Since $f_x$ and $f_y$ are functions of $x$ and $y$, we can take derivatives of these functions to get second derivatives. There are four such second derivatives, since each time we can differentiate with respect to $x$ or $y$.

The notation for second partial derivatives are as follows:
  • $\displaystyle (f_x)_x = f_{xx} = f_{11}= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial ^2 f}{\partial x^2} = \frac{\partial ^2 z}{\partial x^2}$
  • $\displaystyle (f_x)_y = f_{xy} = f_{12} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial ^2 f}{\partial y \partial x} = \frac{\partial ^2 z}{\partial y \partial x}$
  • $\displaystyle (f_y)_x = f_{yx} = f_{21} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial ^2 f}{\partial x \partial y} = \frac{\partial ^2 z}{\partial x \partial y}$
  • $\displaystyle (f_y)_y = f_{yy} = f_{22} =\frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial ^2 f}{\partial y^2} = \frac{\partial ^2 z}{\partial y^2}$
The notation for higher derivatives is similar. $f_{xxyx}$ is what you get by taking a partial derivative of $f$ with respect to $x$, then again with respect to $x$, then with respect to $y$, and finally with respect to $x$.


The main result about higher derivatives is:

Clairaut's Theorem: If $f_{xy}$ and $f_{yx}$ are both defined and continuous in a region containing the point $(a,b)$, then $f_{xy}(a,b)=f_{yx}(a,b)$.

This is often summarized as "Mixed partials are equal".


A consequence of this theorem is that we don't need to keep track of the order in which we take derivatives. We just need to keep track of how many times we differentiate with respect to each variable.

Example: Compute all the first and second-order partial derivatives of $f(x,y)=3x^2-4y^3-7x^2y^3$.

Solution: The first derivatives are $$f_x=6x-14xy^3 \quad \hbox{ and } \quad f_y=-12y^2-21x^2y^2.$$

We have four second derivatives, but as Clairaut's Theorem tells us, $f_{xy}=f_{yx}$, so we really only need to compute three of them.

(But we do all four to illustrate the theorem.)

\begin{eqnarray*} (f_x)_x & = & \frac{\partial}{\partial x}f_x=\frac{\partial}{\partial x}(6x-14xy^3)=6-14y^3\cr (f_x)_y&=&\frac{\partial}{\partial y}f_x=\frac{\partial}{\partial y}(6x-14xy^3)=0-42xy^2=-42xy^2\cr (f_y)_x&=&\frac{\partial}{\partial x}f_y=\frac{\partial}{\partial x}(12y^2-21x^2y^2)=0-42xy^2=-42xy^2\cr (f_y)_y&=&\frac{\partial}{\partial y}f_y=\frac{\partial}{\partial y}(12y^2-21x^2y^2)=24y-42x^2y. \end{eqnarray*}


Higher partial derivatives and Clairaut's theorem are explained in the following video.