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

[eprparadox]

Homework Statement

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

$$(\frac{\partial y}{\partial x})_z$$

Homework Equations

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

The Attempt at a Solution

We know that f(x, y, z) = 0 so using above, I get
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = 0$$

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 $\frac{\partial y}{\partial x}$.

My final answer is
$$(\frac{\partial y}{\partial x})_z = -\frac{\frac{\partial f}{\partial x} }{\frac{\partial f}{\partial y} }$$

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

Thanks!

 

[maajdl]

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.

 

[eprparadox]

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.

 

[Quesadilla]

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 $x$.

 

[eprparadox]

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?