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Implicit differentiation of many variables

  1. Dec 22, 2016 #1
    1. The problem statement, all variables and given/known data

    For the given function z to demonstrate the equality:

    10wl0qp.jpg



    As you see I show the solution provided by the book, but I have some questions on this.

    I don't understand how the partial derivative of z respect to x or y has been calculated.

    Do you think this is correct?

    I think this is a giant errata, I guess the function z is not given implicitly and it simply is:

    z = ln ( x ^2 + y^2)

    The partial derivatives are calculated normally:

    ∂z/∂x= 2 * x/(x^2 + y^2)

    Similar for y, and with this it is straighforward to demonstrate the equality.

    What do you think? There are two options:

    1.- Or the statement and solution of the given problem is correct---> In that case I don't understand anything. Could you explain how to get the partial derivatives?

    2.- Or there is a giant errata, z is not given implicitly and the calculation is easy.

    And forgeting this problem I was wondering in case I found an equation with z given implicitly like:

    z^2 = x * z + y * z^3

    How would we differenciate this equation?

    As we have many variables we should choose which are maintained constant and which are changing. Suppose we differenciate this expression considering x is changing, y is constant but z obviously changes, due to the changes in x.

    The receipt is changing x for x+dx, z changes to z+dz and y doesn't change at all. I get:

    (z+dz)2 - z2 = ( (x+dx) * (z+dz) + y * (z+dz)3 ) - (x * z + y * z3)

    After neglecting diferentials of order two and three I get:

    dz = dx * (z dx / 2x - x -3 y z^2)

    But this differential arose because there was a change on x, so I should call it dzx, then I should do the same calculation for dzy and the total differential of the function z should be:

    dz = dzx + dz y

    Is this correct?

     
    Last edited: Dec 22, 2016
  2. jcsd
  3. Dec 22, 2016 #2

    Mark44

    Staff: Mentor

    Is there information missing from the image above, especially in the upper right corner?
     
  4. Dec 22, 2016 #3

    vela

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This result doesn't match what's given in the solution, so why do you think your guess for ##z## is correct?

    The solution is definitely wrong for the given ##z##, but your guess is wrong too. It looks like the solution used ##z = \ln (x^2+xy+y^2)##.
     
  5. Dec 23, 2016 #4
    Thanks for your answer.

    But with my guess I find:

    ∂z/∂x= 2x/(x^2 + y^2)

    ∂z/∂y= 2y/(x^2 + y^2)

    So if I substitute in the equation I get:

    x * (2x/(x^2 + y^2)) + y * (2y/(x^2 + y^2)) = 2

    In the denominator we have the same functions, so we can simply sum the numerators to get:

    (2 x^2 + 2 y^2 )/ (x^2 + y^2) = 2

    And this equation is true. What am I doing wrong?

    No, there is nothing.
     
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