Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Today, in my lesson, I was introduced to partial derivatives. One of the things that confuses me is the notation. I hope that I am wrong and hope the community can contribute to my learning. In single-variable calculus, we know that, given a function $y =f(x)$, the derivative of $y$ is denoted as $\frac {dy}{dx}$. I understand this as the relative change in $y$, $\delta y$ given a small change in $x$, $\delta x$.

However, in today's lesson on partial derivative, my professor constantly used this notation.

Given a function $z = f(x,y)$, the first derivative with respected to $x$ is written as

$$ \frac{\partial z}{\partial x} $$

So, for example

$$ z = 5x+3y\\ \frac{\partial z}{\partial x} = 5 $$

Why can't I just write it as $$ z = 5x+3y\\ \frac{d z}{d x} = 5 $$

Is it some convention or am I not understanding something in the notation?

share|improve this question
    
I'd say it might be because normal derivatives allow for blind cancellation, whereas partial derivatives can't be cancelled so blindly when chained together. –  Mehrdad Mar 4 '13 at 6:49
1  
Here $\frac{\partial z}{\partial x} = 5 \ne 5 + 3y$. –  Andrew Salmon Mar 4 '13 at 6:50
1  
edited, thanks for spotting the error! I guess fatigue's really kicking in –  bryansis2010 Mar 4 '13 at 6:51

2 Answers 2

up vote 4 down vote accepted

First, rest assured that you're not the only one who's confused by the standard notation for partial derivatives. See this answer for a collection of answers I've written in response to such confusions.

The problem is that the standard notation doesn't indicate which variables are being held constant. It assumes that you've defined a function of a certain set of variables, and that everyone remembers what these are. That's fine if you only introduce a single function and write its partial derivatives as

$$ \frac{\partial f(x,y,z)}{\partial x} $$

and the like, since the arguments for the function evaluation make up for what the notation for the partial derivative is missing, but it becomes a problem when you start writing things like

$$ \frac{\partial f}{\partial x} $$

and especially when you have lots of things like $x,y,z$ floating around that all look like variables and the notation doesn't contain the slightest clue which of these are being treated as functions and which as independent variables being held constant.

In a certain sense, you're right that you could always regard $\dfrac{\partial f}{\partial x}$ as $\dfrac{\mathrm df}{\mathrm dx}$ of a univariate function, namely by regarding all other variables as parameters. That is, given a function $f(x,y)$ of two variables, you can regard $y$ as a fixed parameter and write $g(x)=f(x,y)$, and then $\dfrac{\mathrm dg}{\mathrm dx}=\dfrac{\partial f}{\partial x}$. Then if you feel things aren't quite confusing enough already, you can instead call this new univariate function by the same name as the multivariate function $f$ and write $\vphantom{\dfrac{\partial f}{\partial x}}f(x)=f(x,y)$, and then indeed $\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\partial f}{\partial x}$, but you need to remember what you mean by that: $\dfrac{\mathrm df(x)}{\mathrm dx}=\dfrac{\partial f(x,y)}{\partial x}$, with two different uses of the symbol $f$.

However, this view is rarely very helpful, since the variables of a multivariate function are usually variables on an equal footing for good reason, and one would usually have introduced them as fixed parameters in the first place if that were the natural way to think of them. So usually it's better to view the univariate function that you get by keeping all but one variable fixed as a more temporary construct that's used only for defining and thinking about the partial derivative, but not as something that should appear in the notation as a univariate function in its own right.

share|improve this answer
    
Thanks! both answers are very comprehensive. I think I will spend a few days to digest it and finally apply it. –  bryansis2010 Mar 4 '13 at 9:14

A nicer notion is that of the differential:

$$ \text{If} \qquad z = 5x + 3y \qquad \text{then} \qquad dz = 5\, dx + 3\,dy $$

Then if you decide to hold $y$ constant, that makes $dy = 0$, and you have $dz = 5 \, dx$.


Another notation that works well with function notation is that if we define

$$ f(x,y) = 5x + 3y$$

then $f_i$ means derivative of $f$ with respect to the $i$-th entry; that is

$$ f_1(x,y) = 5 \qquad \qquad f_2(x,y) = 3 $$

This doesn't work well with a common abuse of notation, though; sometimes people write $f(r,\theta)$ when they really mean "evaluate $f$ at the $(x,y)$ pair whose polar coordinates are $(r, \theta)$" rather than the 'correct' meaning of that expression "evaluate $f$ at $(r, \theta)$". So if you're in the habit of doing that, don't try to indicate derivatives by their position.


I confess I really dislike partial derivative notation; when one writes $\partial/\partial x$, one "secretly" means that they intend to hold $y$ constant, then when one passes it through the differential, one gets

$$ \frac{\partial z}{\partial x} = 5 \frac{\partial x}{\partial x} + 3 \frac{\partial y}{\partial x} = 5 \cdot 1 + 3 \cdot 0 = 5$$

However, the suggestive form of Leibniz notation starts becoming very misleading at this point; for example, let's compute other partial derivatives.

  • $\partial z / \partial x = 5$, holding $y$ constant as the notation suggests
  • $\partial x / \partial y = -3/5$, holding $z$ constant as the notation suggests
  • $\partial y / \partial z = 1/3$, holding $x$ constant as the notation suggests

Then putting it together,

$$ \frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = 5 \cdot \left(-\frac{3}{5}\right) \cdot \frac{1}{3} = -1 $$

This is a big surprise if you expect partial derivatives to behave similarly to fractions as their notation suggests!!!

share|improve this answer
    
Thank you! both of your answers have addressed my problem. Anyway, to your 2nd part, is it some sort of Young's Theorem playing out? Coz it seems so to me. –  bryansis2010 Mar 4 '13 at 9:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.