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Homework Help: Implicit differentiation problem!

  1. Oct 14, 2005 #1
    hello everyone i'm stuck!! anyone have any ideas?
    I'm suppose to find dz/dx and dz/dy with implicit differentation. This is calc III!
    http://img221.imageshack.us/img221/4000/lastscan4ou.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 14, 2005 #2


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    Is the question "x - z = arctan(yz). Find dz/dx. Find dz/dy." ?

    Differentiate throughout: dx - dz = (ydz+zdy)arctan'(yz).

    Then manipulate into dz = A + Bdx where A and B are functions of x, y and z. Divide by dx to get dz/dx = A/dx + B.

    Similar for dz/dy.
  4. Oct 14, 2005 #3


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    One problem you have is that you have the wrong derivative for arctangent!

    The derivative of arctan(x) is [tex]\frac{1}{1+ x^2}[/tex]

    if x-z= arctan(yz) then, writing zx and zy for the derivatives of z with respect to x and y respectively, we have
    [tex]1-z_x= \frac{yz_x}{1+ y^2z^}[/tex]
    which you can solve for zx and
    [tex]-z_y= \frac{z+ yz_y}{1+ y^2z^2}[/tex]
    which you can solve for zy.
  5. Oct 16, 2005 #4


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    One thing I couldn't understand here was, what happened to dy/dx and dx/dy.

    We have,

    x - z = arctan(yz)

    differentiating wrt x,

    1 - dz/dx = d/dx{arctan(s)}, where s = yz
    1 - dz/dx = d/ds{arctan(s)}.ds/dx
    1 - dz/dx = 1/(1 + s²) * (y.dz/dx + z.dy/dx)
    1 - dz/dx = (y.dz/dx + z.dy/dx) / (1 + y²z²)

    Adopting HallsofIvy's notation,

    1 - zx = (yzx + zyx)/(1 + y²z²)

    What have I missed out ?
  6. Oct 16, 2005 #5


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    Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "[itex]\partial[/itex]" notation?
    Last edited: Oct 16, 2005
  7. Oct 16, 2005 #6


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    That explains it.
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