# Directional Derivatives Examples 3

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

(1)

For a three variable real-valued function $w = f(x, y, z)$, the directional derivative of $f$ at a point $(x, y, z) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b, c)$ is given by the formula:

(2)

We will now look at some examples of calculating directional derivatives. More examples can be found on the Directional Derivatives Examples 1 and Directional Derivatives Examples 2 pages.

## Example 1

**Find the directional derivative of $w = f(x, y, z) = \sqrt{x + yz}$ at $(1, 3, 1)$ in the direction of $\vec{v} = (2, 3, 6)$.**

We first note that $\vec{v}$ is not a unit vector, so $\| \vec{v} \| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{49} = 7$ and $\vec{u} = \left ( \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right )$ is a unit vector in the direction of $\vec{v}$.

We now want to compute the partial derivatives of $f$. We note that $\frac{\partial w}{\partial x} = \frac{1}{2\sqrt{x + yz}}$, $\frac{\partial w}{\partial y} = \frac{z}{2 \sqrt{x + yz}}$, and $\frac{\partial w}{\partial z} = \frac{y}{2\sqrt{x + yz}}$. Therefore we have that:

(3)

## Example 2

**Find the directional derivative of $w = f(x, y, z) = x \cos y \sin z$ at $\left (1, \pi, \frac{\pi}{4} \right )$ in the direction of $\vec{v} = (2, -1, 4)$.**

Once again, $\vec{v}$ is not a unit vector, so $\| \vec{v} \| =\sqrt{(2)^2 + (-1)^2 + (4)^2} = \sqrt{21}$ and $\vec{u} = \left ( \frac{2}{\sqrt{21}}, \frac{-1}{\sqrt{21}}, \frac{4}{\sqrt{21}} \right )$ is a unit vector in the direction of $\vec{v}$.

We now want to compute the partial derivatives of $f$. We note that $\frac{\partial w}{\partial x} = \cos y \sin z$, $\frac{\partial w}{\partial y} = -x \sin y \sin z$, and $\frac{\partial w}{\partial z} = x \cos y \cos z$. Therefore:

(4)

## Example 3

**Find the directional derivative of $w = f(x, y, z) = \ln(1 + x^2 + y^2 + z^2)$ at $(1, -1, 1)$ in the direction of $\vec{v} = (2, -2, -3)$.**

We note that $\vec{v}$ is not a unit vector, so $\| \vec{v} \| = \sqrt{(2)^2 + (-2)^2 + (-3)^2} = \sqrt{17}$ and $\vec{u} \left ( \frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}} \right )$ is a unit vector in the direction of $\vec{v}$.

We now want to compute the partial derivatives of $f$. We note that $\frac{\partial w}{\partial x} = \frac{2x}{1 + x^2 + y^2 + z^2}$, $\frac{\partial w}{\partial y} = \frac{2y}{1 + x^2 + y^2 + z^2}$, and $\frac{\partial w}{\partial z} = \frac{2z}{1 + x^2 + y^2 + z^2}$. Therefore:

(5)