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Three important theorems in advanced calculus

Sean Fitzpatrick

February 25, 2013

Source: Marsden, J. E. and Tromba, A. J., Vector Calculus, 4th ed. W. H. Freeman andCompany, New York, 1996.

We state three theorems of theoretical importance in multivariable calculus: the chainrule, the implicit function theorem, and the inverse function theorem. The first two arementioned in Stewarts text, but not in their most general form, and the last is not mentioned

at all. Well give a general proof of the chain rule, and state the implicit and inverse functiontheorems. (The proofs can be found in more advanced texts on real analysis.)

1 The Chain Rule

For a general function f : U Rn Rm, the derivative Df(x0) at a point x0 U is themnmatrix whose entries are given by the partial derivatives off. That is, if

f(x1, . . . , xn) = (f1(x1, . . . , xn), . . . , f m(x1, . . . , xn)),

where each of the component functions fiis a real-valued function of the variables x1, . . . , xn,

then the entry in the ith row and jth column ofDf(x0) is fixj

(x0). The chain rule tells us

that the derivative of a composition is given by the product of the derivatives, just as forthe case of single-variable functions. First, we need a fact about linear functions:

Lemma 1.1. LetT : Rn Rm be a linear function given byT(x) =A x, whereA= [aij ]is an m n matrix. Then T is continuous, and in particular, T(x) Mx, where

M=m

i=1

nj=1a

2ij.

Proof. The components ofTare given byTi(x) =n

j=1aijxj =aix, where ai = ai1, . . . , ain.Thus,

T(x)=

(T1(x)2 + +Tm(x)2

=

|a1 x|2 + +|am x|2

a12x2 + am2x2 (Cauchy-Schwartz inequality)

=

(a12 + am2)x2

=Mx.

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Theorem 1.2 (Chain Rule). Let U Rn and V Rm be open. Let g : U Rm andf : V Rp be given functions such that the range ofg is contained inV, so thatfg isdefined. Ifg is differentiable atx0 U andf is differentiable aty0= g(x0) V, thenf gis differentiable atx0 and

D(fg)(x0) =Df(g(x0))Dg(x0). (1)Proof. Using the definition of differentiability, we need to prove that the right-hand side of(1) defines a linear function from Rn to Rp such that

limxx0

f(g(x))f(g(x0)) Df(y0)Dg(x0)(x x0)

x x0 = 0.

The result then follows from the uniqueness of the derivative. By adding and subtractingDf(y0)(g(x)g(x0)) in the numerator and applying the triangle inequality, we get, withy= g(x) and y0 = g(x0),

f(y)f(y0) Df(y0)Dg(x0)(x x0) f(y)f(y0) Df(y0)(y y0)+Df(y0)(g(x)g(x0) Dg(x0)(x x0)).

Let >0 be given. According to the lemma above Df(y0) v Mv for any v Rm,

for some constant M >0. We will apply this for v = g(x) g(x0) Dg(x0) (x x0). Sinceg is differentiable at x0, we can find a 1> 0 such that 0 0 and a constant N such that0 < x x0 < 2 implies g(x) g(x0) Nx x0. Since f is differentiable at

y0 = g(x0), we can find a 3> 0 such that 0< y y0< 3 implies that

f(y)f(y0) Df(y0)(y y0)

2Ny y0=

2Ng(x)g(x0)