We consider the follwing example of the function of two variables

$$f(x,y)=\frac{x}{x+y}$$ The first partial derivatives can be calculatesd in this case by the quotient rule for differentiation and are (dothe calculations yourself!) $$\begin{aligned} \frac{\partial}{\partial x}\left(\frac{x}{x+y}\right) &= \frac{y}{(x+y)^2}\\ \\ \frac{\partial}{\partial y}\left(\frac{x}{x+y}\right) &= -\frac{x}{(x+y)^2}\end{aligned}$$ The partial derivatives are in itselves functions of the two variables $x$ and $y$, and this one can compute partial derivatives of these function again. They are called second partial derivatives.

For example, the partial derivatives of $\displaystyle\frac{\partial}{\partial x}f(x,y)$ are equal to

$$\begin{aligned}\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}f(x,y)\right) &= \frac{\partial}{\partial x}\left(\frac{y}{(x+y)^2}\right) = \frac{\partial}{\partial x}\bigl(y(x+y)^{-2}\bigr)\\ \\ &= -2y(x+y)^{-3}= -2\frac{y}{(x+y)^3} \\ \\ \frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}\! f(x,y)\right) &= \frac{\partial}{\partial y}\left(\frac{y}{(x+y)^2}\right) = \frac{\partial}{\partial y}\bigl(y(x+y)^{-2}\bigr) \\ \\ &=(x+y)^{-2}-2y(x+y)^{-3}\\ \\ &= \frac{1}{(x+y)^2} -\frac{2y}{(x+y)^3}\\ \\ &= \frac{x+y}{(x+y)^3} -\frac{2y}{(x+y)^3}=\frac{x-y}{(x+y)^3}\end{aligned}$$

The partial derivatives of $\displaystyle\frac{\partial}{\partial y}f(x,y)$ are equal (do it yourself after!)

$$\begin{aligned}\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}\! f(x,y)\right) &= \frac{x-y}{(x+y)^3} \\ \\ \frac{\partial}{\partial y}\left(\frac{\partial}{\partial y}f(x,y)\right) &=\frac{2x}{(x+y)^3}\end{aligned}$$

Property of 'mixed' second partial derivatives What is striking in the exmaple is that the 'mixed' derivatives $\displaystyle\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}f(x,y)\right)$ and $\displaystyle\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x} f(x,y)\right)$ are equal. That is no coincidence: this is always true if the first partial derivatives of $f$ exist and are continuous. In other words, in neat functions of two variables, the order in which 'mixed' second partial derivatives are calculated does not matter.

Notations for higher partial derivatives The following notations are used for second partial derivatives:

$$\begin{array}{ccccccc} \dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}\! f(x,y)\right) & = & \dfrac{\partial^2}{\partial x^2} f(x,y) & = & \dfrac{\partial^2f}{\partial x^2} (x,y) & = & f_{xx}(x,y) \\ \\ \dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial y} f(x,y)\right) & = & \dfrac{\partial^2}{\partial x\partial y} f(x,y) & = & \dfrac{\partial^2f}{\partial x\partial y} (x,y) & = & f_{xy}(x,y)\\ \\ \dfrac{\partial}{\partial y}\left(\dfrac{\partial}{\partial x} f(x,y)\right) & = & \dfrac{\partial^2}{\partial y\partial x} f(x,y) & = & \dfrac{\partial^2f}{\partial y\partial x}(x,y) & = & f_{yx}(x,y) \\ \\ \dfrac{\partial}{\partial y}\left(\dfrac{\partial}{\partial y} f(x,y)\right) & = & \dfrac{\partial^2}{\partial y^2} f(x,y) & = & \dfrac{\partial^2f}{\partial y^2} (x,y) & = & f_{yy}(x,y) \end{array}$$ Similar notation is used for higher partial derivatives and partial derivatives of functions of more than two variables.

Note: In most textbooks, the notation $f_{yx}$ for $\frac{\partial^2f}{\partial x\partial y} (x,y)$ is used to stress that you first differentiate w.r.t. $y$ and then w.r.t. $x$. Because we have $f_{xy}=f_{yx}$ for functions with continuous first partial derivatives, it does not matter much. But we prefer look at second partial derivatives as the composition of differential operator $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, and we abbreviate the notation for composition of functions.