16.3. THE DIRECTIONAL DERIVATIVE AND PARTIAL DERIVATIVES 289

Example 16.3.2 Find the directional derivative of the function f (x,y) = x2y in the direc-tion of i+j at the point (1,2).

First you need a unit vector which has the same direction as the given vector. Thisunit vector is v ≡

(1√2, 1√

2

). Then to find the directional derivative from the definition,

write the difference quotient described above. Thus f (x+ tv) =(

1+ t√2

)2(2+ t√

2

)and

f (x) = 2. Therefore,

f (x+ tv)− f (x)t

=

(1+ t√

2

)2(2+ t√

2

)−2

t,

and to find the directional derivative, you take the limit of this as t→ 0. However, this dif-ference quotient equals 1

4

√2(

10+4t√

2+ t2)

and so, letting t→ 0,Dv f (1,2) =(

52

√2).

There is something you must keep in mind about this. The direction vector must alwaysbe a unit vector1.

16.3.2 Partial DerivativesThere are some special unit vectors which come to mind immediately. These are the vectorsei where

ei = (0, · · · ,0,1,0, · · ·0)T

and the 1 is in the ith position.Thus in case of a function of two variables, the directional derivative in the direction

i= e1 is the slope of the indicated straight line in the following picture.

y

z = f (x,y)

xe1

As in the case of a general directional derivative, you fix y and take the derivative ofthe function x→ f (x,y). More generally, even in situations which cannot be drawn, thedefinition of a partial derivative is as follows.

Definition 16.3.3 Let U be an open subset of Rn and let f : U → R. Then letting x=(x1, · · · ,xn)

T be a typical element of Rn,

∂ f∂xi

(x)≡ Dei f (x) .

1Actually, there is a more general formulation of the notion of directional derivative known as the Gateauxderivative in which the length of v is not one but it is not considered here.