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Partial Derivatives

Date: 03/24/2001 at 21:54:33
From: Neeraj
Subject: Partial derivatives meaningless?


We're currently studying partial derivatives. I am unable to determine 
what sense partial derivatives make in the following case: u is given 
as a function of two variables, say x and y. Also, y is itself a 
function of x.

What does the partial derivative D_u/D_y mean? According to my 
reasoning, it is meaningless: To find u/y, we keep the other variables 
of u constant. Here, we have only one other variable, namely x. So, I 
keep x constant. But if x is constant, then y is constant (as y 
depends on x). But if y is constant then what sense does D_u/D_y  
make? After all, D_u/D_y determines the change in u when y is changed.

Please help.

Yours sincerely, 

Date: 03/24/2001 at 22:36:07
From: Doctor Peterson
Subject: Re: Partial derivatives meaningless?

Hi, Neeraj.

In order to understand this, you have to focus on one thing at a time. 

You have been given a function u(x,y), and also a function y = f(x). 
These are separate things. When you take the partial derivatives of u, 
the function f is irrelevant; you are only considering how u depends 
on x and y, not how y may depend on x. As far as u is concerned, x and 
y could have any relation, or none. They are simply independent 
variables when we look at u.

You might want to think of it this way. The equation z = u(x,y) 
defines a surface. The function y = f(x) defines a curve on the x-y 
plane, or a cylinder (in the generalized sense) in space, parallel to 
the z axis. Putting the two together, z = u(x, f(x)) and y = f(x) 
together determine a curve in space, where the cylinder cuts the 
surface. But even if we know that f exists, we can still think about u 
as if x and y were entirely independent, and that is what we do 
whenever we work with the function u alone.

This does suggest some cautions when you work with such situations. 
Suppose z = u(x,y) = x^2 + y^2, and y = f(x) = 3x. Then, ignoring f, 
we can say that D_u/D_y = 2y; and we can say that when x = 2, this is 
2f(2) = 12. But we can't substitute y = f(x) for y and say

    z = x^2 + (3x)^2 = 10x^2

and THEN ask for the partial derivative of z with respect to y. Once 
we have combined the functions, we no longer have any dependence on y; 
we have eliminated that variable. We can only talk about the partial 
derivative of the function u, not of the variable z in the whole 
context. As often happens in calculus, we must keep in mind what 
function and what variables are under consideration.

- Doctor Peterson, The Math Forum   
Associated Topics:
College Calculus
High School Calculus

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