452 CHAPTER 21. FUNCTIONS OF MANY VARIABLES

Definition 21.4.5 Suppose f : U ⊆Rn →R where U is an open set and the partialderivatives of f all exist. Define the gradient of f denoted ∇ f (x) to be the vector

∇ f (x) = ( fx1 (x) , fx2 (x) , · · · , fxn (x))T .

Proposition 21.4.6 In the situation of Definition 21.4.5, if the partial derivatives arecontinuous, then for v a unit vector Dv f (x) = ∇ f (x) ·v.

This proposition will be proved in a more general setting later. For now, you can use itto compute directional derivatives.

Example 21.4.7 Find the directional derivative of the function f (x,y) = sin(2x2 + y3

)at

(1,1) in the direction(

1√2, 1√

2

)T.

First find the gradient. ∇ f (x,y) =(4xcos

(2x2 + y3

),3y2 cos

(2x2 + y3

))T. Therefore,

∇ f (1,1) = (4cos(3) ,3cos(3))T . The directional derivative is therefore,

(4cos(3) ,3cos(3))T ·(

1√2,

1√2

)T

=72(cos3)

√2.

Another important observation is that the gradient gives the direction in which the functionchanges most rapidly. The following proposition will be proved later.

Proposition 21.4.8 In the situation of Definition 21.4.5, suppose ∇ f (x) ΜΈ= 0. Then thedirection in which f increases most rapidly, that is the direction in which the directionalderivative is largest, is the direction of the gradient. Thus v = ∇ f (x)/ |∇ f (x)| is theunit vector which maximizes Dv f (x) and this maximum value is |∇ f (x)|. Similarly, v =−∇ f (x)/ |∇ f (x)| is the unit vector which minimizes Dv f (x) and this minimum value is−|∇ f (x)|.

The concept of a directional derivative for a vector valued function is also easy todefine although the geometric significance expressed in pictures is not.

Definition 21.4.9 Let f : U →Rp where U is an open set in Rn and let v be a unitvector. For x ∈ U, define the directional derivative of f in the direction v, at the point xas

Dvf (x)≡ limt→0

f (x+ tv)−f (x)

t.

Example 21.4.10 Let f (x,y) =(xy2,yx

)T . Find the directional derivative in the direction(1,2)T at the point (x,y).

First, a unit vector in this direction is(

1/√

5,2/√

5)T

and from the definition, thedesired limit is

limt→0

((x+ t

(1/√

5))(

y+ t(

2/√

5))2

− xy2,(

x+ t(

1/√

5))(

y+ t(

2/√

5))

− xy)

t

452 CHAPTER 21. FUNCTIONS OF MANY VARIABLESDefinition 21.4.5 Suppose f :U CR" > R where U is an open set and the partialderivatives of f all exist. Define the gradient of f denoted V f (x) to be the vectorVS (a) = (fay (®) sey (@) 5° sry (@))"Proposition 21.4.6 Jn the situation of Definition 21.4.5, if the partial derivatives arecontinuous, then for v a unit vector Dy f («) = Vf (x) -v.This proposition will be proved in a more general setting later. For now, you can use itto compute directional derivatives.Example 21.4.7 Find the directional derivative of the function f (x,y) = sin (2x? + y>) atT(1,1) in the direction (<5. ss) .First find the gradient. Vf (x,y) = (4xcos (2x* + y3) ,3y? cos (2x? + y3)) " Therefore,Vf (1,1) = (4cos (3) ,3cos (3))” . The directional derivative is therefore,1 1PvAnother important observation is that the gradient gives the direction in which the functionchanges most rapidly. The following proposition will be proved later.(4cos (3) ,3cos(3))’ - ( ) = * (cos3) v2.Proposition 21.4.8 Jn the situation of Definition 21.4.5, suppose V f (x) 4 0. Then thedirection in which f increases most rapidly, that is the direction in which the directionalderivative is largest, is the direction of the gradient. Thus v = Vf (a) /|Vf(a)| is theunit vector which maximizes Dy f (a) and this maximum value is |V f (a)|. Similarly, v =—Vf (x) /|Vf (a)| is the unit vector which minimizes Dy f (x) and this minimum value is—|Vf(#)|-The concept of a directional derivative for a vector valued function is also easy todefine although the geometric significance expressed in pictures is not.Definition 21.4.9 ce f :U —R? where U is an open set in R" and let v be a unitvector. For x € U, define the directional derivative of f in the direction v, at the point x“ Def (x) = lim L2 tM) =F)t0 tExample 21.4.10 Let f (x,y) = (xy’, yx)" Find the directional derivative in the direction(1,2) at the point (x,y).. . . . . . . T eaeFirst, a unit vector in this direction is (1 /V5,2/ V5) and from the definition, thedesired limit islim (Ge (1/v5)) (v+0(2/v3)) —ay*, (x +e (1/v3)) (y +e (2/v3)) »)t0 t