Chapter #6: Differential Calculus
Math Econ (MEF)
Raúl Bajo (rbajo@unav.es)
Single variable calculus
The material on single variable differential calculus was included in Chapter #2
(Real Analysis): differentiability, continuously differentiability, Taylor expan-
sion, etc.
Multivariate differential calculus
• Let X be an open subset of a space Rm , and f a mapping from X into Rn .
We say that f is differentiable at a ∈ X if there is a linear mapping A ∈
L(m, n) s.t. limx→a ||f (x)−f||x−a||
(a)−A(x−a)||
where the || || in the numerator is
a norm in Rn and that in the denominator is one in Rm .
The interpretation is that the mapping x → f (a) + A(x − a) approximates
f in the neighborhood of a.
When the above property holds, the linear mapping A is unique. It is
called the differential of f at a and denoted f 0 (a). In this definition the
choice of the norms in Rn and Rm does not matter.
. If f is differentiable at a, it is continuous at a.
• Properties of differentiation
. (f + g)0 (a) = f 0 (a) + g 0 (a) for all a.
. (λ · f )0 (a) = λ · f 0 (a) for all a, all λ ∈ R.
• Chain rule: let f : Rn ⊃ X → Rm , g : Rm ⊃ Y → Rl and f (X) ⊆
Y , where X and Y are open sets. If f is differentiable at a, and g is
differentiable at f (a), the mapping g ◦ f is differentiable at a and:
(g ◦ f )0 (a) = g 0 (f (a)) ◦ f 0 (a)
• Differentiating the inverse of a function: let X, Y be the open subsets
of Rn and f : X → Y a bijection s.t. f and f −1 are continuous, f is
differentiable at x ∈ X and f 0 (x) ∈ L(n) is non-singular. Then f −1 is
differentiable at f (x) = y and (f −1 )0 (y) = [f 0 (x)]−1 .
• If f ∈ L(Rn1 , Rn2 , · · · , Rnp ; Rn ) is a multilinear function f (x1 , · · · , xp ),
k
Pp in x−k ∈ R
linear nk
for all k = 1, · · · , p, then f 0 (a1 , · · · , ap )(x1 , · · · , xp ) =
k
k=1 f (a , x ).
1
• Differentiation of a (multilinear) product: take f ∈ L(Rn1 , Rn2 ; Rn )
X ⊆ Rm
u : X → Rn 1 , v : X → Rn 2 .
u and v differentiable at a ∈ X.
Define g(x) = f (u(x), v(x)), g : X → Rn . Then g is differentiable at a
and
g 0 (a) · h = f (u0 (a) · h, v(a)) + f (u(a), v 0 (a) · h)
• Partial differentiation (important!): let {e1 , · · · , en } and {u1 , · · · , um }
standard bases of Rn and Rm respectively. Then we can write
be the P
m
f (x) = i=1 fi (x)ui in which fi ’s are the component of f .
∂fi (x) fi (x+hej )−fi (x)
Then the partial derivative is ∂xj = limh→0 h , if this limit
exists.
. If f is differentiable at x ∈ X then the partial derivatives of f at x
exist for all i = 1, · · · , m
The matrix that represents f 0 (x) with respect to the standard basis is
called the Jacobian matrix
∂f
1 (x) ∂f1 (x)
∂x1 ··· ∂xn
[f 0 (x)] =
.. ..
. .
∂fm (x) ∂fm (x)
∂x1 ··· ∂xn
. Note: the existence of partial derivatives does not imply differentia-
bility. One example is f (x, y) = x2xy
+y 2 with f (0, 0) = 0. It is not
even continuous at (0, 0).
• For a scalar function f , the gradient
is just the corresponding
Jacobian
matrix in vector form: 5f = ∂f (x) ∂f (x)
∂x1 , ∂x2 , · · · , ∂f (x)
∂xn .
Local inversion and implicit function theorem
• Assume X, Y ⊆ Rn , both open, f : X → Y is a bijection, f and f −1 are
continuous, f is C 1 regular on X.
Then {f 0 (x) is non singular for all x ∈ X} ⇔ {f −1 is C 1 on Y }.
In this case, we say that f is a C 1 -diffeomorphism from X to Y .
• Local inversion theorem. Assume: X ⊆ Rn is open, a ∈ X, f : X →
Rn is C 1 regular on X and f 0 (a) is non singular. Then there exists an
open subset U of X containing a, and an open subset V of Rn containing
f (a), s.t. f is a C 1 -diffeomorphism from U to V .
• Implicit function theorem. Assume: X × Y ⊆ Rn × Rm is open;
(a, b) ∈ X × Y → Rm is C 1 -regular on X × Y and ∂f
∂y (a, b) ∈ L(m) is non
singular.
2
Then there exists an open subset U × V of X × Y containing (a, b), and a
C 1 -regular function g : U → Rm s.t. {(x, y) ∈ U ×V and f (x, y) = f (a, b)}
⇔ {x ∈ U and y = g(x)}
h i−1 h i
Moreover g 0 (x) = − ∂f ∂y
∂f
∂x (x, y).
Higher order derivatives and Taylor formula
We only consider derivatives of order 2.
• Assume X ⊆ Rn is open and f : X → Rm , is differentiable on X.
We say that f is twice differentiable at a ∈ X iff the mapping f 0 : X →
L(Rn , Rm ) is differentiable at a.
• We denote the second derivative by f 00 (a): f 00 (a) ∈ L(Rn , L(Rn , Rm )).
∂f ∂f
• Assume that f is differentiable on X and f 0 (x) = ∂x 1
(x), · · · , ∂xn (x)
is differentiable at a (f 0 : X → Rn ).
∂2f ∂2f
Then ∂xi ∂xj (a) = ∂xj ∂xi (a) for all i, j = 1, · · · , n.
• The Hessian matrix of f is the matrix of second order partial derivatives
of f :
∂2f ∂2f ∂2f
2 ∂x1 ∂x2 ··· ∂x1 ∂xn
∂∂x2 f1 ∂2f ∂2f
∂x2 ∂x1 ∂x22
··· ∂x2 ∂xn
[H(f )] = .. .. ..
. . .
∂2f ∂2f ∂2f
∂xn ∂x1 ∂xn ∂x2 ··· ∂x2n
. The Hessian matrix is always symmetric.
• Taylor formula of order 2. Assume X ∈ Rn , open and f : X → Rm . If
f is twice differentiable at a ∈ X, then:
||f (x) − f (a) − f 0 (a) · (x − a) 12 f 00 (a)(x − a, x − a)|| = o(||x − a||2 )
Convex functions
• Let ]a, b[ be an open interval of R, and f :]a, b[→ R a convex function.
Then f has a right derivative and a left derivative at every x ∈]a, b[:
0 f (x+t)−f (x) f (x+t)−f (x)
. f+ (x) = limt→0,t>0 t = inf t>0 t
.
0
f− (x) = limt→0,t>0 f (x+t)−f
t
(x)
= inf f (x+t)−f (x)
t
0 0
Moreover, for all x1 , x2 ∈]a, b[ s.t. x1 < x2 ⇒ f− (x1 ) ≤ f+ (x1 ) ≤
0 0
f− (x2 ) ≤ f+ (x2 )
• Let C be an open convex subset of Rn , and f : C → R a convex function.
The directional derivative of f at a ∈ C in the direction v is: Df (a, v) =
limt→0,t>0 1t {f (a + tv) − f (a)} = inf t>0 1t {f (a + tv) − f (a)}.
3
• Characterization of convex, differentiable functions (important!)
. Assume that f is twice differentiable on C. Then f is convex (resp.
strictly convex) on C if and only if its second derivative (the Hessian
matrix) is positive semidefinite (resp. positive definite).
. Assume that f is twice differentiable on C. Then f is concave (resp.
strictly concave) on C if and only if its second derivative (the Hessian
matrix) is negative semidefinite (resp. negative definite).
