482 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

Now we can prove the formula for the directional derivative in terms of the gradient.

Proposition 22.8.1 If f is differentiable at x and for v a unit vector

Dv f (x) = ∇ f (x) ·v. (22.16)

Proof:

f (x+tv)− f (x)t

=1t

(f (x)+

n

∑j=1

∂ f (x)∂xi

tvi +o(tv)− f (x)

)

=1t

(n

∑j=1

∂ f (x)∂xi

tvi +o(tv)

)=

n

∑j=1

∂ f (x)∂xi

vi +o(tv)

t

Now limt→0o(tv)

t = 0 and so

Dv f (x) = limt→0

f (x+tv)− f (x)t

=n

∑j=1

∂ f (x)∂xi

vi = ∇ f (x) ·v

as claimed.

Example 22.8.2 Let f (x,y,z) = x2 + sin(xy)+ z. Find Dv f (1,0,1) where

v =

(1√3,

1√3,

1√3

).

Note this vector which is given is already a unit vector. Therefore, from the above, it isonly necessary to find ∇ f (1,0,1) and take the dot product.

∇ f (x,y,z) = (2x+(cosxy)y,(cosxy)x,1) .

Therefore, ∇ f (1,0,1) = (2,1,1). Therefore, the directional derivative is

(2,1,1) ·(

1√3,

1√3,

1√3

)=

43

√3.

Because of 22.16 it is easy to find the largest possible directional derivative and thesmallest possible directional derivative. That which follows is a more algebraic treatmentof an earlier result with the trigonometry removed.

Proposition 22.8.3 Let f : U → R be a differentiable function and let x ∈U. Then

max{Dv f (x) : |v|= 1}= |∇ f (x)| (22.17)

andmin{Dv f (x) : |v|= 1}=−|∇ f (x)| . (22.18)

Furthermore, the maximum in 22.17 occurs when v = ∇ f (x)/ |∇ f (x)| and the minimumin 22.18 occurs when v =−∇ f (x)/ |∇ f (x)|.