Partial derivatives of higher order.

If the partial derivatives of f are differentiable, we get the second order derivatives of f, and so on. We have:
$$\begin{align*}\frac {\partial^2 f}{\partial x^2} & = \frac {\partial}{\partial x... ...partial}{\partial y} \left( \frac {\partial f}{\partial x} \right) \end{align*}$$

Example 5.13   Let $f(x,y)=x^2y-y^2 \cos x$. Then:
$$\begin{align*}\frac {\partial f}{\partial x} &= 2xy \sin x \\ \frac {\partial f... ...{\partial y \partial x} & = \frac {\partial}{\partial y} (2xy) =2x \end{align*}$$

Notations: $f_{xx} = \frac {\partial^2 f}{\partial x^2}$, $f_{yx}= \frac {\partial^2 f}{\partial y \partial x}$, $f_{xy}= \frac {\partial^2 f}{\partial x \partial y}$and $f_{yy} = \frac {\partial^2 f}{\partial y^2}$.

Theorem 5.14   Let f(x,y) be a function of two variables having partial derivatives f_x, f_y, f_xy and f_yx on an open domain $\mathcal{D}$in $\mathbb{R} ^2$. If $(x_0,y_0) \in \mathcal{D}$and if the partial derivatives are all continuous at (x₀,y₀), then

f_xy(x₀,y₀)=f_yx(x₀,y₀)

As a first example, see Ex.  5.13.

Example 5.15   Let $f(x,y)= xy+ \arctan \frac {y^2}{1+y^2}$. Then:
$$\begin{align*}\frac {\partial f}{\partial x} &= y \\ \frac {\partial^2 f}{\partial y \partial x} &= 1 \end{align*}$$
The computation of the mixed second derivative, in reversed order, is much more complicated:
$$\begin{align*}\frac {\partial f}{\partial y} &= x + \frac {2y}{2y^4+2y^2+1} \\ \frac {\partial^2 f}{\partial x \partial y} &= 1 \end{align*}$$

We can iterate the above description and define the partial derives of order higher than 2, namely $f_{xxx}= \frac {\partial^3 f}{\partial x^3}$, $f_{xxy}=\frac {\partial^3 f}{\partial^2 x \partial y}$, $f_{xyx}=\frac {\partial^3 f}{\partial x \partial y \partial x}$, and so on.