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Implicit Partial Differentiation

  1. Jun 16, 2005 #1
    If there is such a thing. I need to find [itex]\partial z / \partial x[/itex] given [itex]x + y + z = \cosh xyz[/itex]. I've never seen the likes of this before and I haven't a clue where to start. Would a reasonable start be to take [itex]\partial /\partial x[/itex] of both sides? If so, it seems like I'm going to end up with an expression in terms of partial differntials.
     
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  3. Jun 16, 2005 #2

    OlderDan

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    The symbol [tex]\partial z / \partial x[/tex] is incomplete. We do it all the time, but a symbol for a partial derivative is never complete without specifying what is being held constant when the derivative is taken. You need to decide from the context of your problem what is being held constant here. If z is being held constant, then the partial derivative is zero. On the other hand, if z = z(x,y) and only y is being held constant it is a completely different problem.
     
  4. Jun 16, 2005 #3
    All I am given is that z is given implicitly as a function of x and y by the above equation. I take this to mean that y is held constant while x and z are allowed to vary.
     
  5. Jun 16, 2005 #4

    saltydog

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    How about mine:

    [tex]z^2=Sin[xz]+xy[/tex]

    Assuming z is a differentiable function of x and y and I want the partial of z with respect to x, lets take that partial of both sides:

    [tex]\frac{\partial}{\partial x}\{z^2=Sin[xz]+xy\}[/tex]

    [tex]2z\frac{\partial z}{\partial x}=Cos[xz](x\frac{\partial z}{\partial x}+z)+y[/tex]

    Isolating the partial:

    [tex]\frac{\partial z}{\partial x}=\frac{zCos(xz)+y}{2z-xCos[xz]}[/tex]

    Now, I bet you can do yours!
     
  6. Jun 16, 2005 #5
    Thanks for the worked example. I got [itex]\frac{\partial z}{\partial x} = \frac{yz\sinh xyz -1}{1 - xy\sinh xyz}[/itex].
     
  7. Jun 16, 2005 #6

    saltydog

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    Put parenthesis around xyz. It's ambiguous as written. :smile:
     
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