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Implicit Differentiation in Multivariable Calculus

  1. Mar 6, 2006 #1
    I need to compute the partials of z with respect to x and y of:
    xy + z + 3xz^5 = 4 at (1,0).

    I already showed that the equation is solvable for z as a function of (x,y) near (1,0,1) with the special implicit function theorem, but that's the easy part. Could someone explain to me how to begin this?
  2. jcsd
  3. Mar 6, 2006 #2
    Do I start by taking d/dx of xy + z + 3xz^4 - 4 = 0, giving
    y + dz/dx + 3z^5 = 0, or dz/dx = -y - 3z^5..?
  4. Mar 6, 2006 #3


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    You take d/dx of both sides, taking z as a function of x and y, and taking x and y as not functions of each other (treat y as a constant when taking d/dx, treat x as a constant when taking d/dy). Use the chain rule. For example, d(2x^2 * z^2)/dx = 2xz^2 + 4x^2 * z(dz/dx).
  5. Mar 6, 2006 #4
    How do I take z as a function of x and y when I can't solve for it?
  6. Mar 6, 2006 #5
    Alternatively, if I use the fact that the partial of z with respect to x equals -(partial of F with respect to x)/(partial of F with respect to z), don't I still end up with a z (and I need it in terms of x and y so I can plug in my point)?
  7. Mar 6, 2006 #6
    Yeah, that trick works as well to find dz/dx.

    It will give you something in terms of x and y...after all when taking partials with respect to some variable you will get all of those variables in the result.
    Last edited: Mar 6, 2006
  8. Mar 6, 2006 #7
    If you have time, could you walk me through the dz/dx method? I'm not following this... If I try to take dz/dx with the formula mentioned above, I get -(y + dz/dx + 3z^5)/(dxy/dz + 1 + 15xz^4). Do dz/dx and dxy/dz go to 0? What about the extra zs?
  9. Mar 6, 2006 #8
    try just writing in a general sense with letting some function of the variables (x,y) say z=z(x,y) where all the derivative and continuity requirements we need are assumed satisfied. Now from ordinary calc we may write
    [tex] dz = \left( \vec{\nabla} z(x,y) \cdot \hat{x}\right) dx + \left( \vec{\nabla} z(x,y) \cdot \hat{y} \right) dy \Rightarrow
    dz = \frac{\partial z}{\partial x} dx+ \frac{\partial z}{\partial y} dy [/tex]
    would you agree?
    Last edited: Mar 7, 2006
  10. Mar 6, 2006 #9


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    Why would you need to solve for it? Isn't it enough to just say z can be written as a function of x and y?

    I suspect you would profit from reviewing the one-dimensional case of implicit differentiation -- I think you're missing the basic idea behind it, and it might help to work in a simpler setting. (And, presumably, you've learned this once already and it might come back to you. You might even have saved your notes and worked homework problems!)
    Last edited: Mar 6, 2006
  11. Mar 7, 2006 #10
    If you use the formula:

    [tex]\frac {\partial z}{\partial x} = -\frac {\frac {\partial F}{\partial x}}{\frac {\partial F}{\partial z}}[/tex]

    then you will get an answer with only x, y and z in it. The function F has a constant value of zero when you create it by moving everything to one side of the equation. It is still a function of x and y, and z is still the dependant variable.

    If you go through with implicit differentiation, you must solve for [tex]\frac {\partial z}{\partial x}[/tex], not forgetting to do the product rule and chain rule in places where you have both an x and a z.

    In the end you should get the same answer for either method.
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