The Unapologetic Mathematician

Mathematics for the interested outsider

Higher Partial Derivatives

Let’s say we’ve got a function f that’s differentiable within an open region S\subseteq\mathbb{R}^n. In particular, if we pick coordinates on \mathbb{R}^n the function has all partial derivatives \frac{\partial f}{\partial x^i} at each point in S. As we move around within S the value of the partial derivative changes, justifying the functional notation \left[D_{x^i}f\right](x). And if we’re lucky, these functions themselves may be differentiable.

In particular, it makes sense to ask about the existence of so-called “second partial derivatives”, defined as

\displaystyle\left[D_{x^i,x^j}f\right](x)=\left[D_{x^i}\left[D_{x^j}f\right]\right](x)

Or in Leibniz’ notation:

\displaystyle\frac{\partial^2f}{\partial x^i\partial x^j}=\frac{\partial}{\partial x^i}\frac{\partial f}{\partial x^j}=\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}f

If we take the derivative in terms of the same variable twice in a row we sometimes write this as

\displaystyle\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^i}=\frac{\partial^2}{(\partial x^i)^2}

Yes, there’s some dissonance between superscripts as indices and superscripts as powers. But, again, this is pretty much the received notation in many areas. If it seems like it might be confusing we just write out \partial x^i twice in a row.

These, of course, may be defined within the region S, and we can then sensibly ask about third partial derivatives, like

\displaystyle\frac{\partial^3}{\partial x^i\partial x^j\partial x^k}f=\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}\frac{\partial}{\partial x^k}f

and so on.

As an example, let’s consider the function f(x,y)=x^3 - 3xy^2. We can easily calculate the two first partial derivatives.

\displaystyle\begin{aligned}\frac{\partial f}{\partial x}&=3x^2-3y^2\\\frac{\partial f}{\partial y}&=-6xy\end{aligned}

And then we take each derivative of each of these two

\displaystyle\begin{aligned}\frac{\partial^2f}{\partial x^2}&=\frac{\partial}{\partial x}\left(3x^2-3y^2\right)=6x\\\frac{\partial^2f}{\partial y\partial x}&=\frac{\partial}{\partial y}\left(3x^2-3y^2\right)=-6y\\\frac{\partial^2f}{\partial x\partial y}&=\frac{\partial}{\partial x}\left(-6xy\right)=-6y\\\frac{\partial^2f}{\partial y^2}&=\frac{\partial}{\partial y}\left(-6xy\right)=-6x\end{aligned}

where since we’re not using superscripts as indices in these examples its meaning should be clear.

We notice here that the two in the middle — the “mixed” partial derivatives — are the same. This will happen in many cases of interest to us, but not always. As a pathological example, let’s go back and consider the function defined by

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

away from the origin, and patched by f(0,0)=0. Again, we calculate the first partial derivatives (at least away from the origin):

\displaystyle\begin{aligned}\frac{\partial f}{\partial x}&=\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\\\frac{\partial f}{\partial y}&=\frac{x(x^4-4x^2y^2-y^4)}{(x^2+y^2)^2}\end{aligned}

Each partial derivative is {0} at the origin.

Now we can check that \left[D_xf\right](0,y)=-y for all y, and that \left[D_yf\right](x,0)=-x for all x. Thus we can calculate

\displaystyle\begin{aligned}\frac{\partial^2f}{\partial x\partial y}\biggr\vert_{(0,0)}&=1\\\frac{\partial^2f}{\partial y\partial x}\biggr\vert_{(0,0)}&=-1\end{aligned}

and the mixed partial derivatives are not equal.

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October 14, 2009 - Posted by | Analysis, Calculus

5 Comments »

  1. In Leibniz’ notation, it would look better in the background of an episode of NUMB3RS. They normally use read grad students and postdocs to write on the blackboards, normally on the first floor of the Sloan Laboratory (Mathematics & Physics), Caltech. Laboratory. Starting in 1968, I always thought of someone holding a test-tube full of equations up from the Bunsen burner and yelling: εύρηκα

    Comment by Jonathan Vos Post | October 14, 2009 | Reply

  2. I wouldn’t know. The first season was so rife with implausibilities, stereotypes, and hackneyed “applications” clearly cooked up just to say that such a thing existed that I couldn’t bear watching more than the preview they sent to Yale’s department fishing for compliments.

    Comment by John Armstrong | October 14, 2009 | Reply

  3. [...] Now for the most common sufficient condition ensuring that mixed partial derivatives commute. If is a function of variables, we can for the moment hold the values of all but two of [...]

    Pingback by Clairaut’s Theorem « The Unapologetic Mathematician | October 15, 2009 | Reply

  4. [...] Just like we assembled partial derivatives into the differential of a function, so we can assemble higher partial derivatives into higher-order differentials. The differential measures how the function itself changes as we [...]

    Pingback by Higher-Order Differentials « The Unapologetic Mathematician | October 16, 2009 | Reply

  5. [...] on, we consider those functions which have all second partial derivatives, and that these second partials are themselves continuous at each point of . Clairaut’s [...]

    Pingback by Smoothness « The Unapologetic Mathematician | October 21, 2009 | Reply


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