Lectures 26-27: Functions of Several Variables
(Continuity, Differentiability, Increment Theorem and Chain Rule)
The rest of the course is devoted to caleulus of several variables in which we study continuity,
differentiability and integration of functions from R" to R, and their applications.
In calculus of single variable, we had seen that the concept of convergence of sequence played
an important role, especialy, in defining limit and continuity of a function, and deriving some
properties of R and propertics of continuous functions. This motivates us to start with the notion
of convergence of a sequence in R". For simplicity, we consider only R or R. General case is
entirely analogous.
Convergence of a sequence Let Xn = (T1,n, T2,n, T3,n) E RS. We say that the sequence (X,)
is convergent if there exists Xo ¬ R" such that ||X, - Xol| - 0 as n
Xn converges to Xo and we write Xn
oo. In this case we say that
Xo
Note that corresponding to a sequence (XT), Xn = (T1,n, T2,n, F3,n), there are three sequences
can be completely underof
in
and
and vice-versa. Thus the
(T3,n)
(1,n)(T2,n)
properties
R,
(X,)
stood in terms of the properties of the corresponding sequences (T1,n)(T2n) and (T3,n) in R. For
example,
)
X X o in R3the coordinates Tin # i 0 for every i=1,2,3 in R.
(Gi) (X,)
is bounded
(i.e.,
3 M such that
||X,|
bounded.
Using
the
previous idea,
we can
prove the
MVn)
following
cach sequence
(in),i
=
1,2,3,
is
results.
Problem 1: Every convergent sequence R$ is bounded.
Problem 2 (Bolzano-Weierstrass Theorem): Every bounded seguence in Rs has a converyent
subsequence.
In case of a sequence in R, to define the notion of convergence or boundedness, we use|| in
place of ,hence it is clear how we generalized the concept of convergence or boundedness of
a sequence in R to R°. Moreover, it is also now clear how to define the concepts of limit and
continuity of
a
function
f : RS
R at
some
point Xo
E
RS.
Limit and Continuity : ()We say that L is the limit ofa function f: RS--R at Xo E R* (and
we write limx-xo fX) = L) if fX,) -- L whenever a sequence (X,) in R, Xn t Xo, converges
to Xo.
(i) A function f: R3-Ris continuous at Xo eR$ iflimx-Xo f(X) = f(Xo).
Examples 1: () Consider the function f: R2- R, where fla. v) = *
when (,) # (0,0)
and f(0,0) =0. We will show that this function is continuous at (0,0). Note that
Ifv)-f(0,0)|
-y2
Sc|+|u| (or | r-v|)
Therefore, whenever a sequence (Tn Jn)- (0,0), i.e, #n0 and n 0 , we have f(Tns Un)
f0,0). Hence f is continuous at (0,0). In fact, this function is continuous on the entire R.
(i) Consider the function f: R2 -- R, where f(7, v)=
This function is continuous at (0,0), because,I
when ( ,v) # (0,0) and f(0,0)=0.
S
+
0 , as (r,y) -~0.
(ii) Let f(7, y)=
( ,y)
(0,0). Note that f(r, ma)
have a limit at (0,0).
(iv) Let f ( , v)=
(0,0). We will show that this function does not have a limit at
0
1 m
for any m. This shows that the function does not
when ( ,u) # (0,0) and f(0,0) = 0. Note that f(r, ma) 0
But the function is not oontinuous at (0,0) because f(r, a ) that the function f(z, u) defined by f(r,1) =
as z 0 .
as r 0 . Similarly we can show
when (, )
(0,0) and f(0,0) = 0 is not
continuous at (0,0) by taking y = ma and allowingr-0.
Partial derivatives: The partial derivative off with respect to the first variable at Xo =
(O, 0, 20) is defined by
xlimo+o,
20)-f(#o, o,2o)
oj x
h+0
provided the limit exists. Similarly we define 1xo and
Ixo
Example 2: The function f defined by fla,u) =
at ( ,y) # (0,0) and S0,0) = 0 is not
continuous at
at
(0,0), however,
the
partial derivatives exist
Problem 3: Let fr, u) be defined in S
(0,0).
{(a,y) E R2: a <u <b, c<y<d). Suppose that the
partial derivatives of f ezist and are bounded in S. Then show that f is continuous in S.
Solution: Let |fa(r, v)| S M and If,(r, V)] S M for all (z,y) E S. Then
fr+h,y
+
k)
-
f(r,v)
=
=
Hence, 1f( +h,y + k)
-
f(r+h,y +k) - fr +h,) +f(r +h,y) fa, )
kfy(r+h,y+0,k) +hfe(a +02h, ), (for some 1, G2 E R, by the MVT).
-
f(z, v) |s M(|h| + |k|) < 2M V2
+P
Hence, for e> 0, choose 6 = zM or use the sequential argument to show theat the function is
continuous.
It is clear from the previous example that the concept of differentiability of a function of several
variables should be stronger than mere existence of partial derivatives of the function.
Differentiability: When f: R
R, r E R we define
f(a)= lim
h+0
+)- Sa)
()
provided the limit exists. In case f: R$
R and X = (T1, t2, Ta) E R" the above definition of
the differentiability of functions of one variable
(") cannot be generalized as we cannot divide by
an clement of R.
in
order
to
define
the
So,
concept of differentiability, what we do is that we
rearrange the above definition (*) t a form which can be generalized.
Let f: R
R. Then f is differentiable at z if and only if there exists a ER such that
S(
When f is differentiable
scveral variables.
at ,
a
+h)-S(a)-a-hl
o
h
has to be
f (c).
We
ash-0.
generalize
this definition to the
functions
of
3
Let
Definition:
R and X
f: R3
=
(a,T2,#g).
f
We say that
is
differentiable
at X
if there
exIsts a = (a1, a2, as) E R" such that the error function
e(H)
=
JtH)-SOX)
-a
H
tends to 0 as H 0 .
In the above definition
Theorem 26.1:
Proof:
Let
product. Note that
is the scalar
aH
f: R3
R,
If f
X e RS.
is differentiable at X then
Then there exists
Suppose f is differentiable at X.
a
=
as
=
(a1, a2, a3).
is continuous at X.
f
(a1, 02, a3)
and e(H) - 0
1SX+H) -S(X) - a H|= || H| e(H)
Hence
f(X)
the derivative
E
RS such that
H 0.
3
ISX +H)-fX)|E| H I a D+ || H|| e(H
i=1
and e(H)0
at X.
as
H0.
Therefore f(X +H)>f(X)
as
0. This proves that
H
that a given function is differentiable at
helps us to answer this question.
How do
we
verify
a
point
in R$? The
f
is continuous
following
Theorem 26.2: Suppose f is differentiable at X. Then the partial derivatives
x
result
and
Ix exist and the derivative
fx)-(o1,os, os)=
Proof:
Suppose f is
differentiable at X and
)
f'(X)
(a1, a2, a3).
=
Then
H
by taking
=
(t,0,0),
we have
s()=
- aqt
J&+H)-fX)
This implies that a
=x
as
t0, i.e.,
f(X+H)-fX)-at0
Similarly we can show that ag = o
ly and ag =
lx
Example 3: Let
fz,v) = zy
2at (z,1) # (0,0)
= 0 at (0,0)
To
as
verify
H=
f
that
(h, k)
le(E) |
-
is differentiable at
0. In this
case a
=
(0,0), let
(0, 0) and
us
choose
(0,0)-.H,
|J0+H)-S0)=
a=
hk
IHI
( oo
and
verify that e(H)
l s Vh2+k2-0
as
0
H-0.
Hence f is differentiable at (0,0).
Example
2 illustrates that the
partial
derivatives of
that point. The
function need not be differentiable at
a
a point may exist but the
theorem says that if the function is
function at
previous
4
differentiable at X then the derivative f'(X) can be expressed in terms of the partial derivatives
of f at X. Since finding partial derivatives is easy because they are based on one variable and
it is related to the derivative, one naturally asks the following question: Under what additional
assumptions on the partial derivatives the function becomes differentiable. The following criterion
answer this question.
Theorem 26.3: If f : R3
R is such that all its partial derivatives exist in a neighborhood of Xo
and continuous at Xo then f is differentiable at Xo.
We omit the proof of this result. We will see in a tutorial class that the converse of the previous
result is not true.
komplf(xM) = *7)E 21
)E9»)
C
) ofIorD
Chain Rule: We have seen that the chain rule which deals with derivative
a function of a
function is very useful in one variable calculus. In order to derive a similar rule for functions of
several variables we need the following theorem called Increment Theorem. For simplicity we
will state this theorem only for two variables.
We will enploy the notation f =and fy
=
Theorem 26.4: Let f(r, y) be differentiable at (ro, 30). Then we have
f o + Ar,30 + Ag) -
f(ro, vo)
=
fe(To, vo)Ar + fy(ro, vo) Ay +61Ar +62Ay
where e1(Ar, Ay), e2(Ar, Ay) > 0 as Ar -0 and Ay0.
Proof (): Let H = (Az, Ay). Since the function is differentiable at (ro, Jo), we have
fro+ Az,o+Ay) - f(ro, vo) = f=To, vo) Aar + fu(to, 30)Ay + | |H|| e(H),e ( H ) - 0 as H 0 .
We have to show that|
H | E(H) = e1Aw +e2Ay for some functions e and e2. Note that
eH) | H|
e(H)
(Az Ar+(AyTH)Ay
=H(A+A)
=
Defineex(H)= A a n d ea(H) = Av
Note
s(H) s|e(H)|
lea(H)= | Ar
-0 as
H0.
Similarly we can show that eo(H)-0 as H 0 . This proves the result.
In the next result we present the chain rule.
Theorem 26.5: Let fla, ) be differentiable (or f has continuous partial derivatives) and if a=
z(t), y = y(t) are differentiable functions on t, then the function w = f(r(t), v(t)) is differentiable
at t and
df
fCl),u(0))=() +i,at®),v0)W), ie,
4d Oy dt
Proof: By increment theorem we have
Af = fzo, 3o)Ar + fy(To, vo)Ay + 61Ar +e2Ay,
E1,620 as Ar, A y 0
This implies that
Ac
Ay
J
At +fu+E1 AItA
Allow At
f dr +oy
0, which implies that e1,62 0
because Ar,Ay
0. Therefore, we get