Higher-order partial differentiation

Given the function $z=f(x,y)$, suppose that the partial derivatives $f_x(x,y)$ and $f_y(x,y)$ exist, and they also have partial derivatives. For example, the partial derivative of $f_x(x,y)$ with respect to $y$,

$$(f_x{)}_y(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)(x,y),$$

is denoted as $\frac{{\partial }^{2}f}{\partial y\partial x}(x,y)$ or $f_{xy}(x,y)$. Similarly, we may define $f_{xx}(x,y)$, $f_{yx}(x,y)$, $f_{yy}(x,y)$. These are called second partial derivatives.

Example. Let $f(x,y)=\log (x^2+xy+2y^2)$. Then,

$$\begin{array}{rcl}{f}_x(x,y)& =& \frac{2x+y}{x^2+xy+2y^2},\\ f_y(x,y)& =& \frac{x+4y}{x^2+xy+2y^2},\\ f_{xx}(x,y)& =& -\frac{2x^2+2xy-3y^2}{(x^2+xy+2y^2{)}^2},\\ f_{xy}(x,y)& =& -\frac{x^2+8xy+2y^2}{(x^2+xy+2y^2{)}^2},\\ f_{yx}(x,y)& =& -\frac{x^2+8xy+2y^2}{(x^2+xy+2y^2{)}^2},\\ f_{yy}(x,y)& =& \frac{3x^2-4xy-8y^2}{(x^2+xy+2y^2{)}^2}.\end{array}$$ □

In the above example, we have $f_{xy}(x,y)=f_{yx}(x,y)$. Is this generally the case? Short answer: Not always. In general, we have the following result.

Theorem (Changing the order of partial differentiation)

Suppose the function $f(x,y)$ on an open region $U$ has second derivatives $f_{xy}(x,y)$ and $f_{yx}(x,y)$ both of which are continuous. Then $f_{xy}(x,y)=f_{yx}(x,y)$.

Proof. We show that $f_{xy}(a,b)=f_{yx}(a,b)$ for every $(a,b)\in U$. 

Take a sufficiently small $\delta >0$ so that $|x-a|<\delta$ and $|y-b|<\delta$ imply $(x,y)\in U$ (Note: This is always possible because $U$ is open). For all $h,k$ such that $0<|h|<\delta$ and $0<|k|<\delta$, let

$$F(h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b).$$

Define the univariate function $u(y)$ of $y$ by

$$u(y)=f(a+h,y)-f(a,y).$$

Then we can write

$$F(h,k)=u(b+k)-u(b).$$

$u(y)$ is differentiable with respect to $y$ and $u^{\prime }(y)=f_y(a+h,y)-f_y(a,y)$. By the Mean Value Theorem, there exists $\theta \in (0,1)$ such that

$$F(h,k)=u^{\prime }(b+\theta k)k=k[f_y(a+h,b+\theta k)-f_y(a,b+\theta k)].$$

  Next, consider the univariate function of $x$: $f_y(x,b+\theta k)$. Again, by the Mean Value Theorem, there exists $\eta \in (0,1)$ such that

$$\begin{array}{}\text{(Eq:Fhk1)}& F(h,k)=hk{f}_{yx}(a+\eta h,b+\theta k).\end{array}$$

If we repeat the above process by swapping the roles of $x$ and $y$, we can show (exercise!) that there exist ${\theta }^{\prime },{\eta }^{\prime }\in (0,1)$ such that

$$\begin{array}{}\text{(Eq:Fhk2)}& F(h,k)=hk{f}_{xy}(a+{\eta }^{\prime }h,b+{\theta }^{\prime }k).\end{array}$$

From Eqs. (Eq:Fhk1) and (Eq:Fhk2), it follows that

$$f_{yx}(a+\eta h,b+\theta k)=f_{xy}(a+{\eta }^{\prime }h,b+{\theta }^{\prime }k).$$

By the continuity of $f_{xy}(x,y)$ and $f_{yx}(x,y)$, as $(h,k)\to (0,0)$,

$$f_{yx}(a,b)=f_{xy}(a,b).$$ ■

The second derivatives of the function $f(x,y)$ may have derivatives. For example,

$$f_{yxy}(x,y)=\frac{\partial }{\partial y}{f}_{yx}(x,y).$$

If $f_{xy}(x,y)$ and $f_{yx}(x,y)$ are continuous, by the above theorem, $f_{xy}(x,y)=f_{yx}(x,y)$. It follows that $f_{xyy}(x,y)=f_{yxy}(x,y)$, that is,

$$\frac{{\partial }^{3}f}{\partial y^2\partial x}=\frac{{\partial }^{3}f}{\partial y\partial x\partial y}.$$

These are called third derivatives. 

Remark. We usually write

$$\frac{{\partial }^{2}f}{\partial x^2}(x,y)$$

for $f_{xx}(x,y)$, rather than

$$\frac{{\partial }^{2}f}{\partial x\partial x}(x,y).$$

□

Definition (Functions of class $C^n$)

Let $f(x,y)$ be a function on an open region $U$ and $n$ be a non-negative integer.

1. $f(x,y)$ is said to be $n$-times continuously differentiable or of class $C^n$ if it has all the partial derivatives up to the $n$-th order on $U$, all of which are continuous.
2. $f(x,y)$ is said to be infinitely differentiable, smooth, or of class $C^{\mathrm{\infty }}$ if it has the partial derivatives of all orders on $U$, all of which are continuous.

In general, the derivatives of a function of class $C^n$ are determined by the number of partial differentiation with respect to $x$ and $y$ and are independent of the order of differentiation.

Example. If $f(x,y)$ is of class $C^3$, then $f_{xxy}(x,y)=f_{xyx}(x,y)=f_{yxx}(x,y)$.