I ran across the following in a book, but I don't quite understand why we have fx+p fz=0 (and similarly with y). A little guidance would be greatly appreciated.

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- Jan 19th 2013, 01:39 PMHal2001Simple PDE Question
I ran across the following in a book, but I don't quite understand why we have fx+p fz=0 (and similarly with y). A little guidance would be greatly appreciated.

Attachment 26615 - Jan 19th 2013, 06:18 PMTheEmptySetRe: Simple PDE Question
The assumptions is that z is a function of both x and y, that is

$\displaystyle z(x,y) \implies f(x,y,z(x,y),a,b)=0$

Now just use the chain rule

$\displaystyle \frac{\partial }{\partial }f(x,y,z(x,y),a,b)= \frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial f}{\partial b}\frac{\partial b}{\partial x}=0 $

This reduces to

$\displaystyle \frac{\partial }{\partial x}f(x,y,z(x,y),a,b)= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} =0$ - Jan 19th 2013, 06:26 PMHal2001Re: Simple PDE Question
:) got it thanks. Didn't realize at the time that they were defining z as a function of x and y. But it makes sense now.