Section 12.6: Directional Derivatives and the Gradient Vector
Recall that if f is a differentiable function of x and y and z = f (x, y), then the partial
derivatives fx (x, y) and fy (x, y) give the rate of change of z in the direction of x and y,
respectively.
Suppose we want to find the rate of change of z in the direction of an arbitrary unit vector
~u = hu1 , u2 i. Consider the surface S defined by z = f (x, y) and let z0 = f (x0 , y0 ) so that
P = (x0 , y0 , z0 ) lies on S. The vertical plane passing through (x0 , y0 , z0 ) in the direction of
~u intersects S in a curve C. The slope of the tangent line to C at P is the rate of change of
z in the direction of ~u.
Figure 1: Illustration of the directional derivative.
Definition: The directional derivative of f at (x0 , y0 ) in the direction of a unit vector
~u = hu1 , u2 i is
f (x0 + u1 h, y0 + u2 h) − f (x0 , y0 )
D~u f (x0 , y0 ) = lim
h→0
h
provided this limit exists.
Theorem: If f is a differentiable function of x and y, then f has a directional derivative for
any unit vector ~u = hu1 , u2 i and
D~u f (x, y) = fx (x, y)u1 + fy (x, y)u2 .
Example: Find the directional derivative of f (x, y) = x2 y 3 + 2x4 y at (1, −2) in the direction
* √ +
1 3
,
.
of ~u =
2 2
The partial derivatives are
fx (x, y) = 2xy 3 + 8x3 y
fx (1, −2) = −24
fy (x, y) = 3x2 y 2 + 2x4
fy (1, −2) = 14.
Then the directional derivative is
1
D~u f (1, −2) = −24
+ 14
2
√ !
√
3
= 7 3 − 12.
2
Example: Find the directional derivative of f (x, y) =
~v = h12, 5i.
√
x − y at (5, 1) in the direction of
A unit vector in the direction of ~v is
~v
~u =
=
||~v ||
12 5
,
13 13
.
The partial derivatives are
1
fx (x, y) = √
2 x−y
1
fx (5, 1) =
4
1
fy (x, y) = − √
2 x−y
1
fy (5, 1) = − .
4
Then the directional derivative is
1
D~u f (5, 1) =
4
12
13
1
−
4
5
13
=
7
.
52
Note: The vector that specifies the direction must be a unit vector.
Definition: If f is a differentiable function of x and y, then the gradient of f is the vector
function
∇f (x, y) = hfx (x, y), fy (x, y)i.
Example: Find the gradient of f (x, y) = exy + sin(x2 + 2y).
The gradient vector is
∇f (x, y) = hyexy + 2x cos(x2 + 2y), xexy + 2 cos(x2 + 2y)i.
Theorem: (Gradient Formula for the Directional Derivative)
If f is a differentiable function of x and y, then
D~u f (x, y) = ∇f (x, y) · ~u.
Example: Find the directional derivative of f (x, y) = xexy at (−3, 0) in the direction of
~v = h2, 3i.
A unit vector in the direction of ~v is
~v
=
~u =
||~v ||
2
3
√ ,√
13 13
.
The gradient of f is
∇f (x, y) = hexy + xyexy , x2 exy i.
At (−3, 0), ∇f0 = h1, 9i. Then the directional derivative is
2
27
29
D~u f (−3, 0) = ∇f0 · ~u = √ + √ = √ .
13
13
13
Note: The directional derivative and gradient can be extended to functions of three variables.
If f (x, y, z) is differentiable, then
∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i
and the directional derivative of f at (x0 , y0 , z0 ) in the direction of the unit vector ~u is
D~u f (x0 , y0 , z0 ) = ∇f (x0 , y0 , z0 ) · ~u.
Example: Find the directional derivative of f (x, y, z) =
~v = h4, 2, −4i.
√
xyz at (2, 4, 2) in the direction of
A unit vector in the direction of ~v is
~v
1
~u =
= h4, 2, −4i =
||~v ||
6
2 1 2
, ,−
3 3 3
.
The gradient of f is
∇f (x, y, z) =
yz
xz
xy
, √
, √
√
2 xyz 2 xyz 2 xyz
.
At (2, 4, 2), ∇f0 = h1, 21 , 1i. Then the directional derivative is
D~u f (2, 4, 2) = ∇f0 · ~u =
2 1 2
1
+ − = .
3 6 3
6
Given a function f , in which direction does f change most rapidly? What is the greatest
rate of change?
Theorem: (Maximal Property of the Gradient)
Suppose f is a differentiable function of two or three variables. The maximum value of the
directional derivative D~u f is ||∇f || and it occurs in the direction of the gradient ∇f .
Example: Let f (x, y, z) = 3x2 − 5xy + xyz. Find the rate of change of f at (2, 2, 7) in the
direction of ~v = h1, 1, −1i. In which direction does f change most rapidly? What is the
maximum rate of change at (2, 2, 7)?
A unit vector in the direction of ~v is
~v
~u =
=
||~v ||
1 1
1
√ , √ , −√
3 3
3
.
The gradient of f is
∇f (x, y, z) = h6x − 5y + yz, −5x + xz, xyi .
At (2, 2, 7), ∇f0 = h16, 4, 4i. Then the directional derivative is
16
D~u f (2, 2, 7) = ∇f0 · ~u = √ .
3
The greatest change occurs in the direction of ∇f0 = h16, 4, 4i and the greatest change is
√
√
||∇f0 || = 256 + 16 + 16 = 12 2.
Suppose S is a level surface defined by f (x, y, z). That is, f (x, y, z) = k for some constant
k. If (x0 , y0 , z0 ) lies on S, then ∇f (x0 , y0 , z0 ) is a vector that is normal to the tangent plane
at (x0 , y0 , z0 ).
Figure 2: Illustration of a level surface f (x, y, z) = k and the gradient vector ∇f (x0 , y0 , z0 ).
Definition: Let S be the level surface defined by f (x, y, z) = k and let (x0 , y0 , z0 ) lie on S.
The plane with normal vector ∇f0 passing through (x0 , y0 , z0 ) is the tangent plane to the
surface S at (x0 , y0 , z0 ). The line passing through (x0 , y0 , z0 ) in the direction of ∇f0 is the
normal line to S at (x0 , y0 , z0 ).
Example: Find equations for the tangent plane and normal line to the surface defined by
x2 y + y 2 z + xz 2 = 5 at (2, 1, −1).
Let f (x, y, z) = x2 y + y 2 z + xz 2 . The gradient of f is
∇f (x, y, z) = h2xy + z 2 , x2 + 2yz, y 2 + 2xzi.
At (2, 1, −1), ∇f0 = h5, 2, −3i. An equation of the tangent plane is
5(x − 2) + 2(y − 1) − 3(z + 1) = 0
5x + 2y − 3z = 15.
The parametric equations for the normal line are
x = 2 + 5t
y = 1 + 2t
z = −1 − 3t.