2
x + y − 4)
( x , y )→ ( 0 ,0 )
⎡ x − y ⎤
2.
lim
( x, y) (0,0 ⎢
2
)
⎥
→
2
⎣ x + y ⎦
⎡ (x − 2
)
1 ln( x) ⎤
3.
lim
⎢
Consider limits as (x, y) → ,
1
( 0) along
2
⎥
( x , y )→
2
(1,0 )
⎣ (x − )
1 + y ⎦
the three: along x = 1; along y = 0 , and along y = x − 1 . Do you suspect
anything?
Continuity of a function of two variables
In Unit 1 of this Module we came across the concept of continuity of a function
of one independent variable on two occasions.
The first time was in Section 1.4 on Key Concepts where the concept of conti-
nuity at a point x = a was first defined. At that juncture we listed three necessary
and sufficient conditions for a function f ( x) to be continuous at a point x = a
namely:
(i) The function f must have a value (defined) at the point, meaning that f ( a)
must exist;
(ii) The function f must have a limit as x approaches a from right or left,
meaning that lim f (x) must exist (let the limit be L ); and
x→a
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(iii) The value of the function at x = a and the limit at x = a must be equal, meaning L = f ( a) .
In short, we summed up these conditions with the single equation lim f (x) = f (a)
x→a
The second time we came across the concept of continuity was in Section 1.7.2,
where we learnt how to determine whether or not a given function f ( x, y) is
continuous at a point or over an entire interval a ≤ x ≤ b , including what was
referred to as removable discontinuity.
In the current unit (Unit 4) we are dealing with functions of more than one inde-
pendent variable. Specifically, we are considering a function f of two indepen-
dent variables: w = f ( x, y)
We want to define the concept of continuity for a function of two independent
variables. Again we note that we already have defined the concept in Section 1.4
of this Unit. The definition is exactly the same as that for a function of a single
variable, namely, f ( x, y) is said to be continuous at a point ( a, b) in its domain if, and only if
lim
f ( ,
x y) = f (a, )
b . In short,
( x, y)→(a,b)
(i) If either f ( x, y) is not defined at point ( a, b) or
(ii) If
lim
f (x, y) does not exist, or
( x , y )→ ( a ,b)
(iii) If
lim
f (x, y) exists but is different from f ( a, b)
( x , y )→ ( a ,b)
then we can conclude that f ( x, y) is not continuous (discontinuous) at ( a, b) .
In the special case where the function is defined at the point and the limit exists,
but the limit and the function value are different, one speaks of a removable dis-
continuity, for one can under such circumstances define a new function
⎧
f ( ,
x y)
if ( ,
x y) ≠ (a, )
b
⎪⎪
F ( ,
x y) = ⎨
⎪⎪ lim f ( ,x y) if ( ,x y) = (a, )b
⎩(x,y)→(a,b)
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that clearly satisfies all the three stated conditions for continuity at ( a, b) .
Examples
We wish to discuss continuity of each of the following three functions at the
points given. The three examples have purposely been chosen to demonstrate the
three possible situations: continuity, removable discontinuity and non-removable
discontinuity.
1. f (x, y) = xy2 − 2x2 y + 3x at P( )
1
,
2 .
We observe that f is a polynomial of degree two in two variables. A polyno-
mial function is continuous everywhere, and therefore it is continuous at (
)
1
,
2 .
Specifically we note that:
• The function is defined at the point:
f ( )
1
,
2
= 0
• The function has a limit at the point: lim ( 2
xy − 2 2
x y + 3x)= 0
( x, y)→(2 )
1
,
• The limit is equal to the value of the function at the point. Therefore, the
function is continuous at (2,1)
[1 y 2
+
]sin( x)
2. f ( x, y) =
at P(
)
0
,
0
.
x
We note that this function is not defined at point: f (
)
0
,
0
does not exist, and
therefore, the function is not continuous at the origin. However, we note that the
function has a limit at the origin, since
[1 2
+ y ]sin(x)
x
lim
⎡
= lim(1
2
+ y )⎤⎡
sin( )
lim
⎤ =1×1 =1
( x, y)→(0,0)
x
⎢ y→0
⎣
⎥⎢x→0
⎥
⎦⎣
x ⎦
Because the limit exists but the function is discontinuous, we
have a typical case of a removable discontinuity. The derived function
⎧ f ( ,
x y) if ( ,
x y) ≠ ( )
0
,
0
⎪
F ( ,
x y) = ⎨
is continuous at (
)
0
,
0
.
⎪⎩ 1
if ( ,
x y) = ( )
0
,
0
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x
3. f ( x, y) =
at P(
)
0
,
0
.
2
2
x + y
Clearly, this function is not defined at the origin and therefore, it is discontinuous
at the point. As to whether the function has a limit at the origin, we note the
following:
If we approach the point (0,0) along the along the y-axis (x = 0) we see that the
function is identically zero and hence it has zero limit.
If, on the other hand, one approaches the origin along the x-axis (y = 0) we note that
1
the function takes the form f ( x, y) = , which has no limit as x → 0 . We
x
therefore conclude that the function has no limit and hence is not continuous at
(
)
0
,
0
. In this case the discontinuity is not removable.
Now Do This
Discuss the continuity of the following three functions at the points given. If
the function is discontinuous, state giving reasons, whether the discontinuity is
removable or not removable.
2
xy − y
1. f ( ,
x y) =
; (
)
1
,
2
2
1− x − y
y
2. f (x, y) =
; (
)
0
,
0
x − y
1 − x + y
3. f (x, y) =
; (
)
0
,
0
2
2
x + y
Partial Differentiation
One of the key concepts covered in the second learning activity of Unit one is the
derivative of a function f ( x) of a single independent variable. In Section 1.4
of this Unit a key concept of partial derivatives of a function f ( x , x , x ,... x ) 1
2
3
n
of several independent variables is mentioned. We now make a formal definition
of this important concept specifically in connection with functions f ( x, y) of
two independent variables.
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Partial Derivatives
Let w = f ( x, y) be a function of the two variables ( x, y) in a domain 2
D ⊂ ℜ
of the xy − plane and ( a, b) be a point in D . The function g( x) = f ( x, b) , obtained from f ( x, y) by holding y at the constant value y = b is a function of x alone and is defined over an interval on the x– axis that contains the value
x = a . One may therefore wish to find out whether g is differentiable at x = a This involves finding
⎡ g(a + h) − g(a)⎤
f (a + h,b) f (a,b)
⎡
−
⎤
lim
= lim
→
h 0 ⎢
⎥
→
h 0 ⎢
⎥
⎣
h
⎦
⎣
h
⎦
If the limit exists it is called the partial derivative of f ( x, y) with respect to the variable x at point ( a, b) and is denoted by
∂f
∂w
(a, )
b or f ( a, b) or
∂x
x
∂x
f
∂
The partial derivative
(a, )
b is defined similarly by holding x constant at
y
∂
the value x = a and differentiate the function h( y) = f ( a, y) at y = b .
If we allow the point ( a, b) to vary within the domain D then the resulting
partial derivatives will also be functions of the variable point ( x, y) . We there-
fore have
⎡ f (x + ,
h y) − f ( ,
x y)⎤
f
lim
∂
=
( ,
x y)
;
h→0 ⎢
⎣
h
⎥⎦
x
∂
⎡ f ( ,
x y + k) − f ( ,
x y)⎤
f
lim
∂
=
( ,
x y) .
k→0 ⎢
⎣
k
⎥⎦
y
∂
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Worked Example
Find the partial derivatives of the function
3
2
2
3
f ( x, y) = 2 x y + 3 x y + 2 xy
By considering y as a constant and differentiating f as if it were only a
f
∂
function of the variable x we get
2
2
3
( ,
x y) = 6x y + 6xy + 2y
.
x
∂
Similarly, by considering x as a constant and differentiating f as if it were
f
∂
only a function of the variable y one gets
3
2
2
( ,
x y) = 2x + 6x y + 6xy .
x
∂
Now Do This
Find the partial derivatives of the following functions:
⎡ 2
x ⎤
1. f ( x, y) = ln ⎢ 3 ⎥
⎣ y ⎦
sin( 2θ )
2. f (r ,θ ) =
2
r
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Higher Order Partial Derivatives
∂f
f
∂
We have seen that the partial derivatives
and
of a function f ( x, y)
∂x
y
∂
of two independent variables are in turn functions of the same two independent
variables. Therefore, one can consider their partial derivatives in the same way we
did for the original function. One may therefore attempt to find the derivatives
∂ ⎡ ∂f ⎤
∂ ⎡ ∂f ⎤
∂ ⎡ ∂f ⎤
∂ ⎡ ∂f ⎤
⎢
⎥ ;
;
;
.
∂x
⎢
⎥
⎢
⎥
⎢
⎥
⎣ ∂x ⎦
∂x ⎣ ∂y ⎦ ∂x ⎣ ∂y ⎦ ∂y ⎣ ∂y ⎦
If limits of the associated difference quotients exist, then we call them the second
partial derivatives of f and denote them by
∂ ⎡ ∂f ⎤
2
∂ f
∂ ⎡ ∂f ⎤
∂ 2 f
⎢
⎥ =
or f ;
=
or f ;
∂x
xx
xy
⎣ ∂x ⎦
2
x
∂
⎢
⎥
∂y ⎣ ∂x ⎦
∂y∂x
∂ ⎡ ∂f ⎤
∂ 2 f
∂ ⎡ ∂f ⎤
2
∂ f
⎢
⎥ =
or f ;
=
or f .
∂x
yx
⎢
⎥
2
yy
⎣ ∂y ⎦ ∂x∂y
∂y ⎣ ∂y ⎦
y
∂
Note Carefully
The learner should note carefully the notation used for the two mixed second
∂ 2 f
∂ 2 f
∂ 2 f
partial derivatives
and
. The notation
or f means that
∂y∂x
∂x∂y
∂y∂x
xy
one differentiates first with respect to x and then with respect to y , while
∂ 2 f or f means differentiation first with respect to y and then with
∂x∂y
yx
respect to x .
Worked Example
Find all four second partial derivatives of the function f ( x, y) = y cos( xy) .
One finds the following results
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∂f
= f =
2
− y sin( xy) ;
∂x
x
f
∂
= f = cos( )
xy − xysin( )
xy ;
y
∂
y
2
∂ f
= f =
3
− y cos( )
xy ;
2
x
∂
xx
∂ 2 f
= f = − 2ysin( )
2
xy − xy cos( )
xy ;
∂y∂x
xy
∂ 2 f
= f = − xsin( )
xy − ysin( )
2
xy − xy cos( )
xy ;
∂x∂y
yx
2
∂ f
= f = − 2xsin( )
2
xy − x ycos( )
xy
2
y
∂
yy
Now Do This
Find the second partial derivatives of the following functions
1. f ( x, t) =
3
2
2
3
2 x y + 3 x y + 4 xy .
2. f ( x, y) = ln( 2 3
x y ) .
Equality of Second Mixed Partial Derivatives
From the above worked example and from the results the Learner will get by
solving the last two self exercise problems, do you detect any relationship between
the two mixed partial derivatives f and f ? State the relationship and look
xy
yx
back at the key theorem stated in item 5 of Section 1.5 on Key Theorems and/or
Principles.
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What we find is that in all cases the mixed second partial derivatives are equal.
This result is not by accident but a well established result based on the theorem
referred to and which states that if f and f are both continuous over the domain
xy
yx
2
D ⊂ ℜ of f ( x, y) then f and f are equal at all points (x, y) ∈ D .
xy
yx
C l e a r l y, p a r t i a l d e r i v a t i v e s o f o r d e r h i g h e r t h a n t w o c a n be computed in exactly the same way by differentiating second partial derivatives. This will yield eight third partial derivatives, namely
f , f
, f , f
, f
, f
, f
, f
.
xxx
xxy
xyx
xyy
yxx
yxy
yyx
yyy
Do this
1. Find all eight third partial derivatives of the function f ( x, y) = cos(2 x + 3 x) note the equality of the following sets of mixed third partial derivatives: