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Math Help - Tricky Derivative

  1. #1
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    Tricky Derivative

    Hi, I have the following question:
    Given that x = e^{t}, find dy/dx in terms of dy/dt and show that

    \frac{d^{2}y}{dx^{2}} = e^{-2t}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt})

    So first of all, I used the chain rule to get

    dy/dx = dt/dx . dy/dt

    dt/dx = e^{-t}

    so that gives me \frac{dy}{dx} = e^{-t}\frac{dy}{dt}
    but how do I differentiate that a second time. I would have to be the product rule, but it's not implicit? I can't see how

    Thanks
    Last edited by thomas49th; June 15th 2009 at 04:43 AM.
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  2. #2
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    Quote Originally Posted by thomas49th View Post
    Hi, I have the following question:
    Given that x = e^{t}, find dy/dx in terms of dy/dt and show that

    \frac{d^{2}y}{dx^{2}} = e^{-2t}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt})

    So first of all, I used the chain rule to get

    dy/dx = dt/dx . dy/dt

    dt/dx = x = e^{-t}

    so that gives me \frac{dy}{dx} = e^{-t}\frac{dy}{dt}
    but how do I differentiate that a second time. I would have to be the product rule, but it's not implicit? I can't see how

    Thanks
    \frac{d}{dx} \left[ e^{-t}\frac{dy}{dt} \right] = \frac{d}{dt} \left[ e^{-t} \, \frac{dy}{dt} \right] \cdot \frac{dt}{dx}

    = \left( -e^{-t} \, \frac{dy}{dt} + e^{-t} \, \frac{d^2 y}{dt^2}\right) \cdot e^{-t}.
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