Higher Order Partial Derivatives

# Higher Order Partial Derivatives

We are now going to look at partial differentiating partial derivatives of functions of several variables.

## Second Order Partial Derivatives

 Definition: Let $z = f(x, y)$ be a two variable real-valued function with partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. Then the Second Partial Derivatives of $f$ are the partial derivatives of the first partial derivatives of $f$.

We should note that a two variable real-valued function $z = f(x,y)$ has two first partial derivatives $f_x (x,y) = \frac{\partial z}{\partial x}$ and $f_y (x, y) = \frac{\partial z}{\partial y}$, each of which is a two variable real-valued function. Therefore, $f$ will have four second partial derivatives. Be sure to note the differences in notation between these second partial derivatives as order matters.

• The Partial Derivative of $f_x (x, y)$ with respect to $x$: $(f_x)_x = f_{xx} (x, y) = \frac{\partial}{\partial x} \left ( \frac{\partial z}{\partial x} \right ) = \frac{\partial ^2 z}{\partial x^2}$.
• The Partial Derivative of $f_x (x, y)$ with respect to $y$: $(f_x)_y = f_{xy} (x, y) = \frac{\partial}{\partial y} \left ( \frac{\partial z}{\partial x} \right ) = \frac{\partial ^2 z}{\partial y \partial x}$.
• The Partial Derivative of $f_y (x, y)$ with respect to $x$: $(f_y)_x = f_{yx} (x, y) = \frac{\partial}{\partial x} \left ( \frac{\partial z}{\partial y} \right ) = \frac{\partial ^2 z}{\partial x \partial y}$.
• The Partial Derivative of $f_y (x, y)$ with respect to $y$: $(f_y)_y = f_{yy} (x, y) = \frac{\partial}{\partial y} \left ( \frac{\partial z}{\partial y} \right ) = \frac{\partial ^2 z}{\partial y^2}$.

The notation replacing "$z$" with $f$" is also common for second partial derivatives.

 Definition: Let $z = f(x, y)$ be a two variable real-valued function. Then $\frac{\partial ^2 z}{\partial x^2}$ and $\frac{\partial ^2 z}{\partial y^2}$ are called Pure Second Partial Derivatives and $\frac{\partial ^2 z}{\partial y \partial x}$ and $\frac{\partial ^2 z}{\partial x \partial y}$ are called Mixed Second Partial Derivatives.

The term "pure" refers to partial differentiating with respect to the same variable, and the term "mixed" refers to differentiating with respect to more than one variable. Thus, both terms can be applied to higher order partial derivatives and functions of more than two variables

Now let's look at an example. Suppose that we have the function $f(x,y) = 2x^3y^2$ and that we want to find the second partial derivative $\frac{\partial ^2 z}{\partial y \partial x}$. To compute $\frac{\partial ^2 z}{\partial y \partial x}$ we will first compute $\frac{\partial z}{\partial x}$ and then partial differentiate with respect to $y$. We note that $\frac{\partial z}{\partial x} = 6x^2 y^2$. Therefore $\frac{\partial ^2 z}{\partial y \partial x} = 12x^2 y$.

Of course, all of this can be extended to functions of $n$ variables.

### Example 1

Find all of second partial derivatives of the function $f(x,y) = 3x^4 \sin y$.

We must first calculate the first partial derivatives. We note that $\frac{\partial z}{\partial x} = 12x^3 \sin y$ and $\frac{\partial z}{\partial y} = 3x^4 \cos y$, and so the second order partial derivatives are as follows:

(1)
\begin{align} \quad \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left ( \frac{\partial z}{\partial x} \right ) = \frac{\partial}{\partial x} \left ( 12x^3 \sin y \right ) = 36x^2 \sin y \end{align}
(2)
\begin{align} \quad \frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y} \left ( \frac{\partial z}{\partial x} \right ) = \frac{\partial}{\partial x} \left ( 12x^3 \sin y \right ) = 12x^3 \cos y \end{align}
(3)
\begin{align} \quad \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left ( \frac{\partial z}{\partial y} \right ) = \frac{\partial}{\partial x} \left ( 3x^4 \cos y \right ) = 12x^3 \cos y \end{align}
(4)
\begin{align} \quad \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left ( \frac{\partial z}{\partial y} \right ) = \frac{\partial}{\partial y} \left ( 3x^4 \cos y \right ) = -3x^4 \sin y \end{align}

### Example 2

Find all of second partial derivatives of the function $f(x,y,z) = 3x^3e^{-y}\sin z$.

Again, we must first find all of the first partial derivatives of $f$. We note that $\frac{\partial f}{\partial x} = 9x^2 e^{-y} \sin z$, $\frac{\partial f}{\partial y} = -3x^3 e^{-y} \sin z$, and $\frac{\partial f}{\partial z} = 3x^3 e^{-y} \cos z$. Therefore the second partial derivatives are as follows:

(5)
\begin{align} \quad \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left ( \frac{\partial f}{\partial x} \right ) = \frac{\partial}{\partial x} \left ( 9x^2 e^{-y} \sin z \right ) = 18x e^{-y} \sin z \end{align}
(6)
\begin{align} \quad \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial y} \left ( 9x^2 e^{-y} \sin z \right) = -9x^2 e^{-y} \sin z \end{align}
(7)
\begin{align} \quad \frac{\partial^2 f}{\partial z \partial x} = \frac{\partial}{\partial z} \left ( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial z} \left ( 9x^2 e^{-y} \sin z \right) = 9x^2 e^{-y} \cos z \end{align}
(8)
\begin{align} \quad \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial y} \left ( -3x^3 e^{-y} \sin z \right) = 3x^3 e^{-y} \sin z \end{align}
(9)
\begin{align} \quad \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left ( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} \left ( -3x^3 e^{-y} \sin z \right) = -9x^2 e^{-y} \sin z \end{align}
(10)
\begin{align} \quad \frac{\partial^2 f}{\partial z \partial y} = \frac{\partial}{\partial z} \left ( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial z} \left ( -3x^3 e^{-y} \sin z \right) = -3x^3 e^{-y} \cos z \end{align}
(11)
\begin{align} \quad \frac{\partial^2 f}{\partial z^2} = \frac{\partial}{\partial z} \left ( \frac{\partial f}{\partial z} \right) = \frac{\partial}{\partial z} \left ( 3x^3 e^{-y} \cos z \right) = -3x^3 e^{-y} \sin z \end{align}
(12)
\begin{align} \quad \frac{\partial^2 f}{\partial x \partial z} = \frac{\partial}{\partial x} \left ( \frac{\partial f}{\partial z} \right) = \frac{\partial}{\partial x} \left ( 3x^3 e^{-y} \cos z \right) = 9x^2 e^{-y} \cos z \end{align}
(13)
\begin{align} \quad \frac{\partial^2 f}{\partial y \partial z} = \frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial z} \right) = \frac{\partial}{\partial y} \left ( 3x^3 e^{-y} \cos z \right) = -3x^3 e^{-y} \cos z \end{align}

## Higher Order Partial Derivatives

Of course, we can continue the process of partial differentiation of partial derivatives to obtain third, fourth, etc… partial derivatives. Notice though, that the number of partial derivatives increases though. For example, if we wanted to list all fourth partial derivatives of a two variable real-valued function $z = f(x,y)$ then we would have get $16$ fourth order partial derivatives. If e wanted to list all the fourth partial derivatives of a three variable real-valued function $w = f(x,y,z)$ then we would get $81$ fourth order partial derivatives. Often times we will instead be more interested in particular higher order partial derivatives.

### Example 3

Let $f(x,y,z) = x^3y^2z$. Find $\frac{\partial^3 f}{\partial z \partial y \partial x}$.

First we note that $\frac{\partial f}{\partial x} = 3x^2 y^2 z$. If we partial differentiate this with respect to $y$, then we get that $\frac{\partial^2 f}{\partial y \partial x} = 6x^2 y z$. Lastly, if we partial differentiate this with respect to $z$ then we get that $\frac{\partial ^3 f}{\partial z \partial y \partial x} = 6x^2y$.

### Example 4

Let $f(x, y, z) = \cos (xy) \sin (y) e^{2z}$. Find $\frac{\partial ^4 f}{\partial z \partial x \partial z^2}$.

First first have that $\frac{\partial f}{\partial z} = 2\cos (xy) \sin (y) e^{2z}$. If we partial differentiate this with respect to $z$, then we get that $\frac{\partial ^2 f}{\partial z^2} = 4 \cos (xy) \sin (y) e^{2z}$. Now if we partial differentiate this with respect to $x$ then we get that $\frac{\partial ^3 f}{\partial x \partial z^2} = -4y \sin (xy) \sin (y) e^{2z}$. If we partial differentiate this with respect to $z$ then we get that $\frac{\partial ^4 f}{\partial z \partial x \partial z^2} = -8y \sin (xy) \sin (y) e^{2z}$.