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Partial derivatives

  1. Feb 18, 2006 #1

    how do I find [tex]\frac{\partial z}{\partial x}[/tex]?

    if the problem is sin(5x-4y+z)=f(x,y,z), I can find [tex]\frac{\partial f}{\partial x}[/tex] but I dont know what to do when it is just equal to zero.
  2. jcsd
  3. Feb 18, 2006 #2
    Solve for z, take the partial derivative.
  4. Feb 18, 2006 #3


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    Simpler: take the partial derivative with respect to x, assuming that y is independent of x, z a function of x, then solve for zx. Use the chain rule. Remember "implicit differentiation" from Calculus I?
    (sin(5x- 4y+ z))x= cos(5x- 4y+ z)(5- zx)= 0. Solve for zx.
  5. Feb 18, 2006 #4
    HallsofIvy: I think that should be a (5+zx) in your second expression.
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