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Partial derivative of an ordinary derivative?

  1. Sep 4, 2008 #1
    I think the heading says it all. What happens if we take the partial derivative of a rate for example?

    eg [tex]\frac{\delta}{\delta t}(\frac{dx}{dt})[/tex]

    If it was normal differentiation with respect to t we'd get acceleration, or [tex]\ddot{x}[/tex]. I read somewhere that the partial can be treated as an ordinary if the top part tends to zero as the bottom part does, but if this is not the case? (such as if we had a function of cos which would approach 1)...
  2. jcsd
  3. Sep 4, 2008 #2


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    Hi Ultimâ! :smile:

    (have a partial: ∂ :smile:)

    Partial derivative ∂/∂t simply means that you keep all variables constant other than t.

    So if x only depends on t, then ∂/∂t = d/dt.

    If x depends on t and something else, then you shouldn't write d/dt.

    So ∂/∂t(dx/dt) doesn't really make sense. :smile:
    Sorry … not following any of that. :confused:
  4. Sep 4, 2008 #3

    Ben Niehoff

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    This is false. You just need to apply the chain rule. If x is a function of t and some other variable u, then

    [tex]\frac{d}{dt}[x(t, u)] = \frac{\partial x}{\partial t} + \frac{\partial x}{\partial u} \frac{du}{dt}[/tex]

    The result will be some function [itex]\dot x = \dot x(t, u, \dot u)[/itex], and then one can certainly take

    [tex]\frac{\partial}{\partial t}[\dot x(t, u, \dot u)][/tex]
  5. Sep 4, 2008 #4
    Thanks for that. As ben mentions, I had a case of a rate of changes (sounds like a rare disease), where the chain rule needed to be applied, and I thought the outcome seemed a bit weird. Thanks tiny-tim for verifying my suspicions that the partial can be treated as an ordinary differential when x is only dependant on t.

    Given that's the case, I guess [tex]\frac{\delta}{\delta t}(cos\theta)[/tex] can produce [tex]-sin\theta\frac{d\theta}{dt}[/tex]! (of course given dependence on t)

    a [tex]\delta{bye\ for\ now}[/tex]!
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