Messy partial differentials with chain rule.

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phewy13
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Homework Statement


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.


Homework Equations


Im not really sure how to approach this one.


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.
 
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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?