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Messy partial differentials with chain rule.

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data
    the problem asks: Find [tex]\delta[/tex]f/[tex]\delta[/tex]x and [tex]\delta[/tex]f/[tex]\delta[/tex]y at x=1 and y=2 if z=f(x,y) is defined implicitly by 2x[tex]^{}2[/tex]y/z + 3z/xy - xy[tex]\sqrt{}z[/tex] = 3. Note that (1,2,4) is a point on the surface.


    2. Relevant equations
    Im not really sure how to approach this one.


    3. The attempt at a solution

    i started off by saying that [tex]\delta[/tex]f/[tex]\delta[/tex]x is equal to [tex]\delta[/tex]z/[tex]\delta[/tex]x and the same thing for y. i went through and found the partial derivatives of the above equation and it turned out really messy, any help would be greatly appreciated.
     
  2. jcsd
  3. Nov 1, 2007 #2
    You're right, [tex] \frac{\partial f}{\partial x}[/tex] is the same as [tex] \frac{\partial z}{\partial x} [/tex].

    First we want to find the partial derivative with respect to x. Remember in calculus 1 when we differentiated implicitly, we would place a y', or a dy/dx, or something to denote the derivative of y, everytime we had to take the derivative of a term containing y. We pretty much do the same thing when we take the partial derivative of a function of two or more variables. Everytime you take the derivative of a term with z in it, simply take the derivative and place a [tex] \frac{\partial z}{\partial x} [/tex] next to it.

    For example, if I have:

    [tex] x^{2}z + y + z^2 = 10 [/tex]

    I differentiate both sides with respect to x:

    [tex] x^{2}\frac{\partial z}{\partial x} + 2xz + \frac{\partial z}{\partial x}2z = 0 [/tex]

    Get all the [tex]\frac{\partial z}{\partial x}[/tex] terms on one side:

    [tex]\frac{\partial z}{\partial x} = \frac{-2xz}{x^{2} + 2z}[/tex]

    And now you can plug in values. Does this help?
     
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