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Partial Differentiation Question

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    if [itex]z=\frac{1}{x^2+y^2-1} [/itex] . Show that [itex]x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = -2z(1+z) [/itex]

    2. Relevant equations
    n/a

    3. The attempt at a solution
    I am extremely new to partial differentiation, I can get my head around questions where they just give you the function and ask for the partial wrt to any of the variables but have not come accross a problem like this and am a little unsure of the notation, does [itex]x \frac{\partial z}{\partial x} [/itex] litterally mean x multipled with the partial wrt x ?

    First what I did was to find the partial derivatives the question asks for so done them below.
    [tex]
    \frac{\partial z}{\partial x} = \frac{-2}{x^3+y^2-1} \\
    \frac{\partial z}{\partial y} = \frac{-2}{x^2+y^3-1} \\
    [/tex]

    So if it just means x multiplied by.... then wouldn't it just become...
    [tex]
    x \frac{\partial z}{\partial x} = \frac{-2x}{x^3+y^2-1} \\
    y \frac{\partial z}{\partial y} = \frac{-2y}{x^2+y^3-1} \\
    [/tex]

    Or do I need to solve for x and y and substitute that into the above for x and y?
     
    Last edited: Nov 16, 2014
  2. jcsd
  3. Nov 16, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Yes, the notation ##x \frac{\partial z}{\partial x}## means multiply the partial derivative by ##x##. So you have the right idea but you haven't computed ##\partial z/\partial x## or ##\partial z/\partial y## correctly.
     
  4. Nov 16, 2014 #3

    Mark44

    Staff: Mentor

    Yes, that's all it means.
    Both of the above are wrong. It's easier to use the chain rule with z written as (x2 + y2 - 1)-1.
     
  5. Nov 16, 2014 #4

    Zondrina

    User Avatar
    Homework Helper

    The answer to your bolded concern is yes.

    Your derivatives look wrong though. Lets find ##\frac{\partial z}{\partial x} = z_x## first.

    I suggest you re-write ##z## like so:

    $$z = (x^2 + y^2 - 1)^{-1}$$

    Now what is ##z_x##? Don't forget the chain rule!
     
  6. Nov 16, 2014 #5
    Ah right ok thanks, will try again and post in a minute or so
     
  7. Nov 16, 2014 #6
    Ok so using the chain rule for [itex]\frac{\partial z}{\partial x}[/itex]

    if [itex]h=x^2+y^2-1[/itex] and [itex]m=h^{-1} [/itex]

    Then
    [tex]
    h' \cdot m' = 2x \cdot -h^{-2} = -2x(x^2+y^2-1)^{-2}
    [/tex]

    And for [itex]\frac{\partial z}{\partial y}[/itex]
    [tex]
    -2y(x^2+y^2-1)^{-2}
    [/tex]

    EDIT: Oh and is [itex]z_x[/itex] a short hand notation for [itex]\frac{\partial z}{\partial x}[/itex] then? As it does get a bit annoying writing them out all the time.
     
  8. Nov 16, 2014 #7

    Zondrina

    User Avatar
    Homework Helper

    Yes. Your derivatives look okay now too.
     
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