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Confusion about partial derivatives

  1. May 29, 2012 #1
    Dear all,

    I have a confusion about partial derivatives.

    Say I have a function as


    and we know that

    1. Does it make sense to talk about partial derivatives like [itex]\frac{\partial y}{\partial x}[/itex] and [itex]\frac{\partial y}{\partial t}[/itex] ?

    I doubt, because the definition of partial derivative is the change in the function due to the change on the selected variable ( other variables are kept constant).

    2. If it does not make sense, then how in Euler–Lagrange equation we use the partial derivatives with respect to a function(x(t)) and its derivative(x'(t)) while they depend on one another.

    Your help would be appreciated.
  2. jcsd
  3. May 29, 2012 #2


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    Hey Hassan2.

    For 1. Yes it makes sense to do both but you need to factor in things like the chain rule for this particular example: as long as you are taking into account these kinds of factors, then yes it's ok. Your partial with respect to t can take into account your g(t) using the chain rule.

    If the function is differentiable in the the region you are considering, the derivative will make sense: it's guaranteed to as a consequence of differentiability holding.
  4. May 29, 2012 #3
    Thanks Chiro,

    Yes, I checked the derivation of Euler-Largrange equation once again, and i found that the derivatives wit respect to x and x' appear as a result of applying chain rule.

    Thanks again.
  5. May 29, 2012 #4


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    The notation [itex]\partial f/\partial x[/itex] means the derivative of f with respect to x while holding x constant- and ignoring the fact that x is a function of t. We are really dealing with the "form" of f rather than the content.

    The same is true of [itex]\partial f/\partial t[/itex]. However, we can, using x= g(t), think of f as a function of t only- f(x, t)= f(g(t), t). In that case, by the chain rule,
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial t}+ \frac{\partial f}{\partial x}\frac{dx}{dt}[/tex]

    Example: if [itex]f(x,t)= 3tx^2+ e^x[/itex] then [itex]\partial f/\partial x= 6tx+ e^x[/itex] and [itex]\partial f/\partial t= 3x^2[/itex]. That has nothing to do with x being a function of t or vice-versa.

    But if we also know that [itex]x= g(t)= 2t^2+ t[/itex] we can write [itex]f(t)= 3t(2t^2+ t)+ e^{2t^2+ t}= 12t^5+ 12t^4+ 3t^3+ e^{2t^2+ t}[/itex] so that the derivative is
    [tex]\frac{df}{dt}= 60t^4+ 48t^3+ 9t^2+ (4t+ 1)e^{2t^2+ t}[/tex]

    Or, you could use the chain rule as I said:
    [itex]\partial f/\partial t= 3x^2[/itex] and [itex]\partial f/\partial x= 3tx+ e^x[/itex], as above, while [itex]dx/dt= 4t+1[/itex] and so
    [tex]\frac{df}{dx}= 3x^2+ (3tx+ e^x)(4t+ 1)[/tex]
    and, replacing x in that with [itex]2t^2+ t[/itex]
    [tex]\frac{df}{dx}= 3(2t^2+ t)+ (3t(2t^2+ t)+ e^{2t^2+ t})(4t+1)[/tex]
    gives the same thing.
  6. May 29, 2012 #5
    Strictly speaking, one should not write f(t)=f(g(t),t). The function on the left should have gotten a new name. That is the source of some confusion.
  7. May 29, 2012 #6


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    What we should write, is something like:

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