x - \frac{d}{dx} e^{-x}\right)

    \frac{dy}{dx} = \frac{1}{2} \left(\left(\frac{d}{dx} x\right)e^x - \left(\frac{d}{dx} (-x)\right) e^{-x}\right)

    \frac{dy}{dx} = \frac{1}{2} \left((1)e^x - (-1) e^{-x}\right)

    \frac{dy}{dx} = \frac{1}{2} (e^x + e^{-x})
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