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August 4, 2018
CHAPTER 3: PARTIAL DERIVATIVES AND
DIFFERENTIATION
1. Partial Derivatives and Differentiable functions
In all this chapter, D will denote an open subset of Rn.
Definition 1.1. Consider a function f : D → R and let p ∈ D, i = 1,··· ,n. We
define the partial derivative of f with respect to the i-th variable at the point p as
the following limit (if it exists)
∂f (p) = lim f(p+tei)−f(p)
∂x t→0 t
i
where {e1,...,en} is the canonical basis of Rn,
defined as follows
e =(0,...,0,i, 0,...,0)
i | {z } | {z }
i−1terms n−iterms
For example, in R2 the canonical basis is
e1 = (1,0)
e2 = (0,1)
and in R3 the canonical basis is
e = (1,0,0)
1
e = (0,1,0)
2
e = (0,0,1)
3
Remark 1.2. When n=2, in the above definition we let
p = (x,y), f(x,y) : R2 → R
we and use the notation,
∂f(x,y) = lim f(x+t,y)−f(x,y)
∂x t→0 t
∂f(x,y) = lim f(x,y +t)−f(x,y)
∂y t→0 t
Likewise, when n = 3, we let
p = (x,y,z)
and use the notation,
∂f(x,y,z) = lim f(x+t,y,z)−f(x,y,z)
∂x t→0 t
∂f(x,y,z) = lim f(x,y +t,z)−f(x,y,z)
∂y t→0 t
∂f(x,y,z) = lim f(x,y,z +t)−f(x,y,z)
∂y t→0 t
1
2 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION
Example 1.3. In Economics, the partial derivatives of a utility function are called
‘marginal utilities’, the partial derivatives of a production function are called ‘mar-
ginal products’.
Consider, for example the Cobb-Douglas production function
f(K,L)=5K1/3L2/3
where f is the number of units produced, K is the capital and L is labor. That is,
the above formula means that if we use K units of capital and L units of labor, then
we produce f(K,L) = 5K1/3L2/3 units of a good. The constants A = 5, α = 1/3
and β = 2/3 are technological parameters.
The ‘marginal products’ with respect to capital and labor are
∂f = 5K−2/3L2/3
∂K 3
∂f = 10K1/3L−1/3
∂L 3
The marginal product of labor,
∂f(K,L)
∂L
is interpreted in Economics as an approximationtothevariationintheproduction
of the good when we are using K units of capital and L units of labor and we switch
to use an additional unit L+1 of labor and the same units K of capital as before.
Wesee that the marginal product of labor and capital is positive. That is, if we
use more labor and/or more capital, production increases. On the other hand, the
marginal product of labor is decreasing in labor and increasing in capital. We may
interpret this as follows.
• Suppose that we keep constant the amount of capital that we are using K.
If L′ > L then
′ ′
f(K,L +1)−f(K,L) K then
f(K′,L+1)−f(K′,L)>f(K,L+1)−f(K,L)
That is, the increase in the production when we use one additional unit of
labor is larger the more capital we use. Capital and labor are complemen-
tary. In the previous example, hiring and additional worker has a larger
effect on the production the larger is the size of land.
CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 3
Definition 1.4. Consider a function f : D → R. Let p ∈ D and suppose all the
partial derivatives
∂f (p), ∂f (p),··· , ∂f (p)
∂x ∂x ∂x
1 2 n
exist at the point p. We define the gradient of f at p as the following vector
∇f(p) = ∂f (p), ∂f (p),··· , ∂f (p)
∂x ∂x ∂x
1 2 n
Definition 1.5. Consider a function f : D → R. Let p ∈ D and suppose all the
partial derivatives
∂f (p), ∂f (p),··· , ∂f (p)
∂x ∂x ∂x
1 2 n
exist at the point p. We say that f is differentiable at p if
lim f(p+v)−f(p)−∇f(p)·v =0
v→0 kvk
Note that the limit is taken for v ∈ Rn.
Remark 1.6. A function of two variables f : D ⊂ R2 → is differentiable at the
point p = (a,b) if
f(a+v ,b+v )−f(a,b)−∇f(a,b)·(v ,v )
lim 1 2 1 2 =0
(v ,v )→(0,0) k(v ,v )k
1 2 1 2
Letting
x=a+v , y=b+v
1 2
we see that (v ,v ) → (0,0) is equivalent to (x,y) → (a,b), so we may write this
1 2
limit as
lim f(x,y)−f(a,b)−∇f(a,b)·(x−a,y−b) =0
(x,y)→(a,b) k(x−a,y−b)k
Writing this limit explicitly we wee that f is differentiable at the point p = (a,b) if
f(x,y)−f(a,b)− ∂f(a,b)·(x−a)− ∂f(a,b)·(y−b)
(1.1) lim p∂x ∂y =0
(x,y)→(a,b) (x−a)2+(y−b)2
Example 1.7. Consider the function
( xy2 if (x,y) 6= (0,0),
2 2
f(x,y) = x +y
0 if (x,y) = (0,0).
We will show that f is not differentiable at the point p = (0,0). First of all, we
compute ∇f(0,0). Note that
∂f(0,0) = lim f(t,0)−f(0,0) = lim 0 = 0
∂x t→0 t t→0 t3
∂f(0,0) = lim f(0,t)−f(0,0) = lim 0 = 0
∂y t→0 t t→0 t3
4 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION
so, ∇f(0,0) = (0,0). Let us use the notation v = (x,y). Then, f is differentiable
at the point p = (0,0) if and only if
0 = lim f(p+v)−f(p)−∇f(p)·v
v→0 kvk
f ((0,0) + (x,y)) − f(0,0) − ∇f(p)·(x,y)
= lim p 2 2
(x,y)→(0,0) x +y
f(x,y)−f(0,0)−(0,0)·(x,y)
= lim p 2 2
(x,y)→(0,0) x +y
f(x,y)
= lim p 2 2
(x,y)→(0,0) x +y
xy2
= lim 3/2
(x,y)→(0,0) (x2 + y2)
Weprove that the above limit does not exist. Consider the function
xy2
g(x,y) = 3/2
(x2 +y2)
Note that
limg(t,0) = lim 0 =0
t→0 t→0 (2t2)3/2
and note that
t3 1
limg(t,t) = lim 3/2 = 3/2 6= 0
t→0 t→0 (2t2) (2)
so the limit
xy2
lim 3/2
(x,y)→(0,0) 2 2
(x +y )
does not exist and we conclude that f is not differentiable at the point (0,0).
Example 1.8. Consider now the function
( xy3 if (x,y) 6= (0,0),
2 2
f(x,y) = x +y
0 if (x,y) = (0,0).
Wewillshowthatf is differentiable at the point p = (0,0). First of all, we compute
∇f(0,0). Note that
∂f(0,0) = lim f(t,0)−f(0,0) = lim 0 = 0
∂x t→0 t t→0 t3
∂f(0,0) = lim f(0,t)−f(0,0) = lim 0 = 0
∂y t→0 t t→0 t3
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