Higher Order Directional Derivatives

# Higher Order Directional Derivatives

Recall from the Directional Derivatives page that if $z = f(x, y)$ is a two variable real-valued function and $\vec{u} = (a, b)$ is a unit vector, then the directional derivative of $f$ in the direction of $\vec{u}$ is given by:

(1)
\begin{align} \quad D_{\vec{u}} \: f(x, y) = \lim_{h \to 0} \frac{f(x + ha, y + hb) - f(x, y)}{h} \end{align}

We saw that we could compute directional derivatives of $f$ with the following formula:

(2)
\begin{align} \quad D_{\vec{u}} \: f(x, y) = \left ( a, b \right ) \cdot \left ( \frac{\partial z}{\partial x} , \frac{\partial z}{\partial y} \right ) = a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y} \end{align}

We will now look at computing higher order directional derivatives. The process is much the same as computing higher order partial derivatives. Suppose that we have a function $z = f(x, y)$ and a unit vector $\vec{u} = (a, b)$. Then the first directional derivative of $f$ in the direction of $(a, b)$ is $D_{\vec{u}} \: f(x, y) = a \frac{\partial z}{\partial x} + b \frac{\partial z}{\partial y}$ as noted above. The directional derivative in the direction of $\vec{u}$ will define a function, say:

(3)
\begin{align} \quad z^* = D_{\vec{u}} \: f(x, y) \end{align}

If we want to take the second directional derivative of $f$ in the direction of $(a, b)$, then we have that:

(4)
\begin{align} \quad D^2_{\vec{u}} \: f(x, y) = (a, b) \cdot \left ( \frac{\partial z^*}{\partial x} , \frac{\partial z^*}{\partial y} \right ) \\ \quad D^2_{\vec{u}} \: f(x, y) = a \frac{\partial z^*}{\partial x} + b \frac{\partial z^*}{\partial y} \\ \quad D^2_{\vec{u}} \: f(x, y) = a \frac{\partial}{\partial x} \left ( D_{\vec{u}} \: f(x, y) \right ) + b \frac{\partial}{\partial y} \left ( D_{\vec{u}} \: f(x, y) \right ) \\ \quad D^2_{\vec{u}} \: f(x, y) = a \left ( a \frac{\partial^2 z}{\partial x^2} + b \frac{\partial^2 z}{\partial x \partial y} \right ) + b \left ( a \frac{\partial^2 z}{\partial y \partial x} + b \frac{\partial^2 z}{\partial y^2} \right ) \end{align}

If the second partial derivatives of $f$ are continuous, then by Clairaut's theorem we have that:

(5)
\begin{align} \quad D^2_{\vec{u}} \: f(x, y) = a^2 \frac{\partial^2 z}{\partial x^2} + 2ab \frac{\partial^2 z}{\partial y \partial x} + b^2 \frac{\partial^2 z}{\partial y^2} \end{align}

A formula for the second directional derivative of a three variable real-valued function $w = f(x, y, z)$ can be obtained in a similar manner.

Of course, we can take successively higher order directional derivatives if we so choose. It's not practical to remember the formulas for computing higher order direction derivatives of a function of several variables though.

Let's look at an example of finding a higher order directional derivative.

## Example 1

**Find the second order and third order directional derivative of the function $f(x, y) = 2xy^2$ in the direction of $(1, 2)$.**

The vector $(1, 2)$ is not a unit vector. We have that the unit vector in the direction of $(1, 2)$ is given by:

(6)
\begin{align} \quad \vec{u} = \frac{(1, 2)}{\| (1, 2) \|} = \frac{(1, 2)}{\sqrt{5}} = \left ( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right ) \end{align}

Therefore the first directional derivative in the direction of $(1, 2)$ is given by:

(7)
\begin{align} \quad D_{\vec{u}} \: f(x, y) = \frac{1}{\sqrt{5}} 2y^2 + \frac{2}{\sqrt{5}} 4xy \\ \quad D_{\vec{u}} \: f(x, y) = \frac{2}{\sqrt{5}} y^2 + \frac{8}{\sqrt{5}} xy \end{align}

We will now take the second directional derivative in the direction of $(1, 2)$:

(8)
\begin{align} \quad D^2_{\vec{u}} \: f(x, y) = D_{\vec{u}} \left ( \frac{2}{\sqrt{5}} y^2 + \frac{8}{\sqrt{5}} xy \right ) \\ \quad D^2_{\vec{u}} \: f(x, y) = \left ( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right ) \cdot \left (\frac{\partial}{\partial x} \left ( \frac{2}{\sqrt{5}} y^2 + \frac{8}{\sqrt{5}} xy \right ), \frac{\partial}{\partial y} \left ( \frac{2}{\sqrt{5}} y^2 + \frac{8}{\sqrt{5}} xy \right ) \right ) \\ \quad D^2_{\vec{u}} \: f(x, y) = \left ( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right ) \cdot \left ( \frac{8}{\sqrt{5}} y, \frac{4}{\sqrt{5}} y + \frac{8}{\sqrt{5}} x \right ) \\ \quad D^2_{\vec{u}} \: f(x, y) = \frac{1}{\sqrt{5}} \frac{8}{\sqrt{5}} y + \frac{2}{\sqrt{5}} \frac{4}{\sqrt{5}} y + \frac{2}{\sqrt{5}} \frac{8}{\sqrt{5}} x \\ \quad D^2_{\vec{u}} \: f(x, y) = \frac{8}{5} y + \frac{8}{5} y + \frac{16}{5} x \\ \quad D^2_{\vec{u}} \: f(x, y) = \frac{16}{5} (x + y) \end{align}

Lastly, we find the third order directional derivative as follows:

(9)
\begin{align} \quad D^3_{\vec{u}} \: f(x, y) = \left ( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right ) \cdot \frac{16}{5} \left ( \frac{\partial}{\partial x} (x + y) , \frac{\partial}{\partial y} (x + y) \right ) \\ \quad D^3_{\vec{u}} \: f(x, y) = \left ( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right ) \cdot \frac{16}{5} (1, 1) \\ \quad D^3_{\vec{u}} \: f(x, y) = \frac{16}{5\sqrt{5}} + \frac{32}{5\sqrt{5}} \\ \quad D^3_{\vec{u}} \: f(x, y) = \frac{48}{5\sqrt{5}} \end{align}