AS101 - Topic 4 - Partial Derivatives - Revised

AS101 – MATHEMATICS I
TOPIC 4 – PARTIAL DERIVATIVES
KEY QUESTIONS AND KEY IDEAS
1. The formal definition of functions of two variables, and their fundamental properties.
2. The domain and range of functions of two or more variables.
3. The limits and continuity of functions of two or more variables.
4. Partial derivatives of functions of two or more variables and related concepts.
5. The Chain Rule formulae for functions of two or more variables.
6. Implicit differentiation formulae for functions of two or more variables.
Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 1

INTRODUCTION
In the course of your study thus far, we have only dealt with the calculus of functions of a single
variable. However, in the real world, physical quantities often depend on two or more variables.
So in this chapter, you will be introduced to the concept of a function of several variables and
subsequently, the basic ideas of differential calculus for such functions will be studied.
FUNCTIONS OF TWO VARIABLES
Definition 1
A function 𝑓 of two variables is a rule that assigns to each ordered pair of real numbers (𝑥, 𝑦) in a set
𝐷 a unique real number denoted by 𝑓(𝑥, 𝑦). The set 𝐷 is the domain of 𝑓 and its range is the set of
values that 𝑓 takes on, that is, {𝑓(𝑥, 𝑦): (𝑥, 𝑦) ∈ 𝐷}.
Note:
1. We often write 𝑧 = 𝑓(𝑥, 𝑦) to make explicit the value taken on by 𝑓 at the general point
(𝑥, 𝑦).
2. The variables 𝑥 and 𝑦 are the independent variables and 𝑧 is the dependent variable.
3. If the function 𝑓 is given by a formula and no domain is specified, then the domain of 𝑓 is
understood to be the set of all pairs (𝑥, 𝑦) for which the given expression is a well-defined
real number.
For functions of two variables, the domain and range are as defined below:
Domain: The range of 𝑥-values for which the function is defined.
Notation: 𝐷 = {(𝑥, 𝑦): 𝑥 ≠ 1, 𝑦 ≥ 1}
Range: The maximum and minimum values that can be assumed by the function 𝑧 = 𝑓(𝑥, 𝑦).
Notation: 𝑅 = {𝑧: 1 ≤ 𝑧 < 3}

Examples/exercises:
Question 1
For each of the following functions, evaluate 𝑓(3, 2) and find the domain.
√𝑥+𝑦+1
(a) 𝑓(𝑥, 𝑦) = 𝑥−1
(b) 𝑓(𝑥, 𝑦) = 𝑥 ln(𝑦 2 − 𝑥)

Question 2
Find the domain and range of the following functions:
(a) 𝑔(𝑥, 𝑦) = √9 − 𝑥 2 − 𝑦 2 .
(b) ℎ(𝑥, 𝑦) = 4𝑥 2 + 𝑦 2

Question 3
A company makes three sizes od cardboard boxes: small, medium and large. It costs $2.50 to make
a small box, $4.00 to make a medium box and $4.50 to make a large box. Fixed costs are $8000.
(a) Express the cost of making 𝑥 small boxes, 𝑦 medium boxes and 𝑧 large boxes as a function
of three variables: 𝐶 = 𝑓(𝑥, 𝑦, 𝑧).
(b) Find 𝑓(3000, 5000, 4000) and interpret it.
(c) What is the domain of 𝐶?

Question 4
Let 𝑔(𝑥, 𝑦) = cos(2𝑥 + 𝑦).
(a) Evaluate 𝑔(2, −1).

(b) Find the domain of 𝑔.
(c) Find the range of 𝑔.

Question 5
Let 𝐹(𝑥, 𝑦) = 1 + √4 − 𝑦 2 .
(a) Evaluate 𝐹(3, 1).

(b) Find the domain of 𝐹.
(c) Find the range of 𝐹.

Question 6
Let 𝑓(𝑥, 𝑦, 𝑧) = √𝑥 + √𝑦 + √𝑧 + ln(4 − 𝑥 2 − 𝑦 2 − 𝑧 2 ).
(a) Evaluate 𝑓(1, 1, 1).

(b) Find and describe the domain of 𝑓.

Question 7
Let 𝑔(𝑥, 𝑦, 𝑧) = 𝑥 3 𝑦 2 𝑧√10 − 𝑥 − 𝑦 − 𝑧.
(a) Evaluate 𝑔(1, 2, 3).

(b) Find and describe the domain of 𝑔.

Question 8
Find the domain of the following functions.
(a) 𝑓(𝑥, 𝑦) = √𝑥 + 𝑦
(b) 𝑓(𝑥, 𝑦) = √𝑥𝑦
(c) 𝑓(𝑥, 𝑦) = ln(9 − 𝑥 2 − 9𝑦 2 )

(d) 𝑓(𝑥, 𝑦) = √𝑥 2 − 𝑦 2
(e) 𝑓(𝑥, 𝑦) = √1 − 𝑥 2 − √1 − 𝑦 2

(f) 𝑓(𝑥, 𝑦) = √𝑦 + √25 − 𝑥 2 − 𝑦 2
√𝑦−𝑥 2
(g) 𝑓(𝑥, 𝑦) = 1−𝑥 2
(h) 𝑓(𝑥, 𝑦) = arcsin(𝑥 2 + 𝑦 2 − 2)

(i) 𝑓(𝑥, 𝑦, 𝑧) = √1 − 𝑥 2 − 𝑦 2 − 𝑧 2
(j) 𝑓(𝑥, 𝑦, 𝑧) = ln(16 − 4𝑥 2 − 4𝑦 2 − 𝑧 2 )

LIMITS
Similar to the study of single variable calculus, here we study the limits and continuity of functions
of two variables.
The informal definition of the limit of a function of two variables are as given below:
Definition 2
Let 𝑓 be a function of two variables whose domain 𝐷 includes points arbitrarily close to (𝑎, 𝑏). Then
we use the notation
lim 𝑓(𝑥, 𝑦) = 𝐿
(𝑥,𝑦)→(𝑎,𝑏)
to indicate that the values of 𝑓(𝑥, 𝑦) approach the number 𝐿 as the point (𝑥, 𝑦) approaches the point
(𝑎, 𝑏) along any path that stays within the domain of 𝑓.
In other words, we can make the values of 𝑓(𝑥, 𝑦) as close to 𝐿 as we like by taking the point (𝑥, 𝑦)
sufficiently close to the point (𝑎, 𝑏), but not equal to (𝑎, 𝑏).
The formal definition of the limit of a function of two variables are as given below:
Definition 3 (The Epsilon-Delta definition of the limit of a function of two variables)
Let 𝑓 be a function of two variables whose domain 𝐷 includes points arbitrarily close to (𝑎, 𝑏). Then
we say that the limit of 𝑓(𝑥, 𝑦) as (𝑥, 𝑦) approaches (𝑎, 𝑏) is 𝐿 and we write
lim 𝑓(𝑥, 𝑦) = 𝐿
(𝑥,𝑦)→(𝑎,𝑏)
if for every number 𝜀 > 0 there is a corresponding number 𝛿 > 0 such that if (𝑥, 𝑦) ∈ 𝐷 and
0 < √(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 < 𝛿, then |𝑓(𝑥, 𝑦) − 𝐿| < 𝜀.

Note:
1. Other commonly used notations for the limit of functions of two variables given in
Definition 1 are:
i. lim 𝑓(𝑥, 𝑦) = 𝐿
𝑥→𝑎
𝑦→𝑏
ii. 𝑓(𝑥, 𝑦) → 𝐿 as (𝑥, 𝑦) → (𝑎, 𝑏)
2. 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 (𝑎, 𝑏). Therefore
Definition 3 says that the distance between 𝑓(𝑥, 𝑦) and 𝐿 can be made arbitrarily small by
making the distance from (𝑥, 𝑦) and (𝑎, 𝑏) sufficiently small but not 0.
The non-existence of a limit for a function of two variables are as given below:
Definition 4
Let 𝑓 be a function of two variables whose domain 𝐷 includes points arbitrarily close to (𝑎, 𝑏).
If 𝑓(𝑥, 𝑦) → 𝐿1 as (𝑥, 𝑦) → (𝑎, 𝑏) along a path 𝐶1 (i.e. along the 𝑥 -axis) and 𝑓(𝑥, 𝑦) → 𝐿2 as
(𝑥, 𝑦) → (𝑎, 𝑏) along a path 𝐶2 (i.e. along the 𝑦-axis), where 𝐿1 ≠ 𝐿2 , then lim 𝑓(𝑥, 𝑦) = 𝐿
(𝑥,𝑦)→(𝑎,𝑏)
does not exist.

Properties of Limits
Just as for functions of one variable, the calculation and computation of limits can be greatly
simplified by the use of properties of limits. The limit laws that was previously studied in AS100
can be extended to the functions of two variables. The main properties of limits are as given below:
1. The limit of a sum is the sum of the limits:

lim [𝑓(𝑥) + 𝑔(𝑥)] = lim 𝑓(𝑥) + lim 𝑔(𝑥)
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
2. The limit of a difference is the difference of the limits:

lim [𝑓(𝑥) − 𝑔(𝑥)] = lim 𝑓(𝑥) − lim 𝑔(𝑥)
3. The limit of a constant times a function is the constant times the limit of the function:
lim [𝑐𝑓(𝑥)] = 𝑐 [lim 𝑓(𝑥)]

𝑥→𝑎 𝑥→𝑎
4. The limit of a product is the product of the limits:

lim [𝑓(𝑥)𝑔(𝑥)] = lim 𝑓(𝑥) . lim 𝑔(𝑥)
5. The limit of a quotient is the quotient of the limits (provided that the limit of the
denominator is not 0):
𝑓(𝑥) lim 𝑓(𝑥)
lim [ ] = 𝑥→𝑎 , provided that lim 𝑔(𝑥) ≠ 0
𝑥→𝑎 𝑔(𝑥) lim 𝑔(𝑥) 𝑥→𝑎
𝑥→𝑎
6. The limit of a function raised to a power 𝑛 is:

𝑛
lim [𝑓(𝑥)]𝑛 = [lim 𝑓(𝑥)]
7. The limit of the nth root of a function is the nth root of the limit of the function:
𝑛
lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥)

8. The limit of a constant value is the constant value itself:
lim 𝑐 = 𝑐, where 𝑐 is a constant
𝑥→𝑎
9. For functions of two variables, the following properties also hold true:
i. lim 𝑥=𝑎
(𝑥,𝑦)→(𝑎,𝑏)
ii. lim 𝑦=𝑏

(𝑥,𝑦)→(𝑎,𝑏)
iii. lim 𝑐 = 𝑐, where 𝑐 is a constant.

(𝑥,𝑦)→(𝑎,𝑏)

Examples/exercises:
Question 1
Determine whether the limit of the following function exists for (𝑥, 𝑦) → (0, 0).
𝑥2 − 𝑦2
𝑓(𝑥, 𝑦) = 2
𝑥 + 𝑦2
Question 2
Evaluate lim (𝑥 2 𝑦 3 − 𝑥 3 𝑦 2 + 3𝑥 + 2𝑦).

(𝑥,𝑦)→(1,2)

CONTINUITY
Recall that evaluating limits of continuous functions of a single variable function is easy. It can be
accomplished by direct substitution because the defining property of a continuous function is
lim 𝑓(𝑥) = 𝑓(𝑎). Continuity of the functions of two variables are also defined by a similar direct
𝑥→𝑎
substitution property.
The formal definition of the continuity of functions of two variables are as given below:
Definition 5
A function 𝑓 of two variables is said to be continuous at (𝒂, 𝒃) if
lim 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏).

(𝑥,𝑦)→(𝑎,𝑏)
We say that 𝑓 is continuous on an interval/region 𝐷 if 𝑓 is continuous at every point (𝑎, 𝑏) in the

interval/region 𝐷.
Examples:
1. A polynomial function of two variables (polynomials, for short) is a sum of terms of the
form 𝑐𝑥 𝑚 𝑦 𝑛 , where 𝑐 is a constant and 𝑚 and 𝑛 are non-negative integers.
Since any polynomial can be constructed using simple functions by multiplication,
subtraction and addition, all polynomials are continuous on ℝ𝟐 .
2. All rational functions of functions of two variables are only continuous in their respective
domains.

Examples/exercises:
Question 1
Where are the following functions continuous?
(a) 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 3 − 𝑥 3 𝑦 2 + 3𝑥 + 2𝑦
𝑥 2 −𝑦 2
(b) 𝑓(𝑥, 𝑦) = 𝑥 2 +𝑦 2

Examples/exercises (Limits and Continuity):
Question 1
Find the limit for the following functions, if it exists, or show that the limit does not exist.
(a) lim (5𝑥 3 − 𝑥 2 𝑦 2 )

(𝑥,𝑦)→(1,2)
(b) lim 𝑒 −𝑥𝑦 cos(𝑥 + 𝑦)

(𝑥,𝑦)→(1,−1)
4−𝑥𝑦
(c) lim
(𝑥,𝑦)→(2,1) 𝑥 2 +3𝑦 2

1+𝑦 2
(d) lim ln (𝑥 2 +𝑥𝑦)
(𝑥,𝑦)→(1,0)
𝑥 4 −4𝑦 2
(e) lim
(𝑥,𝑦)→(0,0) 𝑥 2 +2𝑦 2
𝑥 2 +sin2 𝑦
(f) lim
(𝑥,𝑦)→(2,1) 2𝑥 2 +𝑦 2

𝑥 4 −𝑦 4
(g) lim
(𝑥,𝑦)→(0,0) 𝑥 2 +𝑦 2
𝑥 2 +𝑦 2
(h) lim
(𝑥,𝑦)→(0,0) √𝑥 2 +𝑦2 +1−1
2
(i) lim 1
𝑒 𝑦 tan(𝑥𝑧)
(𝑥,𝑦,𝑧)→(𝜋,0, )
3

Question 2
Find ℎ(𝑥, 𝑦) = 𝑔(𝑓(𝑥, 𝑦)) and the set on which ℎ is continuous.
(a) 𝑔(𝑡) = 𝑡 2 + √𝑡 , 𝑓(𝑥, 𝑦) = 2𝑥 + 3𝑦 − 6

1−𝑥𝑦
(b) 𝑔(𝑡) = 𝑡 + ln 𝑡 , 𝑓(𝑥, 𝑦) = 1+𝑥 2 𝑦 2

Question 3
Determine the set of points at which the following functions are continuous.
𝑥𝑦
(a) 𝑓(𝑥, 𝑦) = 1+𝑒 𝑥−𝑦
(b) 𝑓(𝑥, 𝑦) = cos(√1 + 𝑥 − 𝑦)

1+𝑥 2 +𝑦 2
(c) 𝑓(𝑥, 𝑦) = 1−𝑥 2 −𝑦 2
𝑒 𝑥 +𝑒 𝑦
(d) ℎ(𝑥, 𝑦) = 𝑒 𝑥𝑦 −1

(e) 𝑔(𝑥, 𝑦) = ln(𝑥 2 + 𝑦 2 − 4)
(f) 𝑓(𝑥, 𝑦, 𝑧) = √𝑦 − 𝑥 2 ln 𝑧

PARTIAL DERIVATIVES
In general, if 𝑓 is a function of two variables 𝑥 and 𝑦, suppose we let only 𝑥 vary while keeping 𝑦
fixed. Then we are in fact finding the derivative of a function of a single variable, say 𝑔(𝑥). If 𝑔
has a derivative at 𝑥 = 𝑎, then we call it the partial derivative of 𝑓 with respect to 𝑥 at point (𝑎, 𝑏)
and denote it as 𝑓𝑥 (𝑎, 𝑏).
By the definition of a derivative, we have
𝑔(𝑎 + ℎ) − 𝑔(𝑎)
𝑔′ (𝑎) = lim
ℎ→0 ℎ
and so the partial derivative of 𝑓 w.r.t. 𝑥 becomes
𝑓(𝑎 + ℎ, 𝑏) − 𝑓(𝑎, 𝑏)
𝑓𝑥 (𝑎, 𝑏) = lim
ℎ→0 ℎ
Similarly, the partial derivative of 𝑓 with respect to 𝑦 at (𝑎, 𝑏), denoted by 𝑓𝑦 (𝑎, 𝑏), is obtained by
keeping 𝑥 fixed and finding the ordinary derivative at 𝑏.
𝑓(𝑎, 𝑏 + ℎ) − 𝑓(𝑎, 𝑏)
𝑓𝑦 (𝑎, 𝑏) = lim
ℎ→0 ℎ

Now if we let the point (𝑎, 𝑏) vary in both of the equations given above, 𝑓𝑥 and 𝑓𝑦 become
functions of two variables.
If 𝑓 is a function of two variables, its partial derivatives are the functions 𝑓𝑥 and 𝑓𝑦 defined by
𝑓(𝑥 + ℎ, 𝑦) − 𝑓(𝑥, 𝑦)
𝑓𝑥 (𝑥, 𝑦) = lim
ℎ→0 ℎ
𝑓(𝑥, 𝑦 + ℎ) − 𝑓(𝑥, 𝑦)
𝑓𝑦 (𝑥, 𝑦) = lim
ℎ→0 ℎ
Notations
There are many alternative notations for partial derivatives. Some of the most commonly used
ones are as given in the table below.
If 𝑧 = 𝑓(𝑥, 𝑦), we can write
𝜕𝑓 𝜕 𝜕𝑧
𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑥 = = (𝑓(𝑥, 𝑦)) = = 𝑓1 = 𝐷1 𝑓 = 𝐷𝑥 𝑓
𝜕𝑥 𝜕𝑥 𝜕𝑥
and
𝜕𝑓 𝜕 𝜕𝑧
𝑓𝑦 (𝑥, 𝑦) = 𝑓𝑦 = = (𝑓(𝑥, 𝑦)) = = 𝑓2 = 𝐷2 𝑓 = 𝐷𝑦 𝑓
𝜕𝑦 𝜕𝑦 𝜕𝑦
The notation 𝑓1 or 𝐷1 𝑓 indicates differentiation with respect to the first variable.

The basic rule for finding partial derivatives of 𝑥 = 𝑓(𝑥, 𝑦):
1. To find 𝑓𝑥 , regard 𝑦 as a constant and differentiate 𝑓(𝑥, 𝑦) with respect to 𝑥.
2. To find 𝑓𝑦 , regard 𝑥 as a constant and differentiate 𝑓(𝑥, 𝑦) with respect to 𝑦.
Note: This rule also applies to functions of more than two variables.

HIGHER DERIVATIVES
If 𝑓 is a function of two variables, then its partial derivatives 𝑓𝑥 and 𝑓𝑦 are also functions of two
variables. Therefore, we can also find the partial derivatives of these derivatives, which are called
the second partial derivatives of 𝑓.
If 𝑧 = 𝑓(𝑥, 𝑦), the following notations are used for the second partial derivatives of 𝑓:
1. The second partial derivative of 𝑓𝑥 with respect to 𝑥:

𝜕 𝜕𝑓 𝜕 2𝑓 𝜕 2𝑧
(𝑓𝑥 )𝑥 = 𝑓𝑥𝑥 = 𝑓11 = ( )= 2= 2
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥
2. The second partial derivative of 𝑓𝑥 with respect to 𝑦:

𝜕 𝜕𝑓 𝜕 2𝑓 𝜕 2𝑧
(𝑓𝑥 )𝑦 = 𝑓𝑥𝑦 = 𝑓12 = ( )= =
𝜕𝑦 𝜕𝑥 𝜕𝑦𝜕𝑥 𝜕𝑦𝜕𝑥
3. The second partial derivative of 𝑓𝑦 with respect to 𝑥:

𝜕 𝜕𝑓 𝜕 2𝑓 𝜕 2𝑧
(𝑓𝑦 )𝑥 = 𝑓𝑦𝑥 = 𝑓21 = ( )= =
𝜕𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦 𝜕𝑥𝜕𝑦
4. The second partial derivative of 𝑓𝑦 with respect to 𝑦:

𝜕 𝜕𝑓 𝜕 2𝑓 𝜕 2𝑧
(𝑓𝑦 )𝑦 = 𝑓𝑦𝑦 = 𝑓22 = ( )= 2= 2
𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦
Note:
𝜕2 𝑓
(g) The notation 𝑓𝑥𝑦 or means that we first differentiate with respect to 𝑥 and then with
𝜕𝑦𝜕𝑥
respect to 𝑦, whereas in computing 𝑓𝑦𝑥 the order is reversed.

(ii) Similar notations also apply for the higher derivatives of functions of two or more
variables.

Examples/Exercises:
Question 1
(a) If 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥 2 𝑦 3 − 2𝑦 2 , find 𝑓𝑥 (2, 1) and 𝑓𝑦 (2, 1).

(b) If 𝑓(𝑥, 𝑦) = 4 − 𝑥 2 − 2𝑦 2 , find 𝑓𝑥 (1, 1) and 𝑓𝑦 (1, 1).

Question 2
𝑥 𝜕𝑓 𝜕𝑓
(a) If 𝑓(𝑥, 𝑦) = sin (1+𝑦), calculate 𝜕𝑥 and 𝜕𝑦.
𝜕𝑧 𝜕𝑧
(b) Find 𝜕𝑥 and 𝜕𝑦 if 𝑧 is defined implicitly as a function of 𝑥 and 𝑦 by the equation
𝑥 3 + 𝑦 3 + 𝑧 3 + 6𝑥𝑦𝑧 = 1.

Question 3
(a) Find 𝑓𝑥 , 𝑓𝑦 and 𝑓𝑧 if 𝑓(𝑥, 𝑦, 𝑧) = 𝑒 𝑥𝑦 ln 𝑧.
(b) Find the second partial derivatives of 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥 2 𝑦 3 − 2𝑦 2 .
(c) Calculate 𝑓𝑥𝑥𝑦𝑧 if 𝑓(𝑥, 𝑦, 𝑧) = sin(3𝑥 + 𝑦𝑧).

Question 4
Find ALL the first partial derivatives of the following functions.
(a) 𝑓(𝑥, 𝑦) = 𝑦 5 − 3𝑥𝑦

(b) 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 3 + 8𝑥 2 𝑦
(c) 𝑓(𝑥, 𝑡) = 𝑒 −𝑡 cos 𝜋𝑥
(d) 𝑓(𝑥, 𝑡) = √𝑥 ln 𝑡

(e) 𝑧 = (2𝑥 + 3𝑦)10
(f) 𝑧 = tan 𝑥𝑦
𝑥
(g) 𝑓(𝑥, 𝑦) = 𝑦

𝑥
(h) 𝑓(𝑥, 𝑦) = (𝑥+𝑦)2
𝑎𝑥+𝑏𝑦
(i) 𝑓(𝑥, 𝑦) = 𝑐𝑥+𝑑𝑦

𝑒𝑣
(j) 𝑤 = 𝑢+𝑣2
(k) 𝑔(𝑢, 𝑣) = (𝑢2 𝑣 − 𝑣 3 )5

(l) 𝑓(𝑥, 𝑡) = arctan(𝑥 √𝑡)
(m) 𝑤 = sin 𝛼 cos 𝛽
(n) 𝑓(𝑥, 𝑦) = 𝑥 𝑦

(o) 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑧 − 5𝑥 2 𝑦 3 𝑧 4
(p) 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 sin(𝑦 − 𝑧)

(q) 𝑤 = ln(𝑥 + 2𝑦 + 3𝑧)
(r) 𝑤 = 𝑧𝑒 𝑥𝑦𝑧

(s) 𝑢 = 𝑥𝑦 arcsin(𝑦𝑧)
𝑦
(t) 𝑢 = 𝑥 𝑧

𝑧
(u) ℎ(𝑥, 𝑦, 𝑧, 𝑡) = 𝑥 2 𝑦 cos ( 𝑡 )

𝛼𝑥+𝛽𝑦 2
(v) 𝜙(𝑥, 𝑦, 𝑧, 𝑡) = 𝛾𝑧+𝛿𝑡 2

(w) 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 3
𝑦
(x) 𝑓(𝑥, 𝑦) = 1+𝑥 2 𝑦 2

(y) 𝑓(𝑥, 𝑦) = 𝑥𝑦 2 − 𝑥 3 𝑦
𝑥
(z) 𝑓(𝑥, 𝑦) = 𝑥+𝑦 2

Question 5
Find the indicated partial derivative of the following functions at the indicated points.
(a) 𝑓(𝑥, 𝑦) = ln(𝑥 + √𝑥 2 + 𝑦 2 ); 𝑓𝑥 (3, 4)

𝑦
(b) 𝑓(𝑥, 𝑦) = arctan (𝑥 ) ; 𝑓𝑥 (2, 3)

𝑦
(c) 𝑓(𝑥, 𝑦, 𝑧) = 𝑥+𝑦+𝑧 ; 𝑓𝑦 (2, 1, −1)
𝜋
(d) 𝑓(𝑥, 𝑦, 𝑧) = √sin2 𝑥 + sin2 𝑦 + sin2 𝑧 ; 𝑓𝑧 (0, 0, 4 )

Question 6
Use the first principles method to find 𝑓𝑥 (𝑥, 𝑦) and 𝑓𝑦 (𝑥, 𝑦).
(a) (𝑥, 𝑦) = 𝑥𝑦 2 − 𝑥 3 𝑦

𝑥
(b) 𝑓(𝑥, 𝑦) = 𝑥+𝑦 2

Question 7
𝜕𝑧 𝜕𝑧
Find 𝜕𝑥 and 𝜕𝑦 .
(a) 𝑧 = 𝑓(𝑥) + 𝑔(𝑦)

(b) 𝑧 = 𝑓(𝑥 + 𝑦)
(c) 𝑧 = 𝑓(𝑥)𝑔(𝑦)
(d) 𝑧 = 𝑓(𝑥𝑦)
𝑥
(e) 𝑧 = 𝑓 (𝑦)

Question 8
Find all the second partial derivatives of the following functions.
(a) 𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 5 + 2𝑥 4 𝑦
(b) 𝑓(𝑥, 𝑦) = sin2 (𝑚𝑥 + 𝑛𝑦)

(c) 𝑤 = √𝑢2 + 𝑣 2
𝑥𝑦
(d) 𝑣 = 𝑥−𝑦

𝑥+𝑦
(e) 𝑧 = arctan (1−𝑥𝑦)

𝑦
(f) 𝑣 = 𝑒 𝑥𝑒

Question 9
Find the indicated partial derivative(s) of the following functions.
(a) 𝑓(𝑥, 𝑦) = 𝑥 4 𝑦 2 − 𝑥 3 𝑦 ; 𝑓𝑥𝑥𝑥 , 𝑓𝑥𝑦𝑥

(b) 𝑓(𝑥, 𝑦) = sin(2𝑥 + 5𝑦) ; 𝑓𝑦𝑥𝑦

2
(c) 𝑓(𝑥, 𝑦, 𝑧) = 𝑒 𝑥𝑦𝑧 ; 𝑓𝑥𝑦𝑧
(d) 𝑔(𝑟, 𝑠, 𝑡) = 𝑒 𝑟 sin(𝑠𝑡) ; 𝑔𝑟𝑠𝑡

𝜕3 𝑢
(e) 𝑢 = 𝑒 𝑟𝜃 sin 𝜃 ; 𝜕𝑟 2 𝜕𝜃

𝜕3 𝑧
(f) 𝑧 = 𝑢√𝑣 − 𝑤 ; 𝜕𝑢𝜕𝑣𝜕𝑤

𝑥 𝜕3 𝑤 𝜕3 𝑤
(g) 𝑤 = 𝑦+2𝑧 ; ,
𝜕𝑧𝜕𝑦𝜕𝑥 𝜕𝑥 2 𝜕𝑦

𝜕6 𝑢
(h) 𝑢 = 𝑥 𝑎 𝑦 𝑏 𝑧 𝑐 ; 𝜕𝑥𝜕𝑦 2 𝜕𝑧 3

Question 10
Verify that the function 𝑧 = ln(𝑒 𝑥 + 𝑒 𝑦 ) is a solution of the following differential equations.
𝜕𝑧 𝜕𝑧
(a) + 𝜕𝑦 = 1
𝜕𝑥

2
𝜕2 𝑧 𝜕2 𝑧 𝜕2 𝑧
(b) − (𝜕𝑥𝜕𝑦) = 0.
𝜕𝑥 2 𝜕𝑦 2

THE CHAIN RULE
In this section, we deal with the chain rules and implicit differentiation for functions of more than
one variable.
The chain rule for functions of a single variable gives the rule for differentiating composite
functions of the form:
If 𝑦 = 𝑓(𝑥) and 𝑥 = 𝑔(𝑡), where 𝑓 and 𝑔 are differentiable functions, then 𝑦 is indirectly a
differentiable function of 𝑡 and
𝑑𝑦 𝑑𝑦 𝑑𝑥
=
𝑑𝑡 𝑑𝑥 𝑑𝑡
Similar chain rules are also available for dealing with functions of more than one variable. There
are several cases for these chain rules, but only the three most important cases are given here.
The Chain Rule (Case 1: For finding the derivative of 𝒛 w.r.t. 𝒕.)
Suppose that 𝑧 = 𝑓(𝑥, 𝑦) is a differentiable function of 𝑥 and 𝑦, where 𝑥 = 𝑔(𝑡) and 𝑦 = ℎ(𝑡)
are both differentiable functions of 𝑡. Then 𝑧 is a differentiable function of 𝑡 and
𝑑𝑧 𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦
= +
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

The Chain Rule (Case 2: For finding the partial derivatives of 𝒛)
Suppose that 𝑧 = 𝑓(𝑥, 𝑦) is a differentiable function of 𝑥 and 𝑦, where 𝑥 = 𝑔(𝑠, 𝑡) and 𝑦 =

ℎ(𝑠, 𝑡) are both differentiable functions of 𝑠 and 𝑡. Then 𝑧 is a differentiable function of 𝑡 and
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
= +
𝜕𝑠 𝜕𝑥 𝜕𝑠 𝜕𝑦 𝜕𝑠
and
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
= +
𝜕𝑡 𝜕𝑥 𝜕𝑡 𝜕𝑦 𝜕𝑡
The general version of the chain rule is as given below:
The Chain Rule (General Version: For finding the partial derivatives of the multi-variable
function 𝒖)
Suppose that 𝑢 is a differentiable function of the 𝑛 variables 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥𝑛 and each 𝑥𝑗 is a

differentiable function of the 𝑚 variables 𝑡1 , 𝑡2 , 𝑡3 , … , 𝑡𝑚 . Then 𝑢 is a function of
𝑡1 , 𝑡2 , 𝑡3 , … , 𝑡𝑚 and
𝜕𝑢 𝜕𝑢 𝜕𝑥1 𝜕𝑢 𝜕𝑥2 𝜕𝑢 𝜕𝑥𝑛
= + +⋯+
𝜕𝑡1 𝜕𝑥1 𝜕𝑡𝑖 𝜕𝑥2 𝜕𝑡𝑖 𝜕𝑥𝑛 𝜕𝑡𝑖
for each 𝑖 = 1, 2, … , 𝑚.

Examples/Exercises
Question 1
𝑑𝑧
If 𝑧 = 𝑥 2 𝑦 + 3𝑥𝑦 4 , where 𝑥 = sin 2𝑡 and 𝑦 = cos 𝑡, find 𝑑𝑡 when 𝑡 = 0.

Question 2
𝜕𝑧 𝜕𝑧
If 𝑧 = 𝑒 𝑥 sin 𝑦, where 𝑥 = 𝑠𝑡 2 and 𝑦 = 𝑠 2 𝑡, find 𝜕𝑠 and 𝜕𝑡 .

Question 3
Write out the Chain Rule for the case where 𝑤 = 𝑓(𝑥, 𝑦, 𝑧, 𝑡) and 𝑥 = 𝑥(𝑢, 𝑣), 𝑦 = 𝑦(𝑢, 𝑣),
𝑧 = 𝑧(𝑢, 𝑣) and 𝑡 = 𝑡(𝑢, 𝑣).

Question 4
𝜕𝑢
If 𝑢 = 𝑥 4 𝑦 + 𝑦 2 𝑧 3 , where 𝑥 = 𝑟𝑠𝑒 𝑡 , 𝑦 = 𝑟𝑠 2 𝑒 −𝑡 and 𝑧 = 𝑟 2 𝑠 sin 𝑡, find the value of when
𝜕𝑠
𝑟 = 2, 𝑠 = 1, 𝑡 = 0.

Question 5
If 𝑔(𝑠, 𝑡) = 𝑓(𝑠 2 − 𝑡 2 , 𝑡 2 − 𝑠 2 ) and 𝑓 is differentiable, show that 𝑔 satisfies the equation
𝜕𝑔 𝜕𝑔
𝑡 +𝑠 = 0.
𝜕𝑠 𝜕𝑡

Question 6
𝑑𝑧 𝑑𝑤
Use the Chain Rule to find 𝑑𝑡 or .
𝑑𝑡
(a) 𝑧 = 𝑥 2 + 𝑦 2 + 𝑥𝑦 ; 𝑥 = sin 𝑡 , 𝑦 = 𝑒 𝑡

1
(b) 𝑧 = cos(𝑥 + 4𝑦) ; 𝑥 = 5𝑡 4 , 𝑦 = 𝑡

(c) 𝑧 = √1 + 𝑥 2 + 𝑦 2 ; 𝑥 = ln 𝑡 , 𝑦 = cos 𝑡

𝑦
(d) 𝑧 = tan−1 (𝑥 ) ; 𝑥 = 𝑒 𝑡 , 𝑦 = 1 − 𝑒 −𝑡

𝑦
(e) 𝑤 = 𝑥 𝑒 𝑧 ; 𝑥 = 𝑡 2 , 𝑦 = 1 − 𝑡, 𝑧 = 1 + 2𝑡

(f) 𝑤 = ln √𝑥 2 + 𝑦 2 + 𝑧 2 ; 𝑥 = sin 𝑡 , 𝑦 = cos 𝑡 , 𝑧 = tan 𝑡

Question 7
𝜕𝑧 𝜕𝑧
Use the Chain Rule to find and .
𝜕𝑠 𝜕𝑡
(a) 𝑧 = 𝑥 2 𝑦 3 ; 𝑥 = 𝑠 cos 𝑡 , 𝑦 = 𝑠 sin 𝑡

(b) 𝑧 = arcsin(𝑥 − 𝑦) ; 𝑥 = 𝑠 2 + 𝑡 2 , 𝑦 = 1 − 2𝑠𝑡

(c) 𝑧 = sin 𝜃 cos 𝜙 ; 𝜃 = 𝑠𝑡 2 , 𝜙 = 𝑠 2 𝑡

𝑠 𝑡
(d) 𝑧 = 𝑒 𝑥+2𝑦 ; 𝑥=𝑡 , 𝑦=𝑠

(e) 𝑧 = 𝑒 𝑟 cos 𝜃 ; 𝑟 = 𝑠𝑡, 𝜃 = √𝑠 2 + 𝑡 2

𝑢
(f) 𝑧 = tan (𝑣 ) ; 𝑢 = 2𝑠 + 3𝑡, 𝑣 = 3𝑠 − 2𝑡

Question 8
If 𝑧 = 𝑓(𝑥, 𝑦), where 𝑓 is differentiable, and
𝑥 = 𝑔(𝑡), 𝑔(3) = 2, 𝑔′ (3) = 5, 𝑓𝑥 (2, 7) = 6 ;
𝑦 = ℎ(𝑡), ℎ(3) = 7, ℎ′ (3) = −4, 𝑓𝑦 (2, 7) = −8 ;
𝑑𝑧
find when 𝑡 = 3.
𝑑𝑡

Question 9
Let 𝑊(𝑠, 𝑡) = 𝐹(𝑢(𝑠, 𝑡), 𝑣(𝑠, 𝑡)), where 𝐹, 𝑢, 𝑣 are differentiable, and
𝑢(1, 0) = 2, 𝑢𝑠 (1, 0) = −2, 𝑢𝑡 (1, 0) = 6, 𝐹𝑢 (2, 3) = −1 ;
𝑣(1, 0) = 3, 𝑣𝑠 (1, 0) = 5, 𝑣𝑡 (1, 0) = 4, 𝐹𝑣 (2, 3) = 10.
Find 𝑊𝑠 (1, 0) and 𝑊𝑡 (1, 0).

Question 10
Use the Chain Rule to find the indicated partial derivatives at the indicated points.
𝜕𝑧 𝜕𝑧 𝜕𝑧
(a) 𝑧 = 𝑥 2 + 𝑥𝑦 3 ; 𝑥 = 𝑢𝑣 2 + 𝑤 3 , 𝑦 = 𝑢 + 𝑣𝑒 𝑤 ; , , 𝜕𝑤 when 𝑢 = 2, 𝑣 = 1, 𝑤 = 0
𝜕𝑢 𝜕𝑣

𝜕𝑢 𝜕𝑢 𝜕𝑢
(b) 𝑢 = √𝑟 2 + 𝑠 2 ; 𝑟 = 𝑥 + 𝑦 cos 𝑡 , 𝑠 = 𝑥 + 𝑦 sin 𝑡 ; , ,
𝜕𝑥 𝜕𝑦 𝜕𝑡
when 𝑥 = 1, 𝑦 = 2, 𝑡 = 0

𝜕𝑤 𝜕𝑤 𝜋
(c) 𝑤 = 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 ; 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 , 𝑧 = 𝑟𝜃 ; , ; when 𝑟 = 2, 𝜃 =
𝜕𝑟 𝜕𝜃 2

𝜕𝑃 𝜕𝑃
(d) 𝑃 = √𝑢2 + 𝑣 2 + 𝑤 2 ; 𝑢 = 𝑥𝑒 𝑦 , 𝑣 = 𝑦𝑒 𝑥 , 𝑤 = 𝑒 𝑥𝑦 ; , ; when 𝑥 = 0, 𝑦 = 2
𝜕𝑥 𝜕𝑦

𝑝+𝑞 𝜕𝑁 𝜕𝑁 𝜕𝑁
(e) 𝑁 = ; 𝑝 = 𝑢 + 𝑣𝑤, 𝑞 = 𝑣 + 𝑢𝑤, 𝑟 = 𝑤 + 𝑢𝑣 ; , 𝜕𝑣 , 𝜕𝑤 ;
𝑝+𝑟 𝜕𝑢
when 𝑢 = 2, 𝑣 = 3, 𝑤 = 4

𝜕𝑢 𝜕𝑢 𝜕𝑢
(f) 𝑢 = 𝑥 𝑒 𝑡𝑦 ; 𝑥 = 𝛼 2 𝛽 , 𝑦 = 𝛽 2 𝛾 , 𝑡 = 𝛾 2 𝛼 ; , , ; when 𝛼 = −1, 𝛽 = 2, 𝛾 = 1
𝜕𝛼 𝜕𝛽 𝜕𝛾

IMPLICIT DIFFERENTIATION
Suppose that we have a function of two variables given by
𝐹(𝑥, 𝑦) = 0
that defines 𝑦 implicitly as a differentiable function of 𝑥.
If 𝐹 is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the
equation 𝐹(𝑥, 𝑦) = 0 with respect to 𝑥. Since both 𝑥 and 𝑦 are functions of 𝑥, we obtain
𝜕𝐹 𝑑𝑥 𝜕𝐹 𝑑𝑦
+ = 0.
𝜕𝑥 𝑑𝑥 𝜕𝑦 𝑑𝑥
𝑑𝑥 𝜕𝐹 𝑑𝑦
But = 1, so if ≠ 0, we can solve for and obtain
𝑑𝑥 𝜕𝑦 𝑑𝑥
𝜕𝐹
𝑑𝑦 ( ) 𝐹𝑥
= − 𝜕𝑥 = −
𝑑𝑥 𝜕𝐹 𝐹𝑦
𝜕𝑦
Similarly, for functions of three variables, the formula for implicit differentiation are as given
below:
𝜕𝐹 𝜕𝐹
𝜕𝑧 ( ) 𝐹𝑥 𝜕𝑧 ( ) 𝐹𝑦
𝜕𝑦
= − 𝜕𝑥 = − & = − =−
𝜕𝑥 𝜕𝐹 𝐹𝑧 𝜕𝑦 𝜕𝐹 𝐹𝑧
𝜕𝑧 𝜕𝑧

Examples/Exercises
Question 1
Find 𝑦′ if 𝑥 3 + 𝑦 3 = 6𝑥𝑦.

Question 2
𝜕𝑧 𝜕𝑧
Find and for the following functions:
𝜕𝑥 𝜕𝑦
(a) 𝑥 3 + 𝑦 3 + 𝑧 3 + 6𝑥𝑦𝑧 = 1
(b) 𝑥 2 + 2𝑦 2 + 3𝑧 2 = 1
(c) 𝑥 2 − 𝑦 2 + 𝑧 2 − 2𝑧 = 4

(d) 𝑒 𝑧 = 𝑥𝑦𝑧
(e) 𝑦𝑧 + 𝑥 ln 𝑦 = 𝑧 2

Question 3
𝑑𝑦
Find 𝑑𝑥 for the following functions.
(a) 𝑦 cos 𝑥 = 𝑥 2 + 𝑦 2
(b) cos(𝑥𝑦) = 1 + sin 𝑦

(c) tan−1(𝑥 2 𝑦) = 𝑥 + 𝑥𝑦 2
(d) 𝑒 𝑦 sin 𝑥 = 𝑥 + 𝑥𝑦


AS101 - Topic 4 - Partial Derivatives - Revised

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AS101 - Topic 4 - Partial Derivatives - Revised

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AS101 – MATHEMATICS I

TOPIC 4 – PARTIAL DERIVATIVES

KEY QUESTIONS AND KEY IDEAS

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 1

FUNCTIONS OF TWO VARIABLES

Domain: The range of 𝑥-values for which the function is defined.

Notation: 𝐷 = {(𝑥, 𝑦): 𝑥 ≠ 1, 𝑦 ≥ 1}

Notation: 𝑅 = {𝑧: 1 ≤ 𝑧 < 3}

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 2

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 3

Find the domain and range of the following functions:

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 4

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 5

Let 𝑔(𝑥, 𝑦) = cos(2𝑥 + 𝑦).

(a) Evaluate 𝑔(2, −1).

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 6

(a) Evaluate 𝐹(3, 1).

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 7

Let 𝑓(𝑥, 𝑦, 𝑧) = √𝑥 + √𝑦 + √𝑧 + ln(4 − 𝑥 2 − 𝑦 2 − 𝑧 2 ).

(a) Evaluate 𝑓(1, 1, 1).

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 8

Let 𝑔(𝑥, 𝑦, 𝑧) = 𝑥 3 𝑦 2 𝑧√10 − 𝑥 − 𝑦 − 𝑧.

(a) Evaluate 𝑔(1, 2, 3).

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 9

Find the domain of the following functions.

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 10

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 11

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 12

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 13

Definition 3 (The Epsilon-Delta definition of the limit of a function of two variables)

0 < √(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 < 𝛿, then |𝑓(𝑥, 𝑦) − 𝐿| < 𝜀.

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 14

ii. 𝑓(𝑥, 𝑦) → 𝐿 as (𝑥, 𝑦) → (𝑎, 𝑏)

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 15

1. The limit of a sum is the sum of the limits:

2. The limit of a difference is the difference of the limits:

lim [𝑐𝑓(𝑥)] = 𝑐 [lim 𝑓(𝑥)]

4. The limit of a product is the product of the limits:

6. The limit of a function raised to a power 𝑛 is:

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 16

ii. lim 𝑦=𝑏

iii. lim 𝑐 = 𝑐, where 𝑐 is a constant.

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 17

Evaluate lim (𝑥 2 𝑦 3 − 𝑥 3 𝑦 2 + 3𝑥 + 2𝑦).

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 18

A function 𝑓 of two variables is said to be continuous at (𝒂, 𝒃) if

lim 𝑓(𝑥, 𝑦) = 𝑓(𝑎, 𝑏).

We say that 𝑓 is continuous on an interval/region 𝐷 if 𝑓 is continuous at every point (𝑎, 𝑏) in the

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 19

Where are the following functions continuous?

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 20

(a) lim (5𝑥 3 − 𝑥 2 𝑦 2 )

(b) lim 𝑒 −𝑥𝑦 cos(𝑥 + 𝑦)

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 21

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 22

Dr. Ganeshsree/AS101_Mathematics I/Topic 4_Partial Derivatives Page 23

Find ℎ(𝑥, 𝑦) = 𝑔(𝑓(𝑥, 𝑦)) and the set on which ℎ is continuous.

(a) 𝑔(𝑡) = 𝑡 2 + √𝑡 , 𝑓(𝑥, 𝑦) = 2𝑥 + 3𝑦 − 6