Chapter 01
Partial Derivatives
1.1 Functions of Several Variables
1.1.1 Function
A function is a mathematical relationship in which the values of a single dependent variable are
determined by the values of one or more independent variables.
1.1.2 Functions of One Variable A function of one variable is a mathematical relationship in
which the values of a single dependent variable are determined by the value of independent variable.
■ y = f (x); x Independent variable, y Dependent variable
1.1.3 Functions of Multiple Variable
• To expand the idea of functions to include functions with more than one independent variable.
• For example, consider the functions below:
(iii) h(x1 , x2 , x3 , x4 ) = 2x1 − 4x2 + 3x3 − 8x4
(i) f (x, y) = 2x3 + y 4
(ii) g(x, y, z) = 5xeyz
• The function z = f (x, y) function of two variables. It has independent variables x and y, and
the dependent variable is z.
• The function w = g(x, y, z) function of two variables. It has independent variables x, y and z,
and the dependent variable is w.
• To find the values of the several variable functions, substitute values for each of the independent variables.
Example 01 Evaluate the function f (x, y) =
2x3 + y 4
5
for (2, 3), (4, 3) and (5, y).
1.1.4 Functions of Two Variables
• A function f of two variables x and y is a rule that assigns to each ordered pair (x, y) in a
given set D(domain), a unique value of f .
• Functions of more variables can be defined similarly.
• The operations we performed with one-variable functions can also be performed with functions
of several variables.
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MA 2103 - Advanced Calculus
Partial Derivatives
1.1.4.1 Algebraic Properties of Functions of Two Variables
• Let f and g be two functions of two variables. Then,
(i) [f ± g](x, y) = f (x, y) ± g(x, y)
(ii) [f · g](x, y) = f (x, y) · g(x, y)
" #
f (x, y)
f
(x, y) =
, provided g(x, y) ̸= 0
(iii)
g
g(x, y)
■ In general, the composition of two multi-variable functions is not considered.
1.2 Limits and Continuity of Function of Multiple Variables
• Recall that if f is a function that takes real numbers to real numbers,
f ′ (x) = lim
f (x + h) − f (x)
h
h→0
.
• To do the same thing in more than one variable, need to take limits in more than one dimension.
1.2.1 Definition : Limits of Function of Multiple Variables
• We write
lim
f (x, y) = L
(x,y)→(a,b)
and we say that the limit of f (x, y) as (x, y) approaches (a, b) is L if we can make the values
of f (x, y) as close to L as we like by taking the point (x, y) to be sufficiently close to (a, b).
• Some easy limits:
(i) lim(x,y)→(a,b) x = a
(ii) lim(x,y)→(a,b) y = b
(iii) lim(x,y)→(a,b) c = c, where c is a constant.
1.2.2 Rules of Limits
• Like regular limits, limits of multi variable functions can be added, subtracted, multiplied,
composed, and divided, provided that the limit of the denominator is not zero.
Example 02 Evaluate the limit
lim
(x5 + 4x3 y − 5xy 2 ).
(x,y)→(5,−2)
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MA 2103 - Advanced Calculus
Partial Derivatives
Example 03 Evaluate the limit
x2
lim
(x,y)→(1,2)
x2 + y 2
.
1.2.3 Complicated Limits
The only real problem is a limit where the denominator goes to zero.
(i) If the numerator goes to some number and the denominator goes to zero then the quotient
cannot have a limit.
(ii) If on the other hand, the numerator and denominator both go to zero we have no clue.
Example:
f (x + h) − f (x)
.
f ′ (x) = lim
h→0
h
■ For a function f (x) of one variable,
lim f (x) = L ⇐⇒ lim+ f (x) = L ⇐⇒ lim− f (x) = L
x→a
x→a
x→a
■ For functions of two variables, “left-hand limits" and “right-hand limits" aren’t enough.
1.2.4 Showing a Limit Does Not Exist
Suppose
lim f (x, y) = L
(x,y)→(a,b)
then, the limit of f as (x, y) → (a, b) is L along all paths through (a,b).
■ There are two contrapositives to this statement:
1. If there is a path through (a, b) along which the limit does not exist, the two-dimensional limit
does not exist.
2. If there are two paths through (a, b) along which the limits exist but disagree, the twodimensional limit does not exist
Example 04 Show that the following limits does not exist.
(i)
(iii)
lim
(x,y)→(0,0)
x
4xy 2
lim
x2 + y 2
(x,y)→(0,0)
x2 + 3y 4
(iv)
(ii)
lim
(x,y)→(0,0)
x
2
lim
x2 + y 2
(x,y)→(2,1)
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(x − 2)(y − 1)
(x − 2)2 + (y − 1)2
MA 2103 - Advanced Calculus
Partial Derivatives
1.2.5 Definition : Continuity of Function of Multiple Variables A function f of two variables
is called continuous at a point (a, b) if
1. f (a, b) exists.
2. lim(x,y)→(a,b) f (x, y) exists.
3. lim(x,y)→(a,b) f (x, y) = f (a, b).
■ If k is a real number and f and g are continuous functions at (a, b) then the functions below are
also continuous at (a, b).
• kf (x, y)
• (f ± g)(x, y)
• (f g)(x, y)
!
f
(x, y)
•
g
Example 05
(i) Show that the function f (x, y) =
3x + 2y
is continuous at point (5, −3).
x+y+1
p
(ii) Show that the function f (x, y) = 26 − 2x2 − y 2 is continuous at point (5, −3).
1.3 Partial Derivatives
1.3.1 Introduction
• The process of differentiating a function of several variables with respect to one of its variables,
while keeping the other variable(s) fixed.
• The resulting derivative is a partial derivative of the function.
1.3.2 Definition : Partial Differentiation of a Function of Two Variables
If z = f (x, y) is a function of two variables, then the partial derivative of f with respect to variables
x and y are the functions fx and fy respectively, defined by,
•
fx (x, y) = lim
f (x + h, y) − f (x, y)
h
h→0
•
fy (x, y) = lim
f (x, y + h) − f (x, y)
h→0
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MA 2103 - Advanced Calculus
Partial Derivatives
1.3.2.1 Notations
If z = f (x, y) is a function of two variables,
• fx (x, y) = fx =
• fy (x, y) = fy =
∂f
∂x
∂f
∂y
∂
=
=
∂x
∂
∂y
∂z
f (x, y) =
∂x
∂z
f (x, y) =
∂y
Example 06 If f (x, y) = xy, find fx and fy .
Example 07 If f (x, y) = x3 y + x2 y 2 , find fx , fy and fx |(1,2) .
Example 08 If f (x, y) = x3 + x2 y 3 − 2y 2 , find fx |(−1,1) and fy |(2,1) .
!
x
, find
Example 09 If z = sin
1+y
(i)
∂z
(ii)
∂x
∂z
∂y
1.4 Higher Order Partial Derivatives
If z = f (x, y), we denote the second order partial derivative by,
!
∂ ∂f
∂ 2f
∂ 2z
• (fx )x = fxx =
=
=
∂x ∂x
∂x2 ∂x2
• (fy )y = fyy =
• (fx )y = fxy =
• (fy )x = fyx =
!
∂
∂f
∂y
∂y
∂
∂f
∂y
∂x
∂
∂f
∂x
∂y
=
!
=
!
=
∂ 2f
∂y 2
=
∂ 2f
∂y∂x
∂ 2f
∂x∂y
∂ 2z
∂y 2
=
=
∂ 2z
∂y∂x
∂ 2z
∂x∂y
Example 10 If f (x, y) = x3 + y 2 , find fxx , fyy ,fxy and fyx .
Example 11 For z = f (x, y) = 5x2 − 2xy + 3y 3 , determine these higher-order partial derivatives
∂ 2z
,
∂ 2z
∂ 2z
.
∂x∂y ∂y∂x
∂x2
Example 12 If f (x, y, z) = exy ln z, find
(i) fx
and
(ii) fy
(iii) fz
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MA 2103 - Advanced Calculus
Partial Derivatives
1.4.1 Equality of mixed partials
If the function f (x, y) has mixed partial derivatives fxy and fyx that are continuous in an open disk
contains (x0 , y0 ) then,
fxy (x0 , y0 ) = fyx (x0 , y0 )
1.5 Total Differential Coefficient
• For the function of one variable, y = f (x), the differential is
dy = f ′ (x)dx
This quantity is used to compute the approximate change in the value of f (x) due to a change
dx in x.
• For the function of two variable, z = f (x, y), the total differential is
df =
∂f
∂x
dx +
∂f
∂y
dy = fx (x, y)dx + fy (x, y)dy
where x and y are independent variables.
• For the function of three variable, w = f (x, y, z), the total differential is
df =
∂f
∂x
dx +
∂f
∂y
dy +
∂f
∂z
dz
where x, y and z are independent variables.
Example 13 Determine the total differential coefficient of the following functions.
(i) f (x, y) = 5x2 y 3
(iii) f (x, y) = yex
(ii) f (x, y) = sin (xy)
(iv) f (x, y, z) = z 2 sin (2x − 3y)
1.6 The Chain Rules
• Consider a function of two variables,z = f (x, y), here
– x can be a function of some other parameter(s).
Example: x = t2 , x = cos (s), x = u2 v
– y can be a function of some other parameter(s).
Example: y = 5t, y = cos (st), y = u2 v 2
then, z can be differentiated with respect to those parameter(s).
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MA 2103 - Advanced Calculus
Partial Derivatives
1.6.1 The chain rule for one independent parameter
Let f (x, y) be a differential function of x and y and let x−x(t) and y = y(t) be differential functions
of t, then z = f (x, y) is a differential function of t, and
dz
dt
=
∂z dx
∂x dt
+
∂z dy
∂y dt
Dependency Tree
Example 14 Let z = f (x, y) = 2xy + y 2 where x = −3t2 and y = 1 + t3 .
(i) write the function f in terms of t and then find
(ii) Use chain rule to find
dz
dt
dz
dt
.
(draw the dependency tree).
1.6.2 The chain rule for two independent parameter
Suppose z = f (x, y) is differentiable at (x, y) and that the partial derivatives of x = x(u, v) and
y = y(u, v) exist at (u, v). Then the composite function z = f (x(u, v), y(u, v)) is differentiable at
(u, v) with
∂z
∂z ∂x ∂z ∂y
=
+
∂u ∂x ∂u ∂y ∂u
and
∂z
∂v
=
∂z ∂x
∂x ∂v
+
∂z ∂y
∂y ∂v
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MA 2103 - Advanced Calculus
Partial Derivatives
Example 15 write out the chain rule for the functions given below.
(i) z = f (x, y) where x = x(s, t) and y = y(s, t)
(ii) w = f (x, y, z) where x = x(s, t), y = y(s, t) and z = z(s, t).
Example 16 Find
1
√
, where w = ln (x + 2y − z 2 ) and x = 2t − 1, y = , z = t.
dt
t
dw
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