What is the formula for finding a partial derivative with constant z?

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eprparadox
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Homework Statement


Given f(x, y, z) = 0, find the formula for

[tex] (\frac{\partial y}{\partial x})_z [/tex]


Homework Equations


Given a function f(x, y, z), the differential of f is
[tex] df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz[/tex]

The Attempt at a Solution



We know that f(x, y, z) = 0 so using above, I get
[tex] df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz<br /> = 0[/tex]

We also know that we are finding the partial with constant z so I set dz = 0. I then divided by dx throughout and solve for [itex]\frac{\partial y}{\partial x}[/itex].

My final answer is
[tex] (\frac{\partial y}{\partial x})_z = -\frac{\frac{\partial f}{\partial x} }{\frac{\partial f}{\partial y} }[/tex]

I just wanted to confirm that I'm doing things correctly in finding this partial derivative.

Thanks!
 
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You are right.
I am just a little bit puzzled about the notations.
How to you switch from the ratio dy/dx to the partial derivative.
This should maybe be explained more explicitly.
 
Hey, thanks for the quick response!

That is a good point about the notation. Do you have any ideas on this? I don't have any good mathematical sense as to why I changed it from dy/dx to a partial derivative.
 
You are correct. As was alluded to in your other recent thread on a similar problem, a rigorous mathematical justification is the implicit function theorem. You write ##y = y(x,z)##, provided ##\frac{\partial f}{\partial y} \neq 0## and then differentiate both sides of ##f(x,y,z) = 0## with respect to ##x##.
 
Although, I suppose if you're given f(x, y, z) = 0, then the equation can be solved for y in terms of x and z in which case the partial derivative notation makes sense. Is that a reasonable explanation?