This document provides an introduction to partial derivatives and several examples of calculating them. It begins by defining partial derivatives as the rate of change of a function with respect to one variable, holding other variables constant. Several examples are then provided of calculating partial derivatives of multivariable functions. The document concludes by stating the chain rule for partial derivatives, which relates the derivative of a composite function to its constituent partial derivatives.

1. 1
PRESENTATION ON INTRODUCTION
TO SEVERAL VARIABLES AND
PARTIAL DERIVATIVES

2. 2
Rokaiya Akter Shorna 171 – 006 – 041
Raju Ahmed 171 - 049 - 041
Rakibul Hasan Rokon 171 – 003 – 041
Masum Parvej 171 – 045 – 041
ALAmin 171 – 021 – 041
Md. Mobasher Hosain 171 – 027 – 041
Rayhan Uddin 171 – 037 – 041
Maria Hossain 171 – 025 – 041
Md. Hossain Refat 171 – 028 – 041
Associates:
Submitted by:
Md. Mazharul Islam ( L ) 171 – 009 – 041
Group 𝛿

3. SUBMITTED TO
3
A.B.M Minhajur Rahman
Lecturer of Mathematics
Department of Basic Science
Primeasia University , Banani

4. Chapter 01
Introduction to several
variables
4

5. Multivariable function: A function f of two variable x and y is a
rule that assigns a unique real number f(x , y) in same domain D
in the x y-plane .
Example : z = x^2 + y^2
Polynomial function : A polynomial function in x and y is a sum
of function of the form Cx^m y^n with nonnegative integers m
and n and C is a constant .
Example:
Level curves : The level curves of a function f of two variables
are the curves with equation f(x , y)=k , where k is a constant (in
the range of f)
113y2xy7xy3x 235

5

6. 6
Application of multivariable function : Techniques of multivariable calculus are used
to study many objects of interest in the material world. In particular
Domain and
codomain
Application Techniques
Curves Length of curves line integrals and
curvature
Surface Areas of surfaces, surface integrals,
flux through surfaces, and curvature.
Scalar field Maxima and minima, Lagrange
multipliers, directional derivatives,
level sets.
Vector field Any of the operations of vector
calculus including gradient,
divergence, and curl.

7. 7
Example 01 : Find the domain and range of g(x , y) = 9 − 𝑥2 − 𝑦2
Solve: Now the domain of this function
D ={(x , y) | 9 − 𝑥2
− 𝑦2
≥ 0 }
= {(x , y) | 𝑥2
+ 𝑦2
≤ 9 }
Which is the disk with center (0 , 0) and radius 3
The range of g is
{ z | z = 9 − 𝑥2 − 𝑦2 , (x , y) ∈ D }
Since z is positive square root , z ≥ 0
So the range is = { z | 0 ≤ z ≤ 3 }
= [0 , 3]

8. 8
Example 02 : Find the domain and range of g(x , y) = 9 − 𝑥2 − 4𝑦2
Solve: Now the domain of this function
D ={(x , y) | 9 − 𝑥2
− 4𝑦2
≥ 0 }
= {(x , y) | 𝑥2
+ 4𝑦2
≤ 9 }
= {(x , y) | 𝑥2/32 + 𝑦2/
3
2
2
= 1 }
Which is the disk with center (3 , 3/2 )
The range of g is
{ z | z = 9 − 𝑥2 − 4𝑦2 , (x , y) ∈ D }
Since z is positive square root , z ≥ 0
So the range is = { z | 0 ≤ z ≤ 3 }
= [0 , 3]

9. 9
Example 03: Sketch the level curves of the function g(x , y) = 9 − 𝑥2 − 𝑦2 𝑓𝑜𝑟 𝑘 =
0 , 1, 2, 3
Solve: The level curves are
9 − 𝑥2 − 𝑦2 = k
Or, 9 − 𝑥2
− 𝑦2
= 𝑘2
Or, 𝑥2
+ 𝑦2
= 9 ─ 𝑘2
This is family of concentric circle with center (0 , 0) and radius 9 − 𝑘2
The cases k = 0 , 1 , 2 , 3
If k = 0 then the equation of the circle 𝑥2
+ 𝑦2
= 3^2 center (0 , 0) radius 3
If k = 1 then the equation of the circle 𝑥2
+ 𝑦2
= 8 center (0 , 0) radius 8
If k = 2 then the equation of the circle 𝑥2
+ 𝑦2
= 5 center (0 , 0) radius 5
If k = 3 then the equation of the circle 𝑥2
+ 𝑦2
= 0 center (0 , 0) radius 0

10. 10
Example 04: Sketch the level curves of the function f(x , y) = 10 ─ 𝑥2
− 𝑦2
Solve: The level curves are
10 ─ 𝑥2
− 𝑦2
= k
Or 𝑥2
+ 𝑦2
= 10 ─ k
This is family of concentric circle with center (0 , 0) and radius 10 − 𝑘
The cases k = 0 , 1 , 2 , 3 are
If k = 0 then the equation of the circle 𝑥2 + 𝑦2 = 10 center (0 , 0) radius 10
If k = 1 then the equation of the circle 𝑥2
+ 𝑦2
= 3^2 center (0 , 0) radius 3
If k = 2 then the equation of the circle 𝑥2
+ 𝑦2
= 8 center (0 , 0) radius 8
If k = 3 then the equation of the circle 𝑥2 + 𝑦2 = 7 center (0 , 0) radius 7

11. Chapter 04
Partial derivatives
11

12. 12
Partial derivatives
The partial derivative of f(x , y) with respect to x at the point (xₒ , yₒ) is ,
𝜕𝑓
𝜕𝑥
⃒ 𝑥ₒ , 𝑦ₒ =
𝑑
𝑑𝑥
f(x , yₒ)⃒x=xₒ =lim
ℎ→0
𝑓 𝑥ₒ+ℎ ,𝑦ₒ −𝑓(𝑥ₒ ,𝑦ₒ)
ℎ
Provided limit exist
The partial derivative of f(x , y) with respect to x at the point (xₒ , yₒ) is ,
𝜕𝑓
𝜕𝑦
⃒ 𝑥ₒ , 𝑦ₒ =
𝑑
𝑑𝑦
f(xₒ , y)⃒y=yₒ =lim
ℎ→0
𝑓 𝑥ₒ ,𝑦ₒ+ℎ −𝑓(𝑥ₒ ,𝑦ₒ)
ℎ
Provided limit exist

13. 13
Application of partial derivatives in our real life:
• Derivatives are constantly used in everyday life to help measure how much
something is changing.
• They're used by the government in population censuses, various types of
sciences, and even in economics .
• Partial Derivatives are used in basic laws of Physics for example- Newton's
law of Linear motion , Maxwell’s equations of electromagnetism and
Einstein’s equation in general relativity .
• Derivatives in chemistry. One use of derivatives in chemistry is when you
want to find the concentration of an element in a product.
• In economics we use Partial Derivative to check what happens to other
variables while keeping one variable constant.

14. 14
Example 01 : Find
𝜕𝑓
𝜕𝑥
where f(x , y) =
2𝑦
𝑦+𝑐𝑜𝑠𝑥
Solve: Differentiating equation partially w . r to x , we have
)cos(
sin2
)cos(
)sin0(20).cos(
)cos(
)cos(2)2()cos(
)
cos
2
(
2
2
xy
xy
xy
xyxy
xy
xy
dx
d
yy
dx
d
xy
x
f
xy
y
dxdx
f















15. 15
Example 02 : Find
𝜕𝑓
𝜕𝑦
where f(x , y) = 𝑦2 𝑒 𝑥 + 𝑥𝑦 sin(𝑥 + 𝑦)
Solve: Differentiating equation partially w . r to y , we have
 
 
)sin()cos(2
).sin()cos(2
)sin()(
)sin(
2
2
yxxyxxyye
xyxyxxyye
yxxy
y
ey
y
yxxyey
yy
f
x
x
x
x
















16. 16
Example 03 :
Find 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑥𝑦 , 𝑓𝑦𝑥 , 𝑓𝑥𝑥𝑥 , 𝑓𝑦𝑥𝑦 , 𝑓𝑥𝑦𝑥 𝑖𝑓 𝑓 𝑥 , 𝑦 =
𝑎𝑥2 + 2ℎ𝑥𝑦 + 𝑏𝑦2
Solve:
hhyax
yx
f
y
fxy
hbyhx
xy
f
x
fyx
byhxbyhxyax
yy
f
fy
hyaxbyhxyax
xx
f
fx
2)22()(
2)22()(
22)2(
22)2(
22
22

































17. 17
 
 hfyxh
x
fxyx
hfxyh
y
fyxy
a
x
hyax
xx
f
x
fxxx
20)2(
20)2(
0)2()22()( 2
2
2
2





















18. 18
Statement of Chain rule : If x=x(t) , y=y(t) are differentiable at t
and if z = f(x , y) is differentiable at the point (x , y) = (x(t) , y(t)
then z=f(x(t) ,Y(t)) is differentiable at t and we write ,
∂𝑧
∂𝑡
=
∂𝑧
∂𝑥
∂𝑥
∂𝑡
+
∂𝑧
∂𝑦
∂𝑦
∂𝑡

19. Example 04 : If 𝑧 = 𝑒 𝑥𝑦 , 𝑥 = 2𝑢 = 𝑣 , 𝑦 =
𝑢
𝑣
, 𝐹𝑖𝑛𝑑
𝑑𝑧
𝑑𝑢
solve:

























v
u
yvux
putting
v
vu
v
u
e
v
x
ye
v
xeye
u
y
y
z
u
x
x
z
xy
xy
xyxy
,2
)
2
2(
)2(
1
2
u
z

20. 20
)
4
1(
)
4
(
)
22
(
v
u
e
v
vu
e
v
vuu
e
xy
xy
xy





Ans

21. 21