Chapter (1)

Partial Differentiation
1.1 Functions of several variables

The functions studied in pervious chapters (ordinary

differentiation), involved only one independent variable.
But, in this chapter we shall define functions of several
variables and some of their properties. we begin with the
definition of a function of two variables.

Definition (1)

Let D be a set of ordered pairs of real numbers. A rule

𝑓 that assigns to each pair in D a unique real number
𝑓(𝑥, 𝑦) is called a function of two variables. The number
𝑓(𝑥, 𝑦) is called the value of 𝑓 at (𝑥, 𝑦). The set D is called
the domain of 𝑓 and the set of all values of 𝑓 is called the
range of 𝑓. The domain D is usually a subset of the
𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒.

Example (1)

Let 𝑓(𝑥, 𝑦) = √𝑥 2 + 𝑦 2 − 4 . Find the value of 𝑓 at points

(2,2) (-1,3) (6,4) and find the domain of 𝑓 𝑎𝑛𝑑 𝑠𝑘𝑒𝑡𝑐ℎ it.
Solution
1
The values of 𝑓 at points (2,2) (-1,3) (6,2) are obtained by
straight substitution

𝑓(2,2) = √4 + 4 − 4 = 2, 𝑓 (−1,3) = √1 + 9 − 4
= √6, 𝑓(6,2) = 6

The domain 𝐷 is the set of all pairs (x, y) with

𝑥 2 + 𝑦 2 − 4 ≥ 0 or 𝑥 2 + 𝑦 2 ≥ 4. these are points lie
outside or on the circle 𝑥 2 + 𝑦 2 = 4

Example (2)

1
Let 𝑓(𝑥, 𝑦) = . Find the domain of 𝑓 𝑎𝑛𝑑 𝑠𝑘𝑒𝑡𝑐ℎ it.
√𝑦−8𝑥 2

Solution

The domain 𝐷 is the set of all pairs (x, y) with 𝑦 − 8𝑥 2 > 0

or 𝑦 > 8𝑥 2 . this is a subset of the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 above the
parabola 𝑦 = 8𝑥 2 Fig (5.2)

1.2 Limits and Continuity

1.2.1 Limits

Let 𝑓(𝑥, 𝑦) be a function of two variables. If the values of

(𝑥, 𝑦) get closer and closer to a fixed point (a ,b), then we
say that the limit of 𝑓 as (𝑥, 𝑦) approaches (a ,b) equal 𝐿,
and we write

2
 lim 𝑓 (𝑥, 𝑦) = 𝐿
(𝑥,𝑦)→(𝑎,𝑏)

Notes,

1 the same concept as the limit of a function of one

variable,

2 the properties the limits function of one variable of

sums, products, and quotients can be extended using the
same arguments.

For example, if

lim 𝑓(𝑥, 𝑦) = 𝐿 , lim 𝑔(𝑥, 𝑦) = 𝑀

(𝑥,𝑦)→(𝑎,𝑏) (𝑥,𝑦)→(𝑎,𝑏)

then
lim (𝑓 + 𝑔)(𝑥, 𝑦) = 𝐿 + 𝑀
(𝑥,𝑦)→(𝑎,𝑏)

In the same way the other properties.

What is the difference between limit of a function of one

variable and two variables?

The difference is that in the one variable, 𝑥 can approach an

essentially in two ways, from the right or the left. However,
in the two variables (𝑥, 𝑦) can approach (𝑎, 𝑏) in infinitely
many ways.

When does limit of the function exist?

3
If the limit is to exist, then 𝑓(𝑥, 𝑦) must get close to the
same number 𝐿 no matter what path is taken by (𝑥, 𝑦) in
approaching (𝑎, 𝑏)

Example (3)

Show that

𝑥 2 − 𝑦2
lim
(𝑥,𝑦)→(0,0) 𝑥 2 + 𝑦 2

does not exist.

Solution

The first path

If (𝑥, 𝑦) approaches (0,0) along the x-axis, then 𝑦 = 0 and

𝑥2 − 0
lim =1
(𝑥,0)→(0,0) 𝑥 2 + 0

The second path

If (𝑥, 𝑦) approaches (0,0) along the y-axis, then 𝑥 = 0 and

0 − 𝑦2
lim = −1
(0,𝑦)→(0,0) 0 + 𝑦 2

The third path

If (𝑥, 𝑦) approaches (0,0) along the line 𝑦 = 𝑚𝑥 , in that

case
4
 𝑥 2 − 𝑚2 𝑥 2 1 − 𝑚2
lim =
(𝑥,𝑦)→(0,0) 𝑥 2 + 𝑚2 𝑥 2 1 + 𝑚2

The limit does not exist, The limit depend on 𝒎 so that the
limit does not exit because its value is different on
difference paths.

Example 4

Evaluate

𝑥𝑦
lim
(𝑥,𝑦)→(−1,2) 𝑥 2 + 𝑦 2

Solution.

𝑦−2
If (𝑥, 𝑦) approaches (−1,2) along the line 𝑚 = or
𝑥+1

𝑦 = 𝑚(𝑥 + 1) + 2, in that case

𝑥𝑦 𝑥 (𝑚(𝑥 + 1) + 2)
lim = lim 2
(𝑥,𝑦)→(−1,2) 𝑥 2 + 𝑦 2 𝑥→−1 𝑥 + (𝑚(𝑥 + 1) + 2)2

−1(𝑚(−1 + 1) + 2) −2
= =
(−1)2 + (𝑚(−1 + 1) + 2)2 5

so that the limit does exit

1.2.2 Continuity

We can define the continuity of 𝑓(𝑥, 𝑦) at the point (𝑎, 𝑏)

as follows. Suppose 𝑓 is a function of two variables with
domain D. Then 𝑓 is continuous at (𝑎, 𝑏) if

5
 lim 𝑓 (𝑥, 𝑦) = 𝑓(𝑎, 𝑏)
(𝑥,𝑦)→(𝑎,𝑏)

All of the continuity theorems for functions of one variable

also hold for functions of two variables.

Example (5)

Discuss the continuity of the following function

𝑥2
, (𝑥, 𝑦) ≠ (0,0)
𝑓(𝑥, 𝑦) = {𝑥 2 + 𝑦 2
0, (𝑥, 𝑦) = (0,0)

Solution

The domain of 𝑓 is the whole plane. We can view 𝑓 as a

quotient of two functions which are continuous, thus 𝑓 is
continuous except at (0,0), which the function is defined by
two formulas. From the study limits we show that

𝑥2
lim
(𝑥,𝑦)→(0,0) 𝑥 2 + 𝑦 2

does not exist. Because, if (𝑥, 𝑦) approaches (0,0) along the

line 𝑦 = 𝑚𝑥 , in that case

𝑥2 1
lim =
𝑥→0 𝑥 2 + 𝑚2 𝑥 2 1 + 𝑚2

The limit depend on 𝒎 so that the limit does not exit which
implies 𝑓(𝑥, 𝑦) is not continuous at (0,0)
6
Theorem
(a) If 𝑔(𝑥) is continuous at 𝑥0 and ℎ(𝑦) is continuous at 𝑦0 ,
then 𝑓(𝑥, 𝑦) = 𝑔(𝑥)ℎ(𝑦) is continuous at (𝑥0 , 𝑦0 ).

(b) If ℎ(𝑥, 𝑦) is continuous at(𝑥0 , 𝑦0 ).and𝑔(𝑢) is continuous

at 𝑢 = ℎ (𝑥0 , 𝑦0 ), then the composition 𝑓 (𝑥, 𝑦) =
𝑔(ℎ(𝑥, 𝑦)) is continuous at (𝑥0 , 𝑦0 ).

(c) If 𝑓 (𝑥, 𝑦) is continuous at (𝑥0 , 𝑦0 )., and if 𝑥(𝑡) and 𝑦(𝑡)

are continuous at t0 with 𝑥 (𝑡0 ) = 𝑥0 and 𝑦(𝑡0 ) = 𝑦0 , then
the composition f (𝑥(𝑡), 𝑦(𝑡)) is continuous at 𝑡0 .

Example 6

Use Theorem to show that the functions

𝑓(𝑥, 𝑦) = 3𝑥 2 𝑦 5 and f(x, y)=sin(3𝑥 2 𝑦 5 ) are continuous

everywhere.
Solution.

The polynomials 𝑔(𝑥) = 3𝑥 2 and ℎ(𝑦) = 𝑦 5 are continuous

at every real number, and therefore by part (a) of Theorem,
the function 𝑓(𝑥, 𝑦) = 3𝑥 2 𝑦 5 is continuous at every point
(𝑥, 𝑦) in the xy-plane. Since 3𝑥 2 𝑦 5 is continuous at every
point in the xy-plane and 𝑠𝑖𝑛 𝑢 is continuous at every real
number 𝑢, it follows from part (b) of theorem, that the
composition f(x, y)=sin(3𝑥 2 𝑦 5 )is continuous everywhere.

7
Recognizing Continuous Functions
• A composition of continuous functions is continuous.
• A sum, difference, or product of continuous functions is
continuous.
• A quotient of continuous functions is continuous, except
where the denominator is zero.

Remark

By using these principles and theorem, you should be able to

confirm that the following functions are all continuous
everywhere:

2 xy
xexy + y 3 , cosh(xy 3 ) − |xy|,
1 + x2 + y2

Example 7

Study the continuous of 𝑓(𝑥, 𝑦) = 𝑥𝑦/(𝑥2 + 𝑦2)

𝑥𝑦
lim
(𝑥,𝑦)→(−1,2) 𝑥 2 + 𝑦 2

Solution.

Since 𝑓(𝑥, 𝑦) = 𝑥𝑦/(𝑥2 + 𝑦2) is continuous at (−1, 2)

(why?), it follows from the definition of continuity for
functions of two variables that
𝑥𝑦 −1(2) −2
lim = =
(𝑥,𝑦)→(−1,2) 𝑥 2 + 𝑦 2 (−1)2 + (2)2 5

8
1.3. Partial Derivatives

Let 𝑓 be a function of two variables, then the first partial

derivatives of 𝑓 with respect to 𝑥 and 𝑦 are the function𝑓𝑥
and 𝑓𝑦 defined by

𝑓 (𝑥 + ∆𝑥, 𝑦) − 𝑓(𝑥, 𝑦)
𝑓𝑥 (𝑥, 𝑦) = lim
∆𝑥→0 ∆𝑥

𝑓 (𝑥, 𝑦 + ∆𝑦) − 𝑓(𝑥, 𝑦)

𝑓𝑦 (𝑥, 𝑦) = lim
∆𝑦→0 ∆𝑦

In the definition of 𝑓𝑥 , 𝑦 is held fixed; only 𝑥 is allowed to

vary. Similarly of 𝑓𝑦 , 𝑥 is held fixed; only 𝑦 is allowed to
vary.

Example 5

If 𝑓 (𝑥, 𝑦) = 𝑥 2 + 𝑥𝑦 2 + 𝑦 3 find

(i) 𝑓𝑥 (𝑥, 𝑦) (ii) 𝑓𝑦 (𝑥, 𝑦) (iii) 𝑓𝑥 (1,2) (iv) 𝑓𝑦 (1,2)

Solution

𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑥𝑦 2 + 𝑦 3

(i) 𝑓𝑥 (𝑥, 𝑦) = 2𝑥 + 𝑦 2
(ii) 𝑓𝑦 (𝑥, 𝑦) = 2𝑥𝑦 + 3𝑦 2
(iii) 𝑓𝑥 (1,2) = 2(1) + (2)2 = 6
(iv) 𝑓𝑦 (1,2) = 2(1)(2) + 3(2)2 = 16

9
Notation

𝑑𝑓
Recall Leibniz notation for the derivative of a function of
𝑑𝑥

one variable. For partial derivatives, the 𝑑 becomes a

script delta 𝜕 thus

𝜕𝑓
𝑓𝑥 can also be written as , and
𝜕𝑥

𝜕𝑓
𝑓𝑦 can also be written as ,
𝜕𝑦

Example (6)

𝜕𝑤
If 𝑤 = 𝑥𝑦 2 𝑒 𝑥𝑦 , find
𝜕𝑦

Solution

We apply the rule of differentiation of product of

functions, we get

𝜕𝑤
= 2𝑥𝑦𝑒 𝑥𝑦 + 𝑥𝑦 2 𝑒 𝑥𝑦 (𝑥),
𝜕𝑦

1.4 Higher Order Partial Derivatives

The second partial derivatives of 𝑓(𝑥, 𝑦) are defined as

follows

𝜕2𝑓
(𝑓𝑥 )𝑥 is defined by 𝑓𝑥𝑥 or
𝜕𝑥 2

10
 𝜕2𝑓
(𝑓𝑥 )𝑦 is defined by 𝑓𝑥𝑦 or
𝜕𝑥𝜕𝑦

𝜕2𝑓
(𝑓𝑦 ) is defined by 𝑓𝑦𝑥 or
𝑥 𝜕𝑦𝜕𝑥

𝜕2𝑓
(𝑓𝑦 ) is defined by 𝑓𝑦𝑦 or
𝑦 𝜕𝑦 2

We usually refer to first partial derivatives as first Partial

and to second partial derivatives as second partials. Second
partials as 𝑓𝑦𝑥 or 𝑓𝑥𝑦 are sometimes called mixed partials.

Example (7)

Let 𝑓 (𝑥, 𝑦) = 𝑥 2 𝑦 + sin(𝑥𝑦). Find all second partials of 𝑓

Solution

The first partials are

𝑓𝑥 = 2𝑥𝑦 + 𝑦 cos(𝑥𝑦) , 𝑓𝑦 = 𝑥 2 + 𝑥 cos(𝑥𝑦)

The second partials

𝑓𝑥𝑥 = 2𝑦 − 𝑦 2 cos(𝑥𝑦)

𝑓𝑥𝑦 = 2𝑥 − 𝑥𝑦 sin(𝑥𝑦) + cos(𝑥𝑦)

𝑓𝑦𝑥 = 2𝑥 − 𝑥𝑦 sin(𝑥𝑦) + cos(𝑥𝑦)

𝑓𝑦𝑦 = −𝑥 2 sin(𝑥𝑦)

11
Notes that in the example above the mixed partials are
equal, 𝑓𝑦𝑥 = 𝑓𝑥𝑦 . This is always true provided the second
partials are continuous.

1.5 The Chain Rule

1. If 𝑧 = 𝑓(𝑥, 𝑦), 𝑥 = 𝑥(𝑡) and 𝑦 = 𝑦(𝑡) are all

differentiable functions, then

𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= +
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

2. If 𝑤 = 𝑓(𝑥, 𝑦, 𝑧), 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡) and 𝑧 = 𝑧(𝑡)

are all differentiable functions, then

𝑑𝑤 𝜕𝑤 𝑑𝑥 𝜕𝑤 𝑑𝑦 𝜕𝑤 𝑑𝑧
= + +
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑧 𝑑𝑡

Similar statements hold for functions of more than three

variables.

Example (8)

Suppose 𝑧 = 𝑥 2 𝑦 + 𝑒 2𝑥−𝑦 and 𝑥 = cos 𝑡 and 𝑦 = 4𝑡 3 .

𝑑𝑧
Find
𝑑𝑡

Solution

𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= +
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡
12
 𝑑𝑧
= (2𝑥𝑦 + 𝑒 2𝑥−𝑦 (2))(− sin 𝑡)
𝑑𝑡
+ (𝑥 2 𝑦 + 𝑒 2𝑥−𝑦 (−1))(12𝑡 2 )

3. If 𝑧 = 𝑓(𝑥, 𝑦), 𝑥 = 𝑥(𝑢, 𝑣), 𝑦 = 𝑦(𝑢, 𝑣)

are all differentiable functions, then

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
(𝑖) = +
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
(𝑖𝑖) = +
𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣

Generally

If 𝑤 = 𝑓(𝑥, 𝑦, 𝑧, … , 𝑡) is a differentiable function of any

number of variables, and each variable, in turn, is
differentiable function of any number of variables

𝑥 = 𝑥(𝑢, 𝑣, … , 𝑠), 𝑦 = 𝑦(𝑢, 𝑣, … , 𝑠) and so on, then for

example

𝜕𝑤 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 𝜕𝑤 𝜕𝑡
= + + ⋯+
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑡 𝜕𝑢

This the most general statement if the chain rule

Example (9)

Suppose 𝑤 = 𝑟 2 + 𝑠𝑣 + 𝑡 3 and 𝑟 = 𝑥 2 + 𝑦 2 + 𝑧 2 , 𝑠=
𝜕𝑤
𝑥𝑦𝑧, 𝑣 = 𝑥𝑒 𝑦 and 𝑡 = 𝑦𝑧 2 . Find
𝜕𝑧

13
Solution

𝜕𝑤 𝜕𝑤 𝜕𝑟 𝜕𝑤 𝜕𝑠 𝜕𝑤 𝜕𝑣 𝜕𝑤 𝜕𝑡
= + + +
𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝑠 𝜕𝑧 𝜕𝑣 𝜕𝑧 𝜕𝑡 𝜕𝑧

= (2𝑟)(2𝑧) + (𝑣)(𝑥𝑦) + (𝑠)(0) + (3𝑡 2 )(2𝑦𝑧)

= 4𝑧(𝑥 2 + 𝑦 2 + 𝑧 2 ) + 𝑥 2 𝑦𝑒 𝑦 + 6𝑦 2 𝑧 5

1.6. Implicit Differentiation

1. If 𝐹 (𝑥, 𝑦) = 0 defines 𝑦 as differentiable function of

𝑥, then

𝑑𝑦 𝐹𝑥 (𝑥, 𝑦)
=
𝑑𝑥 𝐹𝑦 (𝑥, 𝑦)

2. If 𝐹 (𝑥, 𝑦, 𝑧) = 0 defines 𝑧 as differentiable function of

𝑥 and 𝑦, then
𝜕𝑧 𝐹𝑥 (𝑥, 𝑦, 𝑧) 𝜕𝑧 𝐹𝑦 (𝑥, 𝑦, 𝑧)
=− , =−
𝜕𝑥 𝐹𝑧 (𝑥, 𝑦, 𝑧) 𝜕𝑦 𝐹𝑧 (𝑥, 𝑦, 𝑧)

Example (10)

If 𝑦 3 − 3𝑥𝑦 = 5𝑥 2 𝑦 2 , defines 𝑦 as differentiable function

of 𝑥. Find

𝑑𝑦
𝑑𝑥

Solution

First, rewrite the equation with only 0 at right

14
 𝑦 3 − 3𝑥𝑦 − 5𝑥 2 𝑦 2 = 0

Let
𝐹 (𝑥, 𝑦) = 𝑦 3 − 3𝑥𝑦 − 5𝑥 2 𝑦 2 = 0

𝑑𝑦 𝐹𝑥 (𝑥, 𝑦) −3𝑦 − 10𝑥𝑦 2

= =
𝑑𝑥 𝐹𝑦 (𝑥, 𝑦) 3𝑦 2 − 3𝑥 − 10𝑥 2 𝑦

You can see that this is much simpler than our earlier
method of implicit differentiation.

Example (11)

𝜕𝑧 𝜕𝑧
If 𝑥 2 𝑧 2 + 𝑥𝑦 2 + 𝑧 3 − 4𝑦𝑧 − 5 = 0,. Find and
𝜕𝑥 𝜕𝑦

Solution

Let 𝐹 (𝑥, 𝑦, 𝑧) = 𝑥 2 𝑧 2 + 𝑥𝑦 2 − 𝑧 3 − 4𝑦𝑧 − 5 = 0

Then

𝜕𝑧 𝐹𝑥 (𝑥, 𝑦, 𝑧) (2𝑥𝑧 2 + 𝑦 2 )
=− =−
𝜕𝑥 𝐹𝑧 (𝑥, 𝑦, 𝑧) (2𝑥 2 𝑧 − 3𝑧 2 − 4𝑦)

𝜕𝑧 𝐹𝑦 (𝑥, 𝑦, 𝑧) (2𝑥𝑦 − 4𝑧)

=− =−
𝜕𝑦 𝐹𝑧 (𝑥, 𝑦, 𝑧) (2𝑥 2 𝑧 − 3𝑧 2 − 4𝑦)

15
 Exercises

1. Determine the domain of 𝑓 and sketch it

(i) 𝑓 (𝑥, 𝑦) = √1 − 𝑥 2 − 𝑦 2

(ii) 𝑓 (𝑥, 𝑦) = √𝑦 2 − 4𝑥 2 − 16

(iii) 𝑓 (𝑥, 𝑦) = √𝑥 2 + 4𝑦 2 − 25𝑦

(iv) 𝑓 (𝑥, 𝑦) = ln(𝑥 + 𝑦)
2. Evaluate the following limits if it exists

(𝑖) lim 𝑥 3 √𝑦 3 + 2𝑥
(𝑥,𝑦)→(4,−2)

sin 𝑥𝑦
(𝑖𝑖) lim
(𝑥,𝑦)→(0,2) 𝑥
𝑥 2 − 𝑦2
(𝑖𝑖𝑖 ) lim
(𝑥,𝑦)→(4,−2) 𝑦 2 + 𝑥 2

3. Discuss the continuity of the following functions

𝑥+𝑦 𝑦 sin 𝑥
(𝑖) 𝑓(𝑥, 𝑦) = , (𝑖𝑖) 𝑓(𝑥, 𝑦) =
𝑥−𝑦 2𝑥 + 𝑦
(𝑖𝑖𝑖 ) 𝑓(𝑥, 𝑦) = 𝑥 2 +𝑦 2 sin 𝑧 ,
4. Find all the first partials of the given functions
(i) 𝑓 (𝑥, 𝑦) = 3𝑥 3 𝑦 + 2𝑥𝑦 2
(ii) 𝑓 (𝑥, 𝑦) == √𝑥 2 + 𝑦 2
(iii) 𝑔(𝑥, 𝑦) = 𝑢 sin(𝑢𝑣)
(iv) 𝑔(𝑢, 𝑣) = ln(𝑢𝑣)
(v) 𝑓 (𝑥, 𝑦, 𝑧) = sin−1 (𝑥𝑦𝑧)
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5. Evaluate the Partials at the given point
𝜋
(i) 𝑤 = 𝑥 2 + 𝑦 2 − 2𝑥𝑦 cos 𝑧, 𝑤𝑥𝑧 (0,1 )
2
𝑥 2 +𝑦 2
(ii) 𝑢 = , 𝑢𝑦𝑥 (1,3,1)
𝑥𝑧
𝑑𝑧
6. Find for the following functions
𝑑𝑡

(i) 𝑧 = 𝑥 2 𝑦 − 3𝑦 2 , 𝑥 = 3𝑡 + 1, 𝑦 = 𝑡 2
(ii) 𝑧 = 𝑒 𝑥 𝑦, 𝑥 = ln 𝑡, 𝑦 = 3𝑡 2
1
(iii) 𝑧 = tan−1 (𝑥𝑦), 𝑥 = tan 𝑡, 𝑦=
𝑡
𝜕𝑧 𝜕𝑧
7. Find , for the following
𝜕𝑢 𝜕𝑣

(i) 𝑧 = 𝑥 2 𝑦 3 + 𝑥 sin 𝑦, 𝑥 = 𝑢2 , 𝑦 = 𝑢𝑣
(ii) 𝑧 = 𝑒 𝑥 ln 𝑦 , 𝑥 = 𝑢2 − 2𝑣, 𝑦 = 𝑣 2 − 2𝑢
𝑑𝑦
8. Use the partial differentiation find for the following
𝑑𝑥

functions
(i) 𝑥 2 + 6𝑥𝑦 = 5𝑦 2 − 3
(ii) 𝑒 𝑥𝑦 = tan 𝑦
𝜕𝑧 𝜕𝑧
9. Find , for the following
𝜕𝑢 𝜕𝑣

(i) 4𝑧 4 = 2𝑥𝑦 2 − 3𝑧 2 𝑦
(ii) 𝑦𝑥 2 + sin 𝑥𝑦 2 = 6
10. Show that 𝑧 = 𝑒 (𝑥+2𝑦) + ln(𝑥 2 + 4𝑥𝑦 + 4𝑦 2 )
satisfies the equation
𝜕𝑧 𝜕𝑧
=2
𝜕𝑦 𝜕𝑥

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11. Show that any differential function 𝑧 = 𝑓(𝑥, 2𝑦)
will satisfy the partial differential equation
𝜕𝑧 𝜕𝑧
=2
𝜕𝑦 𝜕𝑥

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