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Partial Derivative Question

  1. May 2, 2006 #1
    This is annoying me as i have the answer on the tip of my pen, just can't write it down. I'm not 100% sure i understand what the question is asking me to do.

    Consider the quantity [tex] u = e^{-xy} [/tex] where (x,y) moves in time t along a path:

    [tex]x = \cosh{t}, \mbox{ } y = \sinh{t}[/tex]​

    Use a method based on partial derivatives to calculate [tex] \frac{du}{dt} [/tex] as a function of x, y and t.

    My answer:

    I partially differentiated u, getting:

    [tex] \frac{\delta{u}}{\delta{x}} = -ye^{-xy} [/tex]
    [tex] \frac{\delta{u}}{\delta{y}} = -xe^{-xy} [/tex]
    So does this mean [tex] du = -ye^{-xy} + -xe^{-xy} ? [/tex]

    I though that i would get du from the part iv just explained, then get dt from differentiating x and y. But this ofcourse leaves me with expressions for dx/dt and dy/dt. Where do i go from here?
     
  2. jcsd
  3. May 2, 2006 #2
    Use the chain rule:
    [tex]\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}[/tex]
     
    Last edited: May 2, 2006
  4. May 2, 2006 #3
    Yeah, realised that after some research - just something id never saw. Dead easy tho. Thanks anyway :smile:
     
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