James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

950 Chapter 14 Partial Derivatives
8 Definition If f is a function of two variables x and y, then the gradient of f
is the vector function =f defined by
=f sx, yd − k f x sx, yd, f y sx, yd l − −f
−x i 1 −f
−y j
EXAMPLE 3 If f sx, yd − sin x 1 e x y , then
=f sx, yd − k f x , f y l − kcos x 1 ye x y , xe x y l
and =f s0, 1d − k2, 0 l ■
With this notation for the gradient vector, we can rewrite Equation 7 for the directional
derivative of a differentiable function as
9 D u f sx, yd − =f sx, yd ? u
This expresses the directional derivative in the direction of a unit vector u as the scalar
projection of the gradient vector onto u.
The gradient vector =f s2, 21d in
Example 4 is shown in Figure 6 with
initial point s2, 21d. Also shown is the
vector v that gives the direction of the
directional derivative. Both of these
vectors are superimposed on a contour
plot of the graph of f .
FIGURE 66
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MasterID: 01606
y
±f(2, _1)
(2, _1)
v
x
EXAMPLE 4 Find the directional derivative of the function f sx, yd − x 2 y 3 2 4y at the
point s2, 21d in the direction of the vector v − 2 i 1 5j.
SOLUTION We first compute the gradient vector at s2, 21d:
=f sx, yd − 2xy 3 i 1 s3x 2 y 2 2 4dj
=f s2, 21d − 24 i 1 8 j
Note that v is not a unit vector, but since | v | − s29 , the unit vector in the direction
of v is
u −
Therefore, by Equation 9, we have
v
| v | − 2 i 1 5 j
s29 s29
D u f s2, 21d − =f s2, 21d ? u − s24 i 1 8 jd ?S 2
s29
i 1 5
s29 j D
Functions of Three Variables
− 24 ? 2 1 8 ? 5 − 32
■
s29 s29
For functions of three variables we can define directional derivatives in a similar manner.
Again D u f sx, y, zd can be interpreted as the rate of change of the function in the direction
of a unit vector u.
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