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Homework Help: Multivariable Chain Rule

  1. Feb 6, 2010 #1
    Hello hello. In class we went over the ''mini-chain rule'' once, and haven't gone over the real chain rule yet. I really want to understand how to go about solving this equation, and to really understand what is happening here.

    x=u3-3uv2
    y=3u2v-v3
    z=u2-v2

    Define z as a function of x and y. Determine delZ/delx at the point (u,v) = (2,1) which corresponds to the points (x,y) = (2,11)

    I can see from the last point that points (u,v) = (x(u,v), y(u,v))
    f(x,y) should equal z. I am just confused how to interpret this with z = u2-v2.

    Since I received this as homework, I am guessing it can be done with partial derivatives. Any insight into this would be welcomed!
     
  2. jcsd
  3. Feb 7, 2010 #2

    Mark44

    Staff: Mentor

    I find it helpful to sketch the relationships between the variables.
    ChainRule.JPG
    Here you have z as a function of x and y, and x and y are both functions of u and v. My drawing is pretty small, but it shows z all the way to the left, and then x and y in the center column, and finally, u and v in the right column.

    The chain rule takes different forms for multivariable functions. To find the partial of the variable on the left (z) with respect to either of the variables on the right (say u), you're basically looking at all the ways to get from z to u. There is one path from z to x to u, and another from z to y to u. The first of the formulas below captures this idea.

    Here is the chain rule for these variables.
    [tex]\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}[/tex]
    [tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}[/tex]

    To make the typing simpler, I'll use subscripts to represent partial derivatives, with zu being the partial of z with respect to u.

    You are asked to find zx. From the formulas you are given, you can find zu, xu, and yu. However, there are two unknown quantities in the first equation above: zx and zy. These are unknown because you are not given explicit formulas for z in terms of x and y.

    In the second formula above, for zv, you can calculate zv, xv, and yv. This means that you will have two equations in zx and zy, so you should be able to solve for them algebraically.

    After you have found zx, evaluate it at the given point, and you're done.
     
  4. Feb 7, 2010 #3
    That was a great explanation. Thanks a lot, it makes complete sense now. I can't believe I just discovered this site now, you guys are unbelievably helpful.

    Thanks again!
     
  5. Feb 7, 2010 #4

    Mark44

    Staff: Mentor

    Glad to be of help!
     
  6. Feb 8, 2010 #5
    I have a question:

    1) is it true for chain rule, differentiation only allow with respect to the variable on the right eg.

    [tex] \frac{\partial z}{\partial x}, \frac{\partial y}{\partial u}[/tex] are allowed because the function z and y are differentiated by the variable on the right side of the diagram.

    But [tex] \frac{\partial u}{\partial x}, \frac{\partial v}{\partial y}[/tex] is not allow because the function u and v are differentiated by the variable on the left side of the diagram.

    [tex] \frac{\partial y}{\partial x}, \frac{\partial x}{\partial y}[/tex] is not allowed because there is no path to go from x to y in the diagram.

    Therefore
    [tex] \frac{\partial y}{\partial x} = 0, \frac{\partial x}{\partial y} = 0[/tex]
     
  7. Feb 8, 2010 #6

    Mark44

    Staff: Mentor

    No, these aren't true. x is not a function of y, and y is not a function of x -- x and y are assumed to be independent variables in this problem. It doesn't make sense to take the partial of one independent variable with respect to another.

    For example, if z = f(x, y) = x2 + y2, you can talk about
    [tex] \frac{\partial z}{\partial x}~\text{and}~ \frac{\partial z}{\partial y}[/tex]
    but not
    [tex] \frac{\partial y}{\partial x}~\text{and}~ \frac{\partial x}{\partial y}[/tex]
     
  8. Feb 8, 2010 #7
    Thanks for you reply.

    But since x and y is independent variable, then[tex] \frac{\partial y}{\partial x}=0~\text{and}~ \frac{\partial x}{\partial y}=0[/tex] is true.

    Is [tex] \frac{\partial t}{\partial x}~\text{and}~ \frac{\partial t}{\partial y}[/tex] allow?

    I am very confused what chain rule allow and not allow in differentiation. I have been working on a problem in:

    https://www.physicsforums.com/showthread.php?t=375837

    It turn out to be chain rule problem and I am confused what differentiation is allowed and what is not. I really appreciate if you can help me on that.
     
  9. Feb 8, 2010 #8

    Mark44

    Staff: Mentor

    No it isn't. x and y are independent variables, which means they don't depend on each other. If there is no relationship between them, it's meaningless to talk about the derivative (partial or otherwise) of one with respect to the other.
     
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