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Derivative of multivariable function

  1. Apr 13, 2016 #1
    So this is a problem for microeconomics, but should follow under general calculus:

    The point is x=(x1(p1,p2,u),x2(p1,p2,u)) where u is a constant on the function u(x1,x2). p1 is the price of x1 and p2 for x2. I'm supposed to show that (dx1/dp2)=(dx2/dp1). I've been given the info that for the second derivative of a multivariable function the order of differentation is irrelevant (df(x1,x2)/dx1dx2)=(df(x1,x2)/dx2dx1). All of the mentioned should also be the case for functions of more than two variables.

    I've tried to set second derivatives equal to each other, but obviously ended up with the exact same on both sides. I don't know where to go from here. Hints and tips would be much appreciated.
     
  2. jcsd
  3. Apr 13, 2016 #2

    Ray Vickson

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    Please stop using boldface typing; it looks like you are yelling at us.

    Anyway, I do not think your question can be correct as written, because I can very easily devise two functions ##x_1(p_1,p_2,u)## and ##x_2(p_1,p_2,u)## with ##\partial x_1/\partial p_2 \neq \partial x_2/ \partial p_1##. Are you actually given some specific functions ##x_1## and ##x_2## as functions of ##(p_1,p_2)##? If you are given the right kind of functions then the result could be true.

    Furthermore, if ##u## is really a function of ##(x_1,x_2)## as you state, then you are dealing with functions of the form ##x_1(p_1,p_2,u(x_1,x_2))##, in which the function ##x_1## is also a function of itself. That really makes no sense.
     
  4. Apr 13, 2016 #3
    It is not always relevant. The necessary and sufficient condition for the change of order of differentiation is that all the functions and partial derivatives must be continuous.
     
  5. Apr 13, 2016 #4

    BiGyElLoWhAt

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    Have you applied the product rule and chain rule appropriately for both derivatives?

    The problem I'm running into is I think you need to assume that x is linear in x1 and x2. Or be given that info. Is there any additional information?
     
  6. Apr 13, 2016 #5
    Does this equation talk about rate of substitution?
    I assume whatever you have given.
    x is a function of x1 and x2
    write x =k(x1)m(x2) where k and m are functions of x1 and x2 alone, respectively.
    x1 can similarly be written as V1(p1)U1(p2)
    y1 can similarly be written as V2(p1)U2(p2)
    You have also given function u.
    let u be h(x1)s(x2)


    at this particular point 'x',
    k(x1)m(x2) is constant.
    I think it should be fine now for differentiating and using chain rule.
     
    Last edited: Apr 13, 2016
  7. Apr 13, 2016 #6

    BiGyElLoWhAt

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    So assuming all the necessary conditions are met (continuity)
    ##\frac{\partial x}{\partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial x_1}{\partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial x_2}{\partial p_1}##
    ##\frac{\partial x}{\partial p_2} = \frac{\partial x}{\partial x_1}\frac{\partial x_1}{\partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial x_2}{\partial p_2}##
    ##\frac{\partial}{\partial p_2}\frac{\partial x}{\partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_1 \partial p_2} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_1 \partial p_2} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} ##
    ##\frac{\partial}{\partial p_1}\frac{\partial x}{\partial p_2} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_2 \partial p_1} + \frac{\partial x_1}{\partial p_2}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_2 \partial p_1} + \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1}##
    setting them equal and cancelling:
    ## \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_2 \partial p_1} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_2 \partial p_1} + \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1} = \frac{\partial x}{\partial x_1}\frac{\partial^2 x_1}{\partial p_1 \partial p_2} + \frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_2} + \frac{\partial x}{\partial x_2}\frac{\partial^2 x_2}{\partial p_1 \partial p_2} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} ##
    Should end up with
    ##\frac{\partial x_1}{\partial p_1}\frac{\partial^2 x}{\partial x_1 \partial p_1} + \frac{\partial x_2}{\partial p_1}\frac{\partial^2 x}{\partial x_2 \partial p_2} = \frac{\partial x_2}{\partial p_2}\frac{\partial^2 x}{\partial x_2 \partial p_1} + \frac{\partial x_1}{\partial p_2}\frac{\partial^2 x}{\partial x_1 \partial p_1}##

    Essentially, you need ##\frac{\partial}{\partial p_2} \frac{\partial x}{\partial x_1} = 0## and ##\frac{\partial}{\partial p_1} \frac{\partial x}{\partial x_2} = 0##
    I don't see how you can get here with this information.

    P.S.
    Ctrl+v & Ctrl+c made this post possible...
     
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