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Partial Derivative of Composite Functions

  1. Sep 27, 2012 #1
    Any help would be much appreciated - Is it possible to say the following?

    If z = g(s+at) + f(s-at), let u = s+at and v=s-at, where a is a constant.

    z = g(u) + f(v), [itex]\frac{∂z}{∂u}[/itex] = g'(u), [itex]\frac{∂^{2}z}{∂v∂u}[/itex] = 0?

    or can ∂u and ∂v not even exist because it depends on two variables (a and t), which are the same ones as the ones v depends on.

  2. jcsd
  3. Sep 27, 2012 #2


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    Science Advisor

    "[tex]\partial u[/tex]" does not exist by itself. While we define "differentials", dx and dy, we do NOT define "partial differentials". Rather, if f(x, y) is a function of the two variables, x and y, we define its "total differential":
    [tex]df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}[/tex]

    If If z = g(s+at) + f(s-at), let u = s+at and v=s-at, where a is a constant, and z = g(u) + f(v), then
    [tex]\frac{\partial z}{\partial s}= \frac{\partial f}{\partial s}+ \frac{\partial g}{\partial s}[/tex]
    [tex]= \frac{df}{du}\frac{\partial u}{\partial s}+ \frac{dg}{dv}\frac{\partial v}{\partial s}= \frac{df}{du}(1)+ \frac{dg}{dv}(1)= \frac{df}{du}+ \frac{dg}{dv}[/tex]

    [tex]\frac{\partial z}{\partial t}= a\frac{df}{du}- a\frac{dg}{dv}[/tex]
  4. Sep 27, 2012 #3
    Thanks for your quick reply! I think I get what you mean by ∂u can't exist by itself. Do you think you would be able to take a look at the original equation I posted:

    [itex]z = g(u) + f(v),
    \frac{∂^{2}z}{∂u∂v} = 0[/itex]

    Does that still hold true even when you can't hold v or u constant while changing the other? (they depend on the same two vars)
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