Function of several variables

Faculty of Sciences (Section V)
Lebanese University
Function of Several Variables
Dr. Kamel ATTAR
attar.kamel@gmail.com
F 2020 - 2021 F

2Ú74
Functions of one variable
Functions of SeveraL VariabLes
Limits for Functions of Two Variables
Partial Derivatives of a Function of Two Variables
1 Functions of one variable
Definition-Domain, Range and Graph
Examples
2 Functions of SeveraL VariabLes
Definition
Domains and Ranges
Examples
Functions of two variables
Functions of three variables
Graphs, Level Curves and Level Surfaces
Level Curves and Level Surfaces
Examples
Graph
Graphing with Traces (z-Axis Traces)
Exercises
3 Limits for Functions of Two Variables
Definition
Exercises
Dr. Kamel ATTAR | Function of Several Variables |

3Ú74
Example
Exercises
Two-Path Test for Nonexistence of a Limit
Test for Nonexistence of a Limit
Exercises
Continuity
Definition
Exercises
4 Partial Derivatives of a Function of Two Variables
Partial Derivatives with respect to x
Geometrical interpretation
Partial Derivatives with respect to y
Geometrical interpratation of partial derivatives with respect to x and y
Exercises
The chain rule
Chain Rule for Functions of Two Independent variables
Examples
Chain Rule for Functions of Three Independent variables

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Examples
Exercises

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Examples
Function of one variable
Definition
A function f of one variable is a rule that assigns to each variable x a unique
number denoted by y = f(x). We write
f : D ⊂ R 7−→ R
x 7−→ y = f(x)
I The domain is the set of all possible values of x. (set of inputs).
I The range is the set of all possible value of y from the domain. (set of
outputs)
I The graph is the set of all possible points (x, y) produced by the function.
Geometrically, the graph is a curve in plan.

6Ú74
Examples
Example
Function Domain Range
y = x + 2 R =] − ∞, ∞[ R =] − ∞, ∞[
y = 3x2
− 7 R [−7, +∞[
y = sin x R [−1, 1]
y =
1
x
] − ∞, 0[∪]0, +∞[ ] − ∞, 0[∪]0, +∞[
y =
√
x [0, +∞[ [0, +∞[
y = ln x ]0, +∞[ ] − ∞, +∞[

7Ú74
Examples
Example
Find the domain of the following functions:
f(x) =
p
1 − 2x , g(x) = −
7
x
, h(x) = 4 − x2
and u(x) =
1
x − 2

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Examples

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Examples

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Examples

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Examples

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Examples

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Examples

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Examples

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Definition
Domains and Ranges
Graph
Definition
• A function of two variables is a rule that assigns a real number f(x, y) to
each pair of real numbers (x, y) in the domain D of the function f. we write
f : D ⊂ R2
7−→ R
(x, y) 7−→ z = f(x, y)
• A function of three variables is a rule that assigns a real number f(x, y, z)
to each triple of real numbers (x, y, z) in the domain D. we write
f : D ⊂ R3
7−→ R
(x, y, z) 7−→ w = f(x, y, z)
• A function of n-variables is a rule that assigns a real number
f(x1, x2, · · · , xn) to each n-tuples of real numbers (x1, x2, · · · , xn)
f : D ⊂ Rn
7−→ R
(x1, x2, · · · , xn) 7−→ w = f(x1, x2, · · · , xn)

16Ú74
Definition
Domains and Ranges
Graph
Definition (Domain)
The domain of a function is assumed to be the largest set for which the
defining rule generates real numbers.
Definition (Range)
The range consists of the set of output values for the dependent variable.
f : D ⊂ Rn
7−→ R
(x1, x2, · · · , xn) 7−→ w = f(x1, x2, · · · , xn)
Domain of f : The set D is the function’s domain.
Range of f : The set of w−values taken on by f is the function’s range.
Input variables : We call the xj ’s the function’s input variables.
Output variable : We call w the function’s output variable.

17Ú74
Definition
Domains and Ranges
Graph
Figure: f(x, y) =
p
4 − x2 − y2

18Ú74
Definition
Domains and Ranges
Graph
Figure: f(x, y) =
p
4 − x2 − y2
Figure:
g(x) =
√
x , Dg = [0, +∞[

19Ú74
Definition
Domains and Ranges
Graph
Figure: f(x, y) =
p
4 − x2 − y2
Figure:
g(x) =
√
x , Dg = [0, +∞[
Domain:
4 − x2
− y2
≥ 0
4 ≥ x2
+ y2
Df = {(x, y) ∈ R2
: x2
+ y2
≤ 4}

20Ú74
Definition
Domains and Ranges
Graph
Figure: f(x, y) =
p
4 − x2 − y2
Figure:
g(x) =
√
x , Dg = [0, +∞[
Domain:
4 − x2
− y2
≥ 0
4 ≥ x2
+ y2
Df = {(x, y) ∈ R2
: x2
+ y2
≤ 4}

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Definition
Domains and Ranges
Graph
Figure: f(x, y) = ln(1 − x − y)

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Definition
Domains and Ranges
Graph
Figure:
g(x) = ln(x) , Dg =]0, +∞[

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Definition
Domains and Ranges
Graph
Figure:
g(x) = ln(x) , Dg =]0, +∞[
Domain:
1 − x − y > 0
1 > x + y
Df = {(x, y) ∈ R2
: x+y < 1}

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Definition
Domains and Ranges
Graph
Figure:
g(x) = ln(x) , Dg =]0, +∞[
Domain:
1 − x − y > 0
1 > x + y
Df = {(x, y) ∈ R2
: x+y < 1}

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Definition
Domains and Ranges
Graph
Example (Functions of two variables)
z = ex
cos y R2
Entire plane ] − ∞, ∞[
z =
p
y − x2 y ≥ x2
[0, ∞[
z =
1
xy
xy 6= 0 ] − ∞, 0[∪]0, ∞[
z = sin xy R2
Entire plane [−1, 1]
Example
Find the domain of the function f(x, y) =
ln(x − 3)
p
y + 2
x2 − 4
.

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Definition
Domains and Ranges
Graph
Example (Functions of three variables)
w = x2
cos y sin z R3
Entire space ] − ∞, ∞[
w =
p
x2 + y2 + z2 R3
Entire space [0, ∞[
w =
1
x2 + y2 + z2
(x, y, z) 6= (0, 0, 0) ]0, ∞[
w = xy ln z R2
∪ R∗
+ Half-space z > 0 ] − ∞, ∞[

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Definition
Domains and Ranges
Graph
Definition
Level curve (2 varaibles): The set of points in the plane where a function
f(x, y) has a constant value f(x, y) = c is called a level curve
of f.
Level surface (3 variables): The set of points (x, y, z) in space where a
function of three independent variables has a constant value
f(x, y, z) = c is called a level surface of f.

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Definition
Domains and Ranges
Graph

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Definition
Domains and Ranges
Graph
Example
The level surfaces of f(x, y, z) =
p
x2 + y2 + z2 is described as below

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Definition
Domains and Ranges
Graph
Definition
Graph in R2
: The set of all points (x, y, f(x, y) in space, for (x, y) in the
domain of f, is called the graph of f. The graph of f is also called
the surface z = f(x, y).
Graph in R3
: The set of all points (x, y, z, f(x, y, z) in space, for (x, y, z) in
the domain of f, is called the graph of f.

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Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

32Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

33Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

34Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

35Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

36Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z = f(x, y) =
p
x2 + y2 .
z ∈ {0, 1, 2, 3, 4}

37Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

38Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

39Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

40Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

41Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

42Ú74
Definition
Domains and Ranges
Graph
Sketch the graph of
z =
p
9 − x2 − y2 .
z ∈ {0, 1, 2, 3}

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Definition
Domains and Ranges
Graph
Figure: z =
p
x2 + y2 − 1 Figure: z = e−x2
−y2
+1

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Exercises
¬ Let f(x, y) = ln(x + y − 1)
a) Evaluate f(1, 1) and f(e, 1).
b) Find and sketch the domain of f
Find and sketch the domain of the following functions
a) h(x, y) = x−3y
x+3y
b) g(x, y) =
p
x2 + y2 − 1 + ln(4 − x2
− y2
)
® Draw several level curves of the following functions
a) u(x, y) =
x2
+ y2
y
b) v(x, y) =
p
x2 + y2

45Ú74
Definition
Continuity
Definition
We say that a function f(x, y) approaches the limit L as (x, y) approaches
(x0, y0), if and only if
lim
(x,y)→(x0,y0)
f(x, y) = L .
We simply substitute the x and y values of the point being approached into the
functional expression to find the limiting value.
Example
(a) lim
(x,y)→(0,1)
x − xy + 3
x2y + 5xy − y3
=
0 − (0)(1) + 3
(0)2(1) + 5(0)(1) − (1)3
= −3
(b) lim
(x,y)→(3,−4)
p
x2 + y2 =
q
(3)2 + (−4)2 =
√
25 = 5

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Definition
Continuity
Exercises A:
Find the limits
1. lim
(x,y)→(0,0)
3x2
− y2
+ 5
x2 + y2 + 2
2. lim
(x,y)→(0,4)
x
√
y
3. lim
(x,y)→(0,ln 2)
ex−y
4. lim
(x,y)→(3,4)
p
x2 + y2 − 1
5. lim
(x,y)→(2,−3)

1
x
+
1
y
2
6. lim
(x,y)→(0,0)
cos
x2
+ y3
x + y + 1
!
7. lim
(x,y)→(1,1)
ln |1 + x2
y2
| 8. lim
(x,y)→(0,0)
ey
sin x
x

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Definition
Continuity
Example
Find lim
(x,y)→(0,0)
x2
− xy
√
x −
√
y
Solution

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Definition
Continuity
Example
Find lim
(x,y)→(0,0)
x2
− xy
√
x −
√
y
Solution
Since the denominator
√
x −
√
y approaches 0 as (x, y) → (0, 0), we cannot
calculate the limit. If we multiply numerator and denominator by
√
x +
√
y,
however, we produce an equivalent fraction whose limit we can find:
lim
(x,y)→(0,0)
x2
− xy
√
x −
√
y
= lim
(x,y)→(0,0)
(x2
− xy)(
√
x +
√
y)
(
√
x −
√
y)(
√
x +
√
y)
= lim
(x,y)→(0,0)
x(x − y)(
√
x +
√
y)
x − y
= lim
(x,y)→(0,0)
x(
√
x +
√
y) = 0 .

49Ú74
Definition
Continuity
Exercises B:
Find the limits by rewriting the fractions first
1. lim
(x,y)→(1,1)
x2
− 2xy + y2
x − y
2. lim
(x,y)→(1,1)
x2
− y2
x − y
3. lim
(x,y)→(1,1)
xy − y − 2x + 2
x − 1
4. lim
(x,y)→(2,−4)
y + 4
x2y − xy + 4x2 − 4x
5. lim
(x,y)→(2,2)
x + y − 4
√
x + y − 2
6. lim
(x,y)→(2,0)
p
2x − y − 2
2x − y − 4
7. lim
(x,y)→(4,3)
√
x −
p
y + 1
x − y − 1

50Ú74
Definition
Continuity
Theorem (Two-Path Test for Nonexistence of a Limit)
If a function f(x, y) has different limits along two different paths in the domain
of f as (x, y) approaches (x0, y0), then lim
(x,y)→(x0,y0)
f(x, y) does not exist.
Example
Show that the function
f(x, y) =
2x2
y
x4 + y2
has no limit as (x, y) approaches (0, 0).

51Ú74
Definition
Continuity
Solution
Let’s take the curve y = x2
, the function has the constant value
lim
(x,y=x2)→(0,0)
f(x, y) = lim
(x,y=x2)→(0,0)
2x2
x2
x4 + x2
= lim
(x,y=x2)→(0,0)
2x4
2x4
= 1 .
Now we take y = 2x2
, we obtain
lim
(x,y)→(0,0)
f(x, y) = lim
(x,y=2x2)→(0,0)
2x2
(2x2
)
x4 + 4x4
= lim
(x,y=2x2)→(0,0)
4x4
5x4
= 4/5
Then,
lim
(x,y=2x2)→(0,0)
f(x, y) 6= lim
(x,y=x2)→(0,0)
f(x, y)
By the two-path test, f has no limit as (x, y) approaches (0, 0).

52Ú74
Definition
Continuity
Exercises C:
Show that the limits do not exist.
1. lim
(x,y)→(1,1)
xy2
− 1
y − 1
2. lim
(x,y)→(1,−1)
xy + 1
x2 − y2
.
Answer:
1. y = x and y =
1
x
2. y = −
1
x
and y = −
1
x2

53Ú74
Definition
Continuity
Definition
A function f(x, y) is Continuous at the point (x0, y0) if
1. f is defined at (x0, y0),
2. lim
(x,y)→(x0,y0)
f(x, y) exists
3. lim
(x,y)→(x0,y0)
f(x, y) = f(x0, y0)
A function is continuous if it is continuous at every point of its domain.

54Ú74
Definition
Continuity
Exercises D:
1. Show that
f(x, y) =



x2
− xy
√
x −
√
y
, (x, y) 6= (0, 0)
0 , (x, y) = (0, 0)
Is continuous at every point.
2. Show that
f(x, y) =



2xy
x2 + y2
, (x, y) 6= (0, 0)
0 , (x, y) = (0, 0)
Is continuous at every point except the origin.

55Ú74
Geometrical interpratation of partial derivatives wit
Chain Rule for Functions of Two Independent varia
Chain Rule for Functions of Three Independent va
Definition
The partial derivative of f(x, y) with respect to x at the point (x0, y0) is
d
dx
f(x, y)

(x0,y0)
= lim
h→0
f(x0 + h, y0) − f(x0, y0)
h
,

56Ú74
The slope of the curve z = f(x, y0) at
the point P = (x0, y0, f(x0, y0)) in the
plane y = y0 is the value of the par-
tial derivative of f with respect to x at
(x0, y0). The tangent line to the curve
at P is the line in the plane y = y0 that
passes through P with this slope. The
partial derivative
df
dx
at (x0, y0) gives
the rate of change of f with respect to
x when y is held fixed at the value y0.

57Ú74
Definition
The partial derivative of f(x, y) with respect to y at the point (x0, y0) is
d
dy
f(x, y)

(x0,y0)
= lim
h→0
f(x0, y0 + h) − f(x0, y0)
h
,

58Ú74
The slope of the curve z = f(x0, y) at the
point P = (x0, y0, f(x0, y0)) in the plane
x = x0 is the value of the partial deriva-
tive of f with respect to y at (x0, y0). The
tangent line to the curve at P is the line in
the plane x = x0 that passes through P
with this slope. The partial derivative df
dy
at
(x0, y0) gives the rate of change of f with
respect to y when x is held fixed at the
value x0.

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60Ú74
Example
Find the values of ∂
∂x
f and ∂
∂y
f at the point (4, −5) if
f(x, y) = x2
+ 3xy + y − 1 .
Solution
To find ∂
∂x
f, we treat y as a constant and differentiate with respect to x:
∂
∂x
f =
∂
∂x
(x2
+ 3xy + y − 1) = 2x + 3y .
The value of ∂
∂x
f at (4, −5) is 2(4) + 3(−5) = −7.

61Ú74
Solution
To find ∂
∂y
f, we treat x as a constant and differentiate with respect to y:
∂
∂y
f =
∂
∂y
(x2
+ 3xy + y − 1) = 3x + 1 .
The value of ∂
∂y
f at (4, −5) is 3(4) + 1 = 13.

62Ú74
Example
Find ∂f
∂y
as a function if f(x, y) = y sin xy .
Solution
We treat x as a constant and differentiate with respect to y:
∂f
∂y
=
∂
∂y
(y sin xy)
= y
∂
∂y
sin xy + sin xy
∂
∂y
(y)
= (y cos xy)
∂
∂y
(xy) + sin xy
= xy cos xy + sin xy .

63Ú74
Example
Find ∂f
∂x
and ∂f
∂y
as a functions if f(x, y) =
2y
y + cos x
.
Solution
We treat x as a constant and differentiate with respect to y:
∂f
∂y
=
(y + cos x) ∂
∂y
(2y) − 2y ∂
∂y
(y + cos x)
(y + cos x)2
=
(y + cos x)(2) − 2y
(y + cos x)2
=
2 cos x
(y + cos x)2

64Ú74
Example
The plane x = 1 intersects the paraboloid z = x2
+ y2
in a parabola. Find the
slope of a tangent to the parabola at (1, 2, 5).
Solution
The slope is the value of the partial derivative ∂z
∂y
at (1, 2):
∂z
∂y

Function of several variables

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