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

eprparadox
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Homework Statement


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

<br /> (\frac{\partial y}{\partial x})_z <br />


Homework Equations


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

The Attempt at a Solution



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

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
<br /> (\frac{\partial y}{\partial x})_z = -\frac{\frac{\partial f}{\partial x} }{\frac{\partial f}{\partial y} }<br />

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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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