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Directional
Derivatives

To this point we’ve only looked at the two partial derivatives f x ( x, y ) and f y ( x, y )
...
We now need to discuss how to find the rate of change of f
if we allow both x and y to change simultaneously
...
For instance one could be changing faster than the other and
then there is also the issue of whether or not each is increasing or decreasing
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The main idea that we need to look at is just how are we going to define the changing of x and/or
y
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Let’s also suppose that both x and y are increasing and that, in this case, x is increasing
twice as fast as y is increasing
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To help us see how we’re going to define this change let’s suppose that a particle is sitting at
( x0 , y0 ) and the particle will move in the direction given by the changing x and y
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Now that we’re thinking of this changing x and y as a
direction of movement we can get a way of defining the change
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For our example we will say that we want the rate
of change of f in the direction of v = 2,1
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There is still a small problem with this however
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For instance all of the following vectors point in the same direction
as v = 2,1
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We will
do this by insisting that the vector that defines the direction of change be a unit vector
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This means that for the example that
we started off thinking about we would want to use

v=

2 1
,
5 5

since this is the unit vector that points in the direction of change
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Sometimes we will give the direction of changing x and y as an angle
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The unit vector that points in this
direction is given by,

u = cos θ ,sin θ
Okay, now that we know how to define the direction of changing x and y its time to start talking
about finding the rate of change of f in this direction
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Definition
The rate of change of f ( x, y ) in the direction of the unit vector u = a,b is called the
directional derivative and is denoted by Du f ( x, y )
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However, in practice this can be a very difficult limit to compute so we need an easier way of
taking directional derivatives
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To see how we can do this let’s define a new function of a single variable,

g ( z ) = f ( x0 + az, y0 + bz )

where x0 , y0 , a, and b are some fixed numbers
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Then by the definition of the derivative for functions of a single variable we have,

g ( z + h) − g ( z )
g ′ ( z ) = lim
h →0
h

and the derivative at z = 0 is given by,

g ( h) − g ( 0)
g ′ ( 0 ) = lim
h →0
h

If we now substitute in for g ( z ) we get,

g ( h ) − g ( 0)
f ( x0 + ah, y0 + bh ) − f ( x0 , y0 )
g ′ ( 0 ) = lim
= lim
= Du f ( x0 , y0 )
h →0
h →0
h
h
So, it looks like we have the following relationship
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Let’s rewrite g ( z ) as follows,

g ( z ) = f ( x, y ) where x = x0 + az and y = y0 + bz
We can now use the chain rule from the previous section to compute,

dg ∂f dx ∂f dy
g′ ( z ) =
=
+
= f x ( x , y ) a + f y ( x, y ) b
dz ∂x dz ∂y dz

So, from the chain rule we get the following relationship
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Du f ( x, y ) = f x ( x, y ) a + f y ( x, y ) b
This is much simpler than the limit definition
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There are similar formulas that can be derived by
the same type of argument for functions with more than two variables
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Example 1 Find each of the directional derivatives
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[Solution]
of θ =
3
(b) Du f ( x, y, z ) where f ( x, y, z ) = x 2 z + y 3 z 2 − xyz in the direction of
v = −1, 0,3
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3
We’ll first find Du f ( x, y ) and then use this a formula for finding Du f ( 2, 0 )
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In this case let’s first check to see if the direction vector is a unit vector or not and if it isn’t
convert it into one
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v = 1+ 0 + 9 = 10 ≠ 1
So, it’s not a unit vector
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So, the unit vector that we need is,

1
1
3
u=
−1, 0,3 = −
, 0,
10
10
10
The directional derivative is then,

⎛ 1 ⎞
⎛ 3 ⎞ 2
2
3
Du f ( x, y, z ) = ⎜ −
⎟ ( 2 xz − yz ) + ( 0 ) ( 3 y z − xz ) + ⎜
⎟ ( x + 2 y z − xy )
⎝ 10 ⎠
⎝ 10 ⎠
1
2
3
=
x
+
y
z − 3xy − 2 xz + yz )
3
6
(
10
[Return to Problems]

There is another form of the formula that we used to get the directional derivative that is a little
nicer and somewhat more compact
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Let’s start with the second one and notice that we can write it as follows,

Du f ( x, y, z ) = f x ( x, y,

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