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Math Help - differentiation

  1. #1
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    differentiation

    Hi:
    I have a complicated question here. At least, I think so.
    I have gotten to a stage, but not sure whether I needed to do the last stage. Did I need to get y on its own or was the differentiation fine.
    And now, do I just differentiate again and then put the answers of the two into the equation provided to see if it is equal to zero?

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  2. #2
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    Quote Originally Posted by stealthmaths View Post
    Hi:
    I have a complicated question here. At least, I think so.
    I have gotten to a stage, but not sure whether I needed to do the last stage. Did I need to get y on its own or was the differentiation fine.
    And now, do I just differentiate again and then put the answers of the two into the equation provided to see if it is equal to zero?

    \frac{d}{dx}(e^y = e^x + e^{-x})

    e^y \cdot \frac{dy}{dx} = e^x - e^{-x}<br />

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

    \frac{d}{dx}\left[\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^y}\right]

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

    \frac{d^2y}{dx^2} = \frac{e^{2y} - (e^x - e^{-x})^2}{e^{2y}}<br />



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

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

    \frac{e^{2y}}{e^{2y}} - 1 = 0
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  3. #3
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    Yeah that is a tricky problem. Well, doesn't seem so for you but...
    Thanks skeeter. I'm gonna go over that one after I have done some more easier practice, I think.
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