Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chain rule with multiple variables

  1. Jul 1, 2012 #1
    I was reading over a text book that stated the following, where [itex]y(s,t) = v(x(s,t),t)[/itex]

    [tex]\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial v}{\partial t}[/tex]


    [tex]\frac{\partial^2y}{\partial t^2} = \frac{\partial^2 v}{\partial x^2}\left ( \frac{\partial x}{\partial t} \right )^2 + 2\frac{\partial^2 v}{\partial x \partial t}\frac{\partial x}{\partial t} + \frac{\partial^2 v}{\partial t^2}+ \frac{\partial v}{\partial x}\frac{\partial^2 x}{\partial t^2}[/tex]

    I have been having trouble figuring out how they came to the second statement. My logic proceeds as such, using the product rule:

    [tex]\frac{\partial^2y}{\partial t^2} = \frac{\partial}{\partial t}\left [ \frac{\partial v}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial v}{\partial t} \right] = \frac{\partial^2 v}{\partial t \partial x}\frac{\partial x}{\partial t} + \frac{\partial v}{\partial x}\frac{\partial^2 x}{\partial t^2} + \frac{\partial^2 v}{\partial t^2}[/tex]

    But certainly my result is not equivalent to the correct one. Where did I go astray? Thanks all!
  2. jcsd
  3. Jul 26, 2012 #2
    It looks like the textbook's result is correct. I just came up with a long, messy proof of the formula, and I've attached it as a PDF. You're probably going to need to zoom in to see all the details. Tell me if you don't understand something. It seems like I'm making this proof unnecessarily complicated, so let's see whether I can find a way to make it shorter and more intuitive.

    Attached Files:

  4. Jul 26, 2012 #3
    you have to remember
    [tex] \frac{\partial v}{\partial x},[/tex]
    is a function of t explicitly and a function of t through x.
    so that
    [tex] \frac{\partial}{\partial t}\left( \frac{\partial v(x(t), t)}{\partial x} \right)
    = \frac{\partial}{\partial x}\left( \frac{\partial v(x(t), t)}{\partial x} \right) \frac{\partial x}{\partial t}+
    \frac{\partial^2 v(x(t), t)}{\partial t \partial x}

    this is one of those cases where the notation is kind of sloppy.

    and of course the same thing for
    [tex] \frac{\partial v}{\partial t}( x(t), t) [/tex]
    Last edited: Jul 26, 2012
  5. Jul 26, 2012 #4
    qbert, are you using total derivatives instead of partial derivatives, but denoting them as partial derivatives? If so, why?
  6. Jul 27, 2012 #5
    all derivatives are partial holding s constant.
    you just need to keep the implicit dependence straight.
  7. Jul 27, 2012 #6
    OK, I got it now. I just needed to keep the partial and total derivatives straight. I've attached a PDF with two relatively compehensible proofs of the formula.

    Attached Files:

  8. Jul 27, 2012 #7
    i said something wrong which has been bugging me.
    i was going to let it slide but then there was an answer
    with the phrase "total derivative", which to me needed
    commenting on.

    this is basically nit-picking, but....

    all the derivatives are partial. they're partial and the
    notation is ambiguous as to what's "being held constant"
    which is why i commented on the sloppiness of the notation.

    when you write [itex] \frac{\partial^2 y}{\partial t^2} [/itex]where [itex]y = y(s,t)[/itex] is a function
    of the two variables it means, keep s constant through both derivatives.

    so we need to remember our function always depends on the
    two variables s an t.

    now if we define [itex]y(s,t)=v(x(s,t), t)[/itex] then v is
    a function of the two variables x and t. and x is a function
    of the two variables s and t.

    so that
    [tex]\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x}\frac{\partial x}{\partial t}
    + \frac{\partial v}{\partial t}[/tex]

    The symbol [itex]\frac{\partial }{\partial t}[/itex] means different things on each side of
    the equation! acting on y(s,t) it means differentiate with respect to t keeping s constant.
    When it acts on v(x,t) it means differentiate with respect to t keeping x constant.

    It turns out to be convenient here to introduce
    \frac{\partial v}{\partial x}(x,t) = w(x, t), \text{ and }
    \frac{\partial v}{\partial t}(x, t) = u(x, t).

    So that

    [tex]\frac{\partial y}{\partial t}(s,t) = w(x(s,t), t) \frac{\partial x}{\partial t}(s,t)
    + u(x(s, t), t)[/tex]

    Now doing the derivative holding s constant gives
    [tex] \frac{\partial^2 y}{\partial t^2} =
    \left( \frac{\partial w}{\partial x} \frac{\partial x}{\partial t}
    + \frac{\partial w}{\partial t}
    \right) \frac{\partial x}{\partial t}
    + w \frac{\partial^2 x}{\partial t^2}
    + \left( \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial t} \right)

    now plugging in for w and u the derivatives of v gives the books result.
    notice all the derivatives are partial.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Chain rule with multiple variables
  1. Chain Rule (Replies: 25)

  2. Chain rule. (Replies: 6)

  3. The Chain Rule (Replies: 3)