1
MA 1024 - Methods of Mathematics
Lecturers :
Lecturer
Dr.Bimali Jayasinghe
Mrs.Ravindi Jayasundara
Dr.Miuran Jayaweera
Contacts
Email:
bimalij@uom.lk
Phone: +94 76 227 1560
Email: ravindij@uom.lk
Phone: +94 71 814 6791
Email: miurand@uom.lk
Phone: +94 77 808 1777
Mode of Delivery: Online - Zoom classes each week on scheduled time. Link will be posted in course moodle page.
Learning Outcomes: After the successful completion of this course students should be able to
• Understand the basic theories on several variable calculus and its applications.
Course Content:Multivariate Calculus and Introductions to PDEs
• Limits, Continuity, Partial Derivatives, Mean Value Theorem, Differentiability
• Chain rule-Bivariate
• Gradient, Tangent plain, Directional derivatives
• Jacobian, Hessian
• Inverse Function Theorem, Implicit Function Theorem.
• Maxima, minima and Saddle points, Lagrange multipliers
• Taylor series expansion for two variable, quadratic forms
• Double integrals: Fubini’s theorem, Change of variables, polar coordinates
• Solution of the exact ODE
• Introduction and solve first order PDE, Solution by the method of Characteristics
References:
• Advanced Calculus – David V. Widder
• Mathematical Analysis - Tom M.Apostol,
Assessment Criteria:
Final Exam: 70%
In Class Tests Assignments: 30%
Chapter I
Functions of Several Variables
Definition 1.0.1 (A real valued function with n–variables) Let D be a subset of Rn . A function F of n
variables, also called a function F of several variables, with domain D, is a relation that assigns to every ordered
n−tuple in D a unique real number in R. We denote this by each of the following types of notation.
F :D→R
x→y
(x1 , x2 , ..., xn ) → y
The range of F is the set of all outputs of F . It is a subset of R, not Rn .
Example 1.0.1 Consider
F : R2 → R
(x, y) → x2 + y 2
Find the domain and range of F .
Moreover, If the output of a function consists of multiple numbers, it can also be called multivariable, but
these ones are also commonly called vector-valued functions. i.e. Multivariable Function of m–variables is a
function F : D → Rn , where the domain D ⊂ Rm . So: for each (x1 , x2 , ..., xm ) ∈ D, the value of F is a vector
F (x1 , x2 , ..., xm ) ∈ Rn .
Example 1.0.2 An object rotating around the origin t distance 5 from the origin) in the xy-plane will have its
position described by the function f (t) = (cos t, sin t). This is a function from R to R2 .
1.0.1
Surfaces
Definition 1.0.2 If f is a function of two variables with domain D, then the graph of f is the set of all points
(x, y, z) in R3 such that z = f (x, y) and (x, y) is in D.
Just as the graph of a function f of one variable is a curve with equation y = f (x) so the graph of a function f
of two variables is a surface with equation z = f (x, y). We can visualize the graph of as lying directly above or
below its domain in the xy-plane.
The domain of f is D and the co-domain of f is R. The range of f is
{z ∈ R : z = f (x, y) for some (x, y) ∈ D}.
2
3
Example 1.0.3 Sketch the graph of the function f (x, y) = 6 − 3x − 3y.
Solution 1.0.1 The graph of has the equation z = 6 − 3x − 3y, or3x + 2y + z = 0, which represents a plane. To
graph the plane we first find the intercepts. Putting z = 0 and y = 0 in the equation, we get as the x-intercept.
Similarly, the y-intercept is 3 and the z-intercept is 6. This helps us sketch the portion of the graph that lies in
the first octant.
This is a special case of the function
f (x, y) = ax + by + c
which is called a linear function. The graph of such a function has the equation
z = ax + by + c or ax + by − z + c = 0
so it is a plane. In much the same way that linear functions of one variable are important in single-variable
calculus.
Example 1.0.4 Sketch the graph of the surface z = x2 .
Notice that the equation of the graph, z = x2 , doesn’t involve y. This means that any vertical plane with equation
y = k (parallel to the xz-plane) intersects the graph in a curve with equation z = x2 . So these vertical traces
are parabolas.
The graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola.
Here the rulings of the cylinder are parallel to the y-axis.
Example 1.0.5 Use traces to sketch the quadric surface with equation
x2 +
y2
z2
+
= 1.
9
4
By substituting z = 0, we find that the trace in the xy-plane is x2 +
of an ellipse. In general, the horizontal trace in the plane z = k is
x2 +
y2
k2
=1− ,
9
4
z=k
y2
9
= 1, which we recognize as an equation
4
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
which is an ellipse, provided that k 2 < 4, that is, −2 < k < 2.
Similarly, vertical traces parallel to the yz- and xz-planes are also ellipses:
if −3 < k < 3,
x2 +
z2
k2
=1− ,
4
9
y=k
if −1 < k < 1,
z2
y2
+
= 1 − k2 ,
9
4
It’s called an ellipsoid because all of its traces are ellipses
x=k
Following table shows graphs of the six basic types of quadric surfaces in standard form. All surfaces are
symmetric with respect to the z- axis. If a quadric surface is symmetric about a different axis, its equation
changes accordingly.
5
1.0.2
Level curves
Definition 1.0.3 The level curves of a function f of two variables are the curves with equations f (x, y) = k ,
where k is a constant (in the range of f ).
A level curve f (x, y) = k is the set of all points in the domain of f at which f takes on a given value k. In
other words, it shows where the graph of f has height k. The graph of f can be built up from the level sets:
The slice at height z = k, is the level set f (x, y) = k.
Example 1.0.6 For the elliptic paraboloid z = x2 + y 2 , for example,the level curves will consist of concentric
circles. For, if we seek the locus of all points on the paraboloid for which z = 21 , we solve the equation
1
= x2 + y 2
2
which is of course an equation of a circle. The locus of points 0 units above the xy- plane is just the origin, for
if z = 0,the equation becomes x2 + y 2 = 0, and this implies that x = y = 0.
□
Example 1.0.7 Find the domain of f if
f (x, y, z) = ln(z − y) + xy sin z
Solution 1.0.2 The expression for is defined as long as z − y > 0, so the domain of f is
D = {(x, y, z) ∈ R3 |z > y}
This is a half-space consisting of all points that lie above the plane z = y.
It is very difficult to visualize a function f of three variables by its graph, since that would lie in a fourdimensional space. However, we do gain some insight into f by examining its level surfaces, which are the
surfaces with equations f (x, y, z) = k, where k is a constant.
Example 1.0.8 Find the level surfaces of the function
f (x, y, z) = x2 + y 2 + z 2 .
The level surfaces are x2 + y 2 + z 2 = k, where k ≥ 0 . These form a family of concentric spheres with radius
√
k.
□
6
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
Let D be a subset of the plane R2 and let (a, b) ∈ R2 be any point.
• An ϵ-disk around (a, b) is the set of all points (x, y) ∈ R2 whose distance from (a, b) is less than ϵ.
• (a, b) is an interior point of D iff some ϵ-disk around (a, b) is contained in D.
• (a, b) ∈ D is an isolated point of D iff (a, b) is the only point of D that is contained in some ϵ-disk
around (a, b).
• (a, b) is a boundary point of D iff every ϵ-disk around (a, b) contains points from D and points not from
D.
• R is an open subset of R2 iff all points of D are its interior points and D is a closed subset of R2 iff it
contains all its boundary points.
• D̄ = D ∪ {thesetof boundarypointsof D}; It is the closure of D. D is a bounded subset of R2 iff D is
contained in some ϵ -disk (around some point).
Neighbourhoods
• Circular Neighbourhoods: Let D be the domain of a two variable function f . The set
p
{(x, y)|
(x − a)2 + (y − b)2 < ∆}
is called a δ− neighbourhood of the point (a, b), where (a, b) ∈ D and δ > 0.
• Square Neighbourhoods: Let D be the domain of a two variable function f . The set
{(x, y)| |x − a| < δ and |y − b| < δ}
is called square neighbourhood of the point (a, b), where (a, b) ∈ D and δ > 0.
1.1. LIMIT
1.1
7
Limit
Definition 1.1.1 (Limit) Let f : D → R be a function, where D is a region in the plane. Let (a, b) ∈ D.
Then we say that the limit of f (x,
p y) as (x, y) approaches (a, b) is L iff for each ϵ > 0, there exists δ > 0 such
that for all (x, y) ∈ D with 0 < (x − a)2 + (y − b)2 < δ,we have
|f (x, y) − L| < ϵ.
In this case, we write
lim
f (x, y) = L.
(x,y)→(a,b)
Other notations for the limit are
lim f (x, y) = L and f (x, y) → L as (x, y) → (a, b)
x→a
y→b
p
Notice that |f (x, y) − L| is the distance between the numbers f (x, y) and L , and (x − a)2 + (y − b)2 is the
distance between the point (x, y) and the point (a, b).Thus Definition (1.1.1) says that the distance between
f (x, y) and L can be made arbitrarily small by making the distance from (x, y) to (a, b) sufficiently small (but
not 0). The following Figure illustrates Definition 1.1.1 by means of an arrow diagram. If any small interval
(L − ϵ, L + ϵ) is given around L, then we can find a disk Dδ with center (a, b) and radius δ > 0 such that f
maps all the points in Dδ [except possibly (a, b) ] into the interval (L − ϵ, L + ϵ).
Example 1.1.1 Determine if
4xy 2
exists.
(x,y)→(0,0) x2 + y 2
lim
Observe that the region D of f is R2 \{(0, 0)}. And f (0, y) = 0 for y ̸= 0; f (x, 0) = 0 for x ̸= 0. We guess that
if the limit exists, it would be 0. To see that it is the case, we start with any ϵ > 0. We want to choose a δ > 0
such that the following sentence becomes true: If
0<
p
x2 + y 2 < δ then
4xy 2
< ϵ,
+ y2
x2
Since |y 2 | = y 2 ≤ x2 + y 2 and |x2 | = x2 ≤ x2 + y 2 , we have,
p
4xy 2
≤ 4|x| ≤ 4 x2 + y 2 .
2
+y
x2
So, we choose δ = ϵ/4. Let us verify whether our choice is all right. Assume that 0 <
p
x2 + y 2 < δ. Then
p
4xy 2
−
0
x2 + y 2 < 4δ = ϵ.
≤
4
x2 + y 2
Hence
4xy 2
= 0.
(x,y)→(0,0) x2 + y 2
lim
Example 1.1.2 Determine if
lim
p
1 − x2 − y 2 exists.
(x,y)→(0,0)
p
Consider f (x, y) =
1 − x2 − y 2 where D = {(x, y) : x2 + y 2 ≤ 1}, We guess that limit f (x, y) is 1 as
(x, y) → (0, 0). To show that our guess is right,p
let ϵ > 0. Observe that 0 ≤ f (x, y) ≤ 1 on D. Using our
observation, assume that 0 < ϵ < 1. Choose δ = 1 − (1 − ϵ)2 . Let |(x, y) − (0, 0)| < δ. Then 0 < x2 + y 2 <
1 − (1 − ϵ)2 =⇒ 1 − x2 − y 2 > (1 − ϵ)2 =⇒ f (x, y) > 1 − ϵ. That is, |f (x, y) − 1| = 1 − f (x, y) < ϵ. Therefore,
f (x, y) → 1 as (x, y) → (0, 0).
8
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
Theorem 1.1.1 (Uniqueness of limit) Let f (x, y) be a real valued function defined on a region D ⊆ R2 . Let
(a, b) ∈ D. If limit of f (x, y) as (x, y) approaches (a, b) exists, then it is unique.
Proof 1.1.1 Suppose f (x, y) → l and also f (x, y) → m as (x, y) → (a, b). Let ϵ > 0. For ϵ/2, we have
δ1 > 0, δ2 > 0 such that
0 < (x − a)2 + (y − b)2 < δ12 =⇒ |f (x, y) − l| < ϵ/2, 0 < (x − a)2 + (y − b)2 < δ22 =⇒ |f (x, y) − m| < ϵ/2.
Choose a point (α, β) so that both 0 < (α − a)2 + (β − b)2 < δ12 , 0 < (α − a)2 + (β − b)2 < δ22 hold.
Then
|f (α, β) − l| < ϵ/2and f (α, β) − l| < ϵ/2.
Now, |l − m| ≤ |l − f (α, β)| + |f (α, β) − m| < ϵ/2 + ϵ/2 = ϵ. That is, for each ϵ > 0, we have
|l − m| < ϵ. Hence l = m.
□
For a function of one variable, there are only two directions for approaching a point; from left and from right.
Whereas for a function of two variables, there are infinitely many directions, and infinite number of paths on
which one can approach a point
The limit definition (1.1.1) says that the distance between f (x, y) and L can be made arbitrarily small by
making the distance from (x, y) to (a, b) sufficiently small (but not 0). The definition refers only to the distance
between (x, y) and (a, b). It does not refer to the direction of approach. Therefore, if the limit exists, then
f (x, y) must approach the same limit no matter how (x, y) approaches (a, b). Thus if we can find two different
paths of approach along which the function f (x, y) has different limits, then it follows that
lim
f (x, y)
(x,y)→(a,b)
does not exist.
Theorem 1.1.2 Suppose that f (x, y) → L1 as (x, y) → (a, b) along a path C1 and f (x, y) → L2 as (x, y) →
(a, b) along a path C2 . If L1 ̸= L2 , then the limit of f (x, y) as (x, y) → (a, b) does not exist.
Example 1.1.3 Consider f (x, y) =
Example 1.1.4 Consider f (x, y) =
x2 − y 2
x2 + y 2
x2
for (x, y) ̸= (0, 0). What is its limit at (0, 0)?
xy
for (x, y) ̸= (0, 0). What is its limit at (0, 0)?
+ y2
Example 1.1.5 Consider f (x, y) =
xy 2
for (x, y) ̸= (0, 0). What is its limit at (0, 0)?
x2 + y 4
Example 1.1.6 Consider f (x, y) =
(y − x)(1 + x)
for x + y ̸= 0, −1 < x, y < 1.What is its limit at (0, 0)?
(y + x)(1 + y)
Example 1.1.7 Let f (x, y) = x sin
1.1.1
1
1
+ y sin for x ̸= 0, y ̸= 0. What is its limit at (0, 0)?
y
x
Iterated Limits/ Repeated Limit:
If f (x, y) is defined in a neighbourhood of (a, b) and if lim exists, which is a function of y only and then take
x→a
the limit of this function as y → b, we write
lim lim f (x, y)
y→b x→a
This is called the repeated limit/ iterated limit of f (x, y) as x → a first and then y → b.
Similarly the other possible repeated limit is
lim lim f (x, y)
y→b x→a
Note: The iterated limits lim
x→x0
n
o
o
n
lim f (x, y) and lim
lim f (x, y) also denoted by lim lim f (x, y) and
y→y0
y→y0
x→x0
lim lim f (x, y) respectively are not necessarily equal. Although they must be equal if
y→y0 x→x0
lim
(x,y)→(a,b)
exist, their equality does not guarantee the existence of this last limit.
Example 1.1.8 Find the repeated limits f (x, y) =
x→x0 y→y0
x2 − y 2
x2 + y 2
as (x, y) → (0, 0)
f (x, y) is to
1.2. CONTINUITY
9
Theorem 1.1.3 If the double limit
f (x, y) exist and the lim f (x, y) exists for each fixed y in a
lim
x→a
(x,y)→(a,b)
neighbourhood of (a, b) then the repeated limit lim lim f (x, y) exist and
y→b x→a
lim lim f (x, y) =
y→b x→a
lim
f (x, y)
(x,y)→(a,b)
Note:
• The two repeated limits may or may not exist and when exist these may or may not be equal.
• The existence of the double limit does not imply the existence of either of the two repeated limits
• If a repeated limit exists, along with the double limit, these two would be equal
• If the repeated limit exists and they are not equal, then the double limit cannot exist.
• The existence of double limit is not simply governed by the existence of repeated limits.
The usual operations of addition, multiplication etc (the algebra of limits) have the expected effects as the
following theorem shows. Its proof is analogous to the single variable limits.
Theorem 1.1.4 Let L, M, c ∈ R;
1. Constant Multiple :
lim
f (x, y) = L;
(x,y)→(a,b)
lim
g(x, y) = M. Then,
(x,y)→(a,b)
cf (x, y) = cL.
lim
(x,y)→(a,b)
2. Sum:
lim
(f (x, y) ± g(x, y)) = L ± M.
(x,y)→(a,b)
3. Product:
lim
f (x, y)g(x, y) = LM.
(x,y)→(a,b)
4. Quotient: If M ̸= 0 and g(x, y) ̸= 0 in an open disk around the point (a, b), then
lim
(x,y)→(a,b)
5. Power: If r ∈ R, Lr ∈ R and
1.2
lim
(x,y)→(a,b)
f (x, y)
= L/M.
g(x, y)
f (x, y) = L, then lim(x,y)→(a,b) f (x, y)r = Lr ,
Continuity
Let f (x, y) be a real valued function defined on a subsets D of R2 . We say that f (x, y)pis continuous at a point
(a, b) ∈ D iff for each ϵ > 0, there exists δ > 0 such that for all points (x, y) ∈ D with (x − a)2 + (y − b)2 < δ
we have
|f (x, y) − f (a, b)| < ϵ.
Observe that if (a, b) is an isolated point of D, then f is continuous at (a, b). When D is a region, (a, b) is not
an isolated point of D; and then f is continuous at (a, b) ∈ D iff the following are satisfied:
a) f (a, b) is well defined, i.e. (a, b) ∈ D;
b)
f (x, y) exists; and
lim
(x,y)→(a,b)
c)
lim
f (x, y) = f (a, b).
x,y)→(a,b)
The function f (x, y) is said to be continuous on a subset of D iff f (x, y) is continuous at all points in the subset.
Note:
Constant multiples, sum, difference, product, quotient, and rational powers of continuous functions are continuous whenever they are well defined.
Polynomials in two variables are continuous functions.
Rational functions, i.e., ratios of polynomials, are continuous functions provided they are well defined.

2
 3x y
2
Example 1.2.1 f (x, y) = x + y 2

0
if (x, y) ̸= (0, 0)
if (x, y) = (0, 0)
is continuous on R2 .
10
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
At any point other than the origin, f (x, y) is a rational function; therefore,p
it is continuous. To see that f (x, y)
is continuous at the origin, let ϵ > 0 be given. Take δ = ϵ/3. Assume that x2 + y 2 < δ. Then
p
3(x2 + y 2 )y
3x2 y
≤ 3|y| ≤ 3 x2 + y 2 < 3δ = ϵ.
− f (0, 0) ≤
2
2
2
+y
x +y
(
xy ln x2 + y 2
if x2 + y 2 ̸= (0, 0)
is continuous at (0,0).
Example 1.2.2 f (x, y) =
0
if (x, y) = (0, 0)
x2
Let ϵ > 0 be given. Take δ = (ϵ/2)(1/4) . Assume that |x|, |y| < δ. Then
|f (x, y) − f (0, 0)| = |xy ln x2 + y 2 − 0| ≤ |x||y| ln
 3
3
x − y
Example 1.2.3 f (x, y) =
x+y

0
if x ̸= y ̸= 0
1
|
(x2 + y 2 )
discuss the continuity at (0,0).
if (x, y) = (0, 0)
Hint: y = −x + mx3

xy
p
x2 + y 2
Example 1.2.4 f (x, y) =

0
if x2 + y 2 ̸= 0
discuss the continuity at (0,0).
if x2 + y 2 = (0, 0)
p
Let ϵ > 0 be given. Take δ = ϵ. Assume that x2 + y 2 < δ. Then
|f (x, y) − f (0, 0)| = p
(
Example 1.2.5 f (x, y) =
xy
(x2
+
y2 )
(x + y) sin(x + y)
0
p
|y|
− 0 ≤ |x||y| x2 + y 2 | ≤ |x| p
≤ |x| < ϵ
x2 + y 2
if (x, y) ̸= (0, 0)
if (x, y) = (0, 0)
discuss the continuity at (0, 0).
1.3. PARTIAL DERIVATIVES
1.3
11
Partial Derivatives
If we obtain a derivative of a function f of several variables x, y, ..., with respect to one variable, say x, while
keeping the other variables as constants, this derivative is called the partial derivative of the function f with
respect to the variable x. This is denoted by fx or ∂f
∂x
In particular, if f = f (x, y), the partial derivative of f with respect to x is denoted by
fx , fx (x, y),
∂f
∂x
,
∂f (x, y)
∂x
,
∂f (x, y)
∂y
(x,y)
Similarly,
fy , fy (x, y),
∂f
∂y
(x,y)
Definition 1.3.1 The partial derivative of f (x, y) with respect to x at the point (a, b) is
fx (a, b) =
∂f
∂x
=
(a,b)
df (x, b)
dx
f (a + h, b) − f (a, b)
,
h→0
h
= lim
x=a
provided this limit exists. Notice that f (x, b) must be continuous at x = a. The partial derivative of f (x, y) with
respect to y at the point (a, b) is
fy (a, b) =
∂f
∂y
=
(a,b)
df (a, y)
dx
= lim
k→0
y=b
f (a, b + k) − f (a, b)
,
k
provided this limit exists. Again, f (a, y) must be continuous at y = b.
Similarly, for a function f of n variables x1 , ...xn we can define partial derivatives,
∂f
∂f
∂f
= fx1 ,
= fx2 , ...,
= fxn .
∂x1
∂x2
∂xn
12
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
1.4
Higher partial derivatives
∂f
Notice that ∂f
∂x and ∂x are themselves functions of two variables, so they can also be partially differentiated.
For a function of two variables f : D → R there are four possibilities:
∂2f
∂x2
∂2f
∂y 2
∂2f
∂y∂x
∂2f
∂x∂y
=
=
=
=
∂ ∂f
= fxx
∂x ∂x
∂ ∂f
= fyy
∂y ∂y
∂ ∂f
= fxy
∂y ∂x
∂ ∂f
= fyx
∂x ∂y
Higher order partial derivatives are defined similarly:
∂3f
∂
∂ ∂f
=
= fxyx .
∂x∂y∂x
∂x ∂y ∂x
Example 1.4.1 Find all the second partial derivatives of
f (x, y) = x3 e−2y + y −2 cos(x)
Example 1.4.2 Find fx and fy , where f (x, y) = y sin(xy).
 3
3
x − y
if x + y ̸= 0
Example 1.4.3 f (x, y) =
x+y

0
if x + y = 0
find fx and fy at every point.
When x + y ̸= 0,
∂f
3x2 (x + y) − (x3 − y 3
=
.
∂x
(x + y)2
(−3y 2 )(x + y) − (x3 − y 3 )
∂f
=
.
fy =
∂y
(x + y)2
fx =
When x + y = 0, for example when (x, y) = (0, 0)
• Let’s find fx (0, 0)
f (0 + h, 0) − f (0, 0)
h
f (h, 0) − f (0, 0)
= lim
h→0
h
h3
−
0
= lim h
h→0
h
= lim h
fx (0, 0) = lim
h→0
h→0
fx (0, 0) = 0
• Let’s find fy (0, 0)
f (0, 0 + k) − f (0, 0)
k
f (0, k) − f (0, 0)
= lim
k→0
k
k3
− k −0
= lim
k→0
k
= lim −k
fy (0, 0) = lim
k→0
k→0
fy (0, 0) = 0
1.4. HIGHER PARTIAL DERIVATIVES
Example 1.4.4 Consider f (x, y) =
13
xy(x2 − y 2 )
for (x, y) ̸= (0, 0), and f (0, 0) = 0.
x2 + y 2
f (x, 0) = f (0, y) = f (0, 0) = 0.
fx (x, 0) = fy (0, y) = fxx (0, 0) = fyy (0, 0) = 0.
f (h, y) − f (0, y)
f (x, k) − f (x, 0)
= −y, fy (x, 0) = lim
= x.
k→0
h
k
fx (0, k) − fx (0, 0)
−k − 0
fxy (0, 0) = lim
= lim
= −1.
k→0
k→0
k
k
h−0
fy (h, 0) − fy (0, 0)
= lim
= 1.
fyx (0, 0) = lim
h→0
h→0
h
h
fx (0, y) = lim
h→0
That is, fxy ̸= fyx . But continuity of both of fxy and fyx implies their equality.
Theorem 1.4.1 (Mean Value Theorem)
Let f : D → R where D = {(x, y) : (x − a)2 + (y − b)2 ≤ δ 2 } and fx and fy exists on D. Then,
f (a + ∆x, b + ∆y) − f (a, b) = ∆xfx (a + θ∆x, b) + ∆yfy (a + ∆x, b + α∆y)
where ∆x2 + ∆y 2 < δ 2 and 0 < θ, α < 1
Proof 1.4.1 Set
∆f = f (a + ∆x, b + ∆y) − f (a, b)
Then we may rewrite it as
∆f = [f (a + ∆x, b) − f (a, b)] + [f (a + ∆x, b + ∆y) − f (a + ∆x, b)]
Now by MVT
f (a + ∆x, b) − f (a, b) = ∆xfx (a + θ∆x, b)
where 0 < θ < 1
Applying the same theorem
f (a + ∆x, b + ∆y) − f (a + ∆x, b) = ∆yfy (a + ∆x, b + α∆y)
where 0 < α < 1
Hence we have
∆f = ∆xfx (a + θ∆x, b) + ∆yfy (a + ∆x, b + α∆y)
where 0 < θ, α < 1 and hence the proof
□
Theorem 1.4.2 (Clairaut) Let D be a region in R2 . Let the function f : D → R have continuous first and
second order partial derivatives on D. Then fxy = fyx .
Proof 1.4.2 Let (a, b) ∈ D. Let h ̸= 0. Write g(x) = f (x, b + h) − f (x, b). Then
ϕ(f ) := g(a + h) − g(a) = [f (a + h, b + h) − f (a + h, b)] − [f (a, b + h) − f (a, b)].
By MVT, we have c between a and a + h such that
ϕ(f ) = g ′ (c)h = h[fx (c, b + h) − fx (c, b)].
Again, by MVT (on fx with the second variable), we have d between b and b + h such that
ϕ(f ) = h.hfxy (c, d) = h2 fxy (c, d).
Due to continuity of fxy , we have
lim
h→0
ϕ(f )
=
lim
fxy (c, d) = fxy (a, b).
h2
(c,d)→(a,b)
Similarly, write
ϕ(f ) = [f (a + h, b + h) − f (a, b + h)] − [f (a + h, b) − f (a, b)]
and apply MVT twice as above to get fyx (a, b) = limh→0
So, fxy (a, b) = fyx (a, b).
ϕ(f )
h2 .
But the two limits with
ϕ(f )
h2
are equal.
14
1.4.1
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
Differentiable Functions
Let f be a differentiable at x0 Then the graph of f has a tangent line at x = x0 .
The equation of this tangent line is
y − y0 = f ′ (x0 )(x − x0 ).
The change in f near x0 is well approximated by a linear function:
∆f = f ′ (x0 )∆x + error,
error
→ 0 as ∆x → 0. These ideas can be
∆x
generalized to a function of two variables. A function f is differentiable at (x0 , y0 ) means that f has a well
defined tangent plane at (x0 , y0 ).
where the error is small compared with ∆x or more precisely:
Definition 1.4.1 If z = f (x, y), then f is differentiable at (a, b) if ∆z can be expressed in the form
∆z = fx (a, b)∆x + fy (a, b)∆y + ϵ1 ∆x + ϵ2 ∆y,
where ϵ1 and ϵ2 → 0 as (∆x, ∆y) → (0, 0)
In otherwards: A function f (x, y) is differentiable at (a, b) if the change in f near (a, b) is well approximated
by a linear function:
∆z = fx (a, b)∆x + fy (a, b)∆y
The equation of the tangent plane to z = f (x, y) at (x0 , y0 ) is
z − z0 =
∂f (x0 , y0 )
∂f (x0 , y0 )
∆x +
∆y.
∂x
∂y
Example 1.4.5 Find the cartesian equation of the tangent plane to the surface
z = 1 − x2 − y 2
at the point (1, 2, 4).
1.4. HIGHER PARTIAL DERIVATIVES
15
Theorem 1.4.3 If f is differentiable then it is continuous.
Proof 1.4.3 If f is differentiable then by the definition
∆z = f (x + ∆x, y + ∆y) − f (x, y) = fx (a, b)∆x + fy (a, b)∆y + ϵ1 ∆x + ϵ2 ∆y,
as f is differentiable, ϵ1 and ϵ2 → 0 as (∆x, ∆y) → (0, 0).
Therefore, when (∆x, ∆y) → (0, 0) we have
f (x + ∆x, y + ∆y) = f (x, y)
lim
(x,y)→(0,0)
Hence f is continuous.
But Continuity does not imply the differentiablity.
 3
3
x + y
if (x, y) ̸= (0, 0)
2
2
Example 1.4.6 f (x, y) = x + y

0
if (x, y) = (0, 0).
Discuss the differentiability of f at (0,0).
Solution 1.4.1
∂f
∂y
∂f
∂x
f (0 + h, 0) − f (0.0)
=1
h→0
h
= fx (0, 0) = lim
(0,0)
f (0, 0 + k) − f (0.0)
=1
k→0
k
= fy (0, 0) = lim
(0,0)
If f is differentiable at (0, 0), then we must have
∆z = fx (0, 0)∆x + fy (0, 0)∆y + ϵ1 ∆x + ϵ2 ∆y,
f (0 + ∆x, 0 + ∆y) − f (0, 0) = 1.∆x + 1.∆y + ϵ1 ∆x + ϵ2 ∆y,
∆x3 + ∆y 3
= 1.∆x + 1.∆y + ϵ1 ∆x + ϵ2 ∆y
∆x2 + ∆y 2
Let ∆x = r cos(θ), ∆y = r sin(θ) then,
cos3 (θ) + sin3 (θ) = cos(θ) + sin(θ) +
ϵ1 rcos(θ) + ϵ2 r sin(θ)
r
when r → 0
cos3 (θ) + sin3 (θ) = cos(θ) + sin(θ)
if the function is differentiable at (0, 0) then the equation must be true for all θ. But this is not true for all θ.
Therefor this is not differentiable at (0, 0)
(
x sin(1/x) + y sin(1/y)
Example 1.4.7 f (x, y) =
0
Discuss the differentiability of f at (0, 0).
if (x, y) ̸= 0
if (x, y) = 0.
Solution 1.4.2 This is continuous, but not differentiable.
Theorem 1.4.4 If the partial derivatives fx and fy exist near (a, b) and are continuous at (a, b), then f is
differentiable at (a, b). .
Example 1.4.8 Show that f (x, y) = xexy is differentiable at (1, 0).
16
CHAPTER I. FUNCTIONS OF SEVERAL VARIABLES
1.4.2
Errors and approximations
If the function z = f (x, y) is differentiable at (x0 , y0 ), then for (x, y) near (x0 , y0 )
f (x, y) ≈ f (x0 , y0 ) + fx (x0 , y0 )∆x + fy (x0 , y0 )∆y.
where ∆x = x − x0 , ∆y = y − y0 . The right hand side of this equation is called the linear approximation to f
near (x0 , y0 ). This equation may also be written as
∆f ≈ fx ∆x + fy ∆y
which is sometimes written as
df = fx dx + fy dy
which is called the differential of f . Some applications of this linear approximation to f are
• Estimating values of functions.
• Estimating errors in measurements.
Example 1.4.9
• Estimate the value of f (x, y) =
√
1 − x + 2y when x = 0.01, y = 0.02.
• Show that f (x, y) = xexy is differentiable at (1, 0). Then use it to approximate f (1.1, 0.1)
1.4.3
Differentials
For a differentiable function of one variable, we define the differential dx to be an independent variable; that is,
dx can be given the value of any real number. The differential of is then defined as
dy = f ′ (x)dx
For a differentiable function of two variables,z = f (x, y) , we define the differentials dx and dy to be independent
variables; that is, they can be given any values. Then the differential dz, also called the total differential, is
defined by
dz = fx (x, y)dx + fy (x, y)dy =
∂z
∂z
dx +
dy
∂x
∂y
Also,
∆z = dz + ϵ1 ∆x + ϵ2 ∆y
where ϵ1 , ϵ2 → 0 as ∆x, ∆y → 0
• If z = f (x, y) = x2 + 3xy − y 2 , find the differential dz.
p
• Use differentials to find an approximate value for 9(1.95)2 + (8.1)2
Example 1.4.10