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

SectION 14.6 Directional Derivatives and the Gradient Vector 951
10 Definition The directional derivative of f at sx 0 , y 0 , z 0 d in the direction of
a unit vector u − ka, b, c l is
D u f sx 0 , y 0 , z 0 d − lim
h l 0
f sx 0 1 ha, y 0 1 hb, z 0 1 hcd 2 f sx 0 , y 0 , z 0 d
h
if this limit exists.
If we use vector notation, then we can write both definitions (2 and 10) of the directional
derivative in the compact form
11 D u f sx 0 d − lim
h l 0
f sx 0 1 hud 2 f sx 0 d
h
where x 0 − kx 0 , y 0 l if n − 2 and x 0 − kx 0 , y 0 , z 0 l if n − 3. This is reasonable because
the vector equation of the line through x 0 in the direction of the vector u is given by
x − x 0 1 tu (Equation 12.5.1) and so f sx 0 1 hud represents the value of f at a point on
this line.
If f sx, y, zd is differentiable and u − ka, b, c l, then the same method that was used to
prove Theorem 3 can be used to show that
12 D u f sx, y, zd − f x sx, y, zd a 1 f y sx, y, zd b 1 f z sx, y, zd c
For a function f of three variables, the gradient vector, denoted by =f or grad f , is
or, for short,
=f sx, y, zd − k f x sx, y, zd, f y sx, y, zd, f z sx, y, zd l
13 =f − k f x , f y , f z l − −f
−x i 1 −f
−y j 1 −f
−z k
Then, just as with functions of two variables, Formula 12 for the directional derivative
can be rewritten as
14 D u f sx, y, zd − =f sx, y, zd ? u
EXAMPLE 5 If f sx, y, zd − x sin yz, (a) find the gradient of f and (b) find the directional
derivative of f at s1, 3, 0d in the direction of v − i 1 2 j 2 k.
SOLUTION
(a) The gradient of f is
=f sx, y, zd − k f x sx, y, zd, f y sx, y, zd, f z sx, y, zd l
− ksin yz, xz cos yz, xy cos yz l
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