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Partial derivative and chain rule

  1. Feb 9, 2013 #1
    How is the double derivative equal to that in the equation 2 in the attachment? =|

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  3. Feb 9, 2013 #2

    Simon Bridge

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    $$\dot{y} = \frac{\partial f_i}{\partial x_j}\dot{x_j} + \frac{\partial f_i}{\partial t}\qquad \text{...(1)}\\

    \ddot{y} = \frac{\partial f_i}{\partial x_j}\ddot{x_j} + \frac{\partial^2 f_i}{\partial x_j \partial x_k}\dot{x_j}\dot{x_k} + 2\frac{\partial^2 f_i}{\partial x_j \partial t}\dot{x_j} + \frac{\partial^2 f_i}{\partial t^2}\qquad \text{...(2)}$$​

    2 follows from 1 (and the definition of y - what is this?) by the chain rule ... so apply the chain rule and show where you get stuck.
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