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Homework Help: Partial derivatives and chain rule

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    express [tex](\frac{\partial u}{\partial s})_{v}[/tex] in terms of partial derivatives of u(s,t) and t(s,v)

    2. Relevant equations



    3. The attempt at a solution

    I'm pretty stuck with this problem. I know that

    dv = [tex](\frac{\partial v}{\partial s})_{t} ds + (\frac{\partial v}{\partial t})_{s} dt[/tex]

    and similarly for u(s,t) and t(s,v). But problem is that where do I get [tex](\frac{\partial u}{\partial s})_{v}[/tex] from, and how to express it?
     
  2. jcsd
  3. May 22, 2010 #2

    UD1

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    d/(dv)[du/ds]=

    d/(dv)[du(s,t(s,v))/ds]=

    d/(dv)[du/ds + du/dt * dt/ds]=

    d/(dv)[du/ds+du(s,t(s,v))/dt * dt(s,v)/ds]=

    d^2u/dsdt * dt/dv + d^2u/dt^2 * dt/dv * dt/ds + du/dt* d^t/dsdv

    Write it down in usual partial derivative notation and you'll see where it all comes from.
     
  4. May 22, 2010 #3

    gabbagabbahey

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    Please don't give out complete solutions (even if they are grievously incorrect like this one), it violates forum rules and doesn't help the student learn the material.
     
  5. May 22, 2010 #4

    gabbagabbahey

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    For starters, [itex]u(s,t)[/itex] is a function of both [itex]s[/itex] and [itex]t[/itex], so use the chain rule.
     
  6. May 22, 2010 #5

    UD1

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    I apologize. The solution is correct though.
     
  7. May 22, 2010 #6

    gabbagabbahey

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    No, it isn't. You've misinterpreted the question/notation.

    [tex]\left(\frac{\partial u}{\partial s}\right)_v\neq\frac{\partial}{\partial v}\left(\frac{\partial u}{\partial s}\right)[/tex]

    Instead, [itex]\left(\frac{\partial u}{\partial s}\right)_v[/itex] is defined as the derivative of [itex]u[/itex] w.r.t [itex]s[/itex], while [itex]v[/itex] is held constant. It's a specific type of partial derivative.
     
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