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Chain Rule and Partials

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that if
    [tex]z(x,y)=e^y f(ye^{\frac{x^2}{2y^2}}) [/tex]
    is differentiable, then
    [tex] (x^2-y^2) \frac{\partial z}{\partial x} + xy\frac{\partial z}{\partial y} = xyz [/tex]


    2. Relevant equations

    Chain Rule.

    3. The attempt at a solution
    A similar question is solved like this:
    Have this:
    z(x,y) = f(x² - y²)

    Prove this:
    y(∂z/∂x) + x(∂z/∂y) = 0

    I define
    z = f(u)
    then
    u = x² - y²
    ∂z/∂x = ∂z/∂u ∂u/∂x = ∂z/∂u 2x
    ∂z/∂y = ∂z/∂u ∂u/∂y = ∂z/∂u -2y
    adding it up:
    y(∂z/∂x) + x(∂z/∂y) = y*∂z/∂u*2x - x*∂z/∂u*2y = 0

    When I try the same for the mentioned problem, I run into problems and just can't get it right. I guess it's something with the multipilication in there - but can't figure it out.
    What's the correct way?
     
    Last edited: Sep 4, 2009
  2. jcsd
  3. Sep 4, 2009 #2

    Office_Shredder

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    Gold Member

    So for example, to start off
    [tex] \frac{\partial z}{\partial y} = \frac{\partial (e^y)}{\partial y} f(ye^{\frac{x^2}{2y^2}}) + e^y \frac{\partial f(ye^{\frac{x^2}{2y^2}})}{\partial y}[/tex]

    Why don't you work through the two partial derivatives and let us know what you get?
     
  4. Sep 4, 2009 #3
    ok so we got:
    [tex]\frac{\partial z}{\partial y} = \frac{\partial (e^y)}{\partial y} f(ye^{\frac{x^2}{2y^2}}) + e^y \frac{\partial f(ye^{\frac{x^2}{2y^2}})}{\partial y}[/tex]
    for shortness:
    [tex]f(ye^{\frac{x^2}{2y^2}}) = f(u) = f[/tex]
    define:
    [tex]u = ye^{\frac{x^2}{2y^2}}
    [/tex]
    let's do the derivative:
    [tex]\frac{\partial z}{\partial y} = e^yf + e^y\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}[/tex]
    noticing that eyf is the same as z:
    [tex]\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}[/tex]
    and doing the derivative:
    [tex]\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}+y\frac{x^2}{2}\frac{-2}{y^3}e^{\frac{x^2}{2y^2}})
    [/tex]
    let's make it look better:
    [tex]\frac{\partial z}{\partial y} = z + e^y\frac{\partial f}{\partial u}(e^{\frac{x^2}{2y^2}}-\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}) = z + \frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y}(1-\frac{x^2}{y^2})
    [/tex]
    I'll make this substitution for readbility:
    [tex]
    \frac{\partial f}{\partial u}e^{\frac{x^2}{2y^2}+y} = a
    [/tex]
    so I end up with:
    [tex]\frac{\partial z}{\partial y} = z + a(1-\frac{x^2}{y^2})[/tex]

    doing for the other derivative now:
    [tex]\frac{\partial z}{\partial x} = a\frac{x}{y}[/tex]
    now the proof:
    [tex] (x^2-y^2)a\frac{x}{y} + xy(z+a-a\frac{x^2}{y^2} = a\frac{x^3}{y}-axy+xya-a\frac{x^3}{y} + xyz = xyz[/tex]

    Problem solved. Thanks a lot!
    Was also using this for a bit of tex practice.
    Looks so easy now.
     
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