# Messy partial differentials with chain rule.

1. Nov 1, 2007

### phewy13

1. The problem statement, all variables and given/known data
the problem asks: Find $$\delta$$f/$$\delta$$x and $$\delta$$f/$$\delta$$y at x=1 and y=2 if z=f(x,y) is defined implicitly by 2x$$^{}2$$y/z + 3z/xy - xy$$\sqrt{}z$$ = 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 $$\delta$$f/$$\delta$$x is equal to $$\delta$$z/$$\delta$$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. Nov 1, 2007

### hotcommodity

You're right, $$\frac{\partial f}{\partial x}$$ is the same as $$\frac{\partial z}{\partial x}$$.

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 $$\frac{\partial z}{\partial x}$$ next to it.

For example, if I have:

$$x^{2}z + y + z^2 = 10$$

I differentiate both sides with respect to x:

$$x^{2}\frac{\partial z}{\partial x} + 2xz + \frac{\partial z}{\partial x}2z = 0$$

Get all the $$\frac{\partial z}{\partial x}$$ terms on one side:

$$\frac{\partial z}{\partial x} = \frac{-2xz}{x^{2} + 2z}$$

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