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Math Help - Differentiating yields different answers

  1. #1
    Member Jones's Avatar
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    Differentiating yields different answers

    Hi,

    I have the following:
    y = Ate^{-2x}
    y ^{\prime} = Ae^{-2x}-2Ate^{-2x}

    For y^{\prime\prime} i seem to get different answers, depending on the method.

     \frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)

    1. \frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}
    2. If i factor out the e term. \frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}

    One of these is more correct than the other, which is it?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Jones View Post
    Hi,

    I have the following:
    y = Ate^{-2x}
    y ^{\prime} = Ae^{-2x}-2Ate^{-2x}

    For y^{\prime\prime} i seem to get different answers, depending on the method.

     \frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)

    1. \frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}
    2. If i factor out the e term. \frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}

    One of these is more correct than the other, which is it?
    You are mixing the x and t when calculating the derivative.
    If the power of "e" is -2t, then you have a product of 2 functions of "t".
    If you differentiate with respect to x, you do not know what the derivative of "t" is, if t is itself a function of x.
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  3. #3
    Member mathemagister's Avatar
    Joined
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    Quote Originally Posted by Jones View Post
    Hi,

    I have the following:
    y = Ate^{-2x}
    y ^{\prime} = Ae^{-2x}-2Ate^{-2x}

    For y^{\prime\prime} i seem to get different answers, depending on the method.

     \frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)

    1. \frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}
    2. If i factor out the e term. \frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}

    One of these is more correct than the other, which is it?
    The mistake you are doing, is that you are differentiating with respect to t sometimes and x the others. \frac{d}{dx} means you are differentiating with respect to x, so, \frac{d}{dx}At = A\frac{dt}{dx}. What you didn't notice is that \frac{d}{dx}At \neq A
    Hope that helps
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