The Unapologetic Mathematician

Partial Derivatives

Okay, we want to move towards some analogue of the derivative of a function that applies to functions of more than one variable. For the moment we’ll stick to single real outputs. As a goal, we want “differentiability” to be a refinement of the idea of smoothness started with “continuity“, so an important check is that it’s a stronger condition. That is, a differentiable function should be continuous.

For functions with a single real input we defined the derivative of the function $f$ at the point $a$ by the limit of the difference quotient

$\displaystyle f'(a)=\lim\limits_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

The problem here is that for vector inputs we can’t “divide” by the vector $x-a$. So we need some other way around this problem.

Our first attempt may be familiar from calculus classes: we’ll just look at one variable at a time. That is, if we have a function of $n$ real variables and we keep all of them fixed except the $i$th one, we can try to take the limit

$\displaystyle f_{i}(a_1,\dots,a_n)=\lim\limits_{x_i\rightarrow a_i}\frac{f(a_1,\dots,x_i,\dots,a_n)-f(a_1,\dots,a_i,\dots,a_n)}{x_i-a_i}$

That is, we fix down the values of all the other variables and get a function of the single remaining variable. We then take the single-variable derivative as normal.

The first problem here is that it having these partial derivatives — even having a partial derivative for each variable — doesn’t make a function continuous. Let’s look at the first pathological example of a limit we discussed:

$\displaystyle f(x,y)=\frac{x^2-y^2}{x^2+y^2}$

If we consider the point $(0,0)$, we can calculate both partial derivatives here. First we fix $y=0$ and find $f(x,0)=\frac{x^2}{x^2}=1$. Thus it’s easy to check that $f_1(0,0)=0$. Similarly, we can fix $x=0$ to find $f(0,y)=\frac{-y^2}{y^2}=-1$, and thus that $f_2(0,0)=0$. So both partial derivatives exist at $(0,0)$, but the function doesn’t even have a limit there, much less one which equals its value.

The problem is the same one we saw in the case of multivariable limits: we can’t take a limit as one input point approaches another along a single path and just blithely expect that it’s going to mean anything. Here we’re just picking out two paths towards the same point and establishing that the function is continuous when we restrict to those paths, which doesn’t establish continuity in general.

There’s a deeper problem with partial derivatives, though. Implicit in the whole set-up is choosing a basis of our space. To write $f$ as a function of $n$ real variables instead of one $n$-dimensional vector variable means picking a basis. In practice we often have no problem with this. Indeed, many problems come to us in terms of a collection of variables which we bind together to make a single vector variable. But in principle, anything with any geometric meaning should be independent of artificial choices of coordinates. We can’t even talk about partial derivatives without making such a choice, and so they clearly don’t get to the heart of any sensible notion of “differentiability”.

September 21, 2009 - Posted by | Analysis, Calculus

1. “We can’t even talk about partial derivatives without making such a choice, and so they clearly don’t get to the heart of any sensible notion of “differentiability”.”

Well now I want to know how we can talk about differentiability?

Comment by David | September 21, 2009 | Reply

2. Patience, grasshopper. We’ve got one more blind alley before we get to the right answer.

Comment by John Armstrong | September 21, 2009 | Reply

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Pingback by Uniqueness of the Differential « The Unapologetic Mathematician | September 29, 2009 | Reply

7. So according to the definition of partial derivative you need a value for f(0,0) to compute its partial derivatives at (0,0). But you do not have any such value.

Comment by Johan Richter | September 29, 2009 | Reply

8. Sorry, Johan. Easiest way around is to patch it at that point by defining $f(0,0)=0$. Then the partial derivatives exist, but the limit doesn’t. I did something similar when talking about directional derivatives.

Comment by John Armstrong | September 29, 2009 | Reply

9. With that definition I do not think either limit defining the partial derivatives exist.

For the x-derivative for example I get:

$\lim_{x \rightarrow 0} \frac{1}{x}$

which does not exist.

Comment by Johan Richter | September 29, 2009 | Reply

10. Okay, Johan, here’s one that’s a lot more hacky but should finally settle this: let $f(x,y)=x+y$ if either $x$ or $y$ are $0$, and $f(x,y)=1$ otherwise. Then both partials exist but the limit doesn’t.

Or go back to the original function and use the definition of the derivative that takes a sample point approaching the given point from either side, which is a common alternative to handle the nonexistence of a function at the point in question

$\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$

Enough.

Comment by John Armstrong | September 29, 2009 | Reply

11. [...] we can say “the” differential), we showed that given an orthonormal basis we have all partial derivatives. We even have all directional derivatives, with pretty much the same proof. We replace with an [...]

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