MATH - 1107
TOPIC:
PARTIAL DIFFERENTIATION
&
EXTREMA OF FUNCTIONS
Md. Farhad Uddin
Assistant Professor
Department of Mathematics
Chittagong University of Engineering & Technology
Chittagong-4349, Bangladesh
Partial Differentiation
&
Extrema of Functions
 4.1 Function of Several Variables
 4.2 Partial Derivatives
 4.3 Euler’s Theorem on Homogeneous
Function
 4.4 Maxima and Minima of Functions
4.1
Functions of Several Variables
z
f(x, y) = x2 + y2
x2 + y2 = 16
x2 + y2 = 9
x2 + y2 = 4
x2 + y2 = 1
x2 + y2 = 0
x
y
Functions of Two Variables
A real-valued function of two variables f, consists of
1. A set A of ordered pairs of real numbers (x, y)
called the domain of the function.
2. A rule that associates with each ordered pair in
the domain of f one and only one real number,
denoted by z = f(x, y).
4.2
Partial Derivatives

x
f
x

x
  f   22 f
 
x  x  x 22

y
  f   2 f


y  x  y x

x
  f   2 f
 
x  y  xy
f

y
f
y

y
  f   22 f
 
y  y  y 22
 22 f
 22 f

y x xy
When both are
continuous
First Order Partial Derivatives
First Partial Derivatives of f(x, y)
 Suppose f(x, y) is a function of two variables x and y.
 Then, the first partial derivative of f with respect to x
at the point (x, y) is
f
f ( x  h, y )  f ( x, y )
 lim
x h  0
h
provided the limit exists.
 The first partial derivative of f with respect to y at the
point (x, y) is
f
f ( x, y  k )  f ( x, y )
 lim
y k  0
k
provided the limit exists.
Geometric Interpretation of the Partial Derivative
z
What does
f
x
f(x, y)
mean?
y
x
Geometric Interpretation of the Partial Derivative
z
What does
f
x
x
f(x, y)
mean?
f
 slope of f ( x , b )
f(x, b)
y = b plane
b
a
(a, b)
x
y
Geometric Interpretation of the Partial Derivative
z
f(x, y)
f
What does
mean?
y
y
x
Geometric Interpretation of the Partial Derivative
z
f(x, y)
f
What does
mean?
y
f
f(c, y)
y
 slope of f (c , y )
x = c plane
d
c
(c, d)
x
y
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x , y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2) .
Solution
 To compute ∂f/∂x, think of the variable y as a constant and
differentiate the resulting function of x with respect to x:
f ( x, y )  x 2  y 2 x  y 3
f
 2x  y2
x
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x , y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2).
Solution
 To compute ∂f/∂y, think of the variable x as a constant and
differentiate the resulting function of y with respect to y:
f ( x , y )  x 2  xy 2  y 3
f
 2 xy  3 y 2
y
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x , y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2).
Solution
 The rate of change of f in the x-direction at the point (1, 2)
is given by
f
x
 2(1)  2 2  2
(1,2)
 The rate of change of f in the y-direction at the point (1, 2)
is given by
f
y
  2(1)(2)  3(2) 2  8
(1,2 )
Examples
 Find the first partial derivatives of the function
w( x , y ) 
xy
x2  y2
Solution
 To compute ∂w/∂x, think of the variable y as a constant and
differentiate the resulting function of x with respect to x:
xy
w( x , y )  2
x  y2
w ( x 2  y 2 ) y  x y (2 x )

x
( x 2  y 2 )2
y( y 2  x2 )
 2
( x  y 2 )2
Examples
 Find the first partial derivatives of the function
w( x , y ) 
xy
x2  y2
Solution
 To compute ∂w/∂y, think of the variable x as a constant and
differentiate the resulting function of y with respect to y:
xy
w( x , y )  2
x  y2
w ( x 2  y 2 ) x  x y (2 y )

y
( x 2  y 2 )2
x( x 2  y 2 )
 2
( x  y 2 )2
Examples
 Find the first partial derivatives of the function
w  f ( x , y , z )  xyz  xe yz  x ln y
Solution
 Here we have a function of three variables, x, y, and z, and
we are required to compute
f f f
,
,
x y z
 For short, we can label these first partial derivatives
respectively fx, fy, and fz.
Examples
 Find the first partial derivatives of the function
w  f ( x , y , z )  xyz  xe yz  x ln y
Solution
 To find fx, think of the variables y and z as a constant and
differentiate the resulting function of x with respect to x:
w  f ( x , y , z )  xyz  xe yz  x ln y
f x  yz  e yz  ln y
Examples
 Find the first partial derivatives of the function
w  f ( x , y , z )  xyz  xe yz  x ln y
Solution
 To find fy, think of the variables x and z as a constant and
differentiate the resulting function of y with respect to y:
w  f ( x, y , z )  xyz  xe yz  x ln y
x
f y  xz  xze 
y
yz
Examples
 Find the first partial derivatives of the function
w  f ( x , y , z )  xyz  xe yz  x ln y
Solution
 To find fz, think of the variables x and y as a constant and
differentiate the resulting function of z with respect to z:
w  f ( x, y , z )  xyz  xe yz  x ln y
f z  xy  xye yz
Second Order Partial Derivatives
 The first partial derivatives fx(x, y) and fy(x, y) of a function
f(x, y) of two variables x and y are also functions of x and y.
 As such, we may differentiate each of the functions fx and
fy to obtain the second-order partial derivatives of f.
Second Order Partial Derivatives
 Differentiating the function fx with respect to x leads to the
second partial derivative
f xx
2 f



( fx )
2
x
x
 But the function fx can also be differentiated with respect
to y leading to a different second partial derivative
f xy
2 f



( fx )
y x y
Second Order Partial Derivatives
 Similarly, differentiating the function fy with respect to y
leads to the second partial derivative
f yy
2 f



( fy)
2
y
y
 Finally, the function fy can also be differentiated with
respect to x leading to the second partial derivative
f yx
2 f



( fy)
xy x
Second Order Partial Derivatives
 Thus, four second-order partial derivatives can be
obtained of a function of two variables:

x
f
x

x
  f   2 f
 
x  x  x 2

y
  f   2 f


y  x  y x

x
  f   2 f
 
x  y  xy
f

y
f
y

y
  f   2 f
 
y  y  y 2
2 f
2 f

y x xy
When both are
continuous
Examples
 Find the second-order partial derivatives of the function
f ( x , y )  x 3  3 x 2 y  3 xy 2  y 2
Solution
 First, calculate fx and use it to find fxx and fxy:
fx 
 3
( x  3 x 2 y  3 xy 2  y 2 )
x
 3 x 2  6 xy  3 y 2
f xx


(3 x 2  6 xy  3 y 2 )
x
f xy


(3 x 2  6 xy  3 y 2 )
y
 6x  6 y
 6 x  6 y
 6( x  y )
 6( y  x )
Examples
 Find the second-order partial derivatives of the function
f ( x , y )  x 3  3 x 2 y  3 xy 2  y 2
Solution
 Then, calculate fy and use it to find fyx and fyy:
 3
fy 
( x  3 x 2 y  3 xy 2  y 2 )
y
 3 x 2  6 xy  2 y
f yx


(  3 x 2  6 xy  2 y )
x
f yy


( 3 x 2  6 xy  2 y )
y
 6 x  6 y
 6x  2
 6( y  x )
 2(3 x  1)
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  e
xy 2
Solution
 First, calculate fx and use it to find fxx and fxy:
 xy 2
fx 
(e )
x
 y e
2
f xx

2 xy 2

(y e )
x
 y e
4
xy 2
xy 2
f xy

2 xy 2

(y e )
y
 2 ye
xy 2
3 xy 2
 2 xy e
 2 ye
xy 2
(1  xy 2 )
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  e
xy 2
Solution
 Then, calculate fy and use it to find fyx and fyy:
 xy 2
fy 
(e )
y
 2 xye
f yx

xy 2

(2 xye )
x
xy 2
f yy
 2 ye
xy 2
3 xy 2
 2 xy e
 2 ye
xy 2
(1  xy )
2

xy 2

(2 xye )
y
 2 xe
 2 xe
xy 2
xy 2
 (2 xy )(2 xy ) e
(1  2 xy 2 )
xy 2
Problems
4.3
Euler’s Theorem on
Homogeneous Functions
𝝏𝒇
𝝏𝒇
𝒙
+𝒚
= 𝒏𝒇
𝝏𝒙
𝝏𝒚
Homogeneous Functions
Euler’s Theorem on Homogeneous Functions
Statement
If 𝒇 is a homogeneous function of degree 𝒏 in 𝒙, 𝒚 , then
𝒙
𝝏𝒇
+
𝝏𝒙
𝒚
𝝏𝒇
𝝏𝒚
= 𝒏𝒇
Euler’s Theorem on Homogeneous Functions
Statement
If 𝒇 is a homogeneous function of degree 𝒏 in 𝒙, 𝒚 , then
𝒙
𝝏𝒇
+
𝝏𝒙
𝒚
𝝏𝒇
𝝏𝒚
= 𝒏𝒇
Euler’s Theorem on Homogeneous Functions
Statement
If 𝒇 is a homogeneous function of degree 𝒏 in 𝒙, 𝒚 , then
𝟐
𝝏 𝒇
𝟐
𝒙
𝝏𝒙𝟐
𝝏𝟐 𝒇
+ 𝟐𝒙𝒚
𝝏𝒙 𝝏𝒚
𝝏𝟐 𝒇
𝟐
+𝒚
𝝏𝒚𝟐
=𝒏 𝒏−𝟏 𝒇
Euler’s Theorem on Homogeneous Functions
Statement
If 𝒇 is a homogeneous function of degree 𝒏 in 𝒙, 𝒚 , then
𝟐
𝝏 𝒇
𝟐
𝒙
𝝏𝒙𝟐
𝝏𝟐 𝒇
+ 𝟐𝒙𝒚
𝝏𝒙 𝝏𝒚
𝝏𝟐 𝒇
𝟐
+𝒚
𝝏𝒚𝟐
=𝒏 𝒏−𝟏 𝒇
Problems
Example If 𝒖 = sin−1
𝑥+ 𝑦
𝑥− 𝑦
, show by Euler’s theorem that
𝝏𝒖
𝒚 𝝏𝒖
=−
.
𝝏𝒙
𝒙 𝝏𝒚
Problems
Problems
Example If 𝒖 = log
𝑥 4 +𝑦 4
𝑥+𝑦
, show that
𝝏𝒖
𝝏𝒖
𝒙
+𝒚
= 𝟑.
𝝏𝒙
𝝏𝒚
Problems
Example If 𝒖 = log
𝑥 4 +𝑦 4
𝑥+𝑦
, show that
𝝏𝒖
𝝏𝒖
𝒙
+𝒚
= 𝟑.
𝝏𝒙
𝝏𝒚
Problems
Example If 𝒖 =
3
3
−1 𝑥 +𝑦
tan
𝑥−𝑦
, prove that
𝝏𝒖
𝒙 +
𝝏𝒙
𝝏𝒖
𝒚
𝝏𝒚
= sin 2𝑢.
Problems
 If 𝒖 = 𝑙𝑛
𝑥 3 +𝑦 3
𝑥2 + 𝑦2
, prove that
 If 𝒖 =
𝑥 3 +𝑦 3
−1
tan
𝑥+𝑦
 If 𝒖 =
sin−1
 If 𝒖 =
cosec −1
𝑥 +𝑦
𝑥+ 𝑦
𝝏𝒖
𝝏𝒖
𝒙 +𝒚
𝝏𝒙
𝝏𝒚
= 𝟏.
, prove that
𝝏𝒖
𝝏𝒖
𝒙 +𝒚
𝝏𝒙
𝝏𝒚
, prove that
𝝏𝒖
𝒙 +
𝝏𝒙
𝑥 1/2 +𝑦 1/2
,
𝑥 1/3 +𝑦 1/3
prove that
𝝏𝒖
𝒚
𝝏𝒚
𝝏𝒖
𝒙 +
𝝏𝒙
= sin 2𝑢.
=
𝟏
tan
𝟐
𝝏𝒖
𝒚
𝝏𝒚
=
𝑢.
𝟏
tan 𝑢.
𝟏𝟐
4.4
Maxima and Minima of Functions
z
x
(g, h)
(a, b)
(c, d)
(e, f )
y
Maxima & Minima of Functions
In a smoothly changing function a low point (a
minimum) or high point (a maximum) are found where
the function flattens out
But not all flat parts are maxima and minima, we can also have saddle point.

When slope is zero then we get maxima and minima

If the second derivative is less than zero then we get maximum value

If the second derivative is greater than zero then we get minimum value
Critical Point of a Function
Test for Maxima & Minima of Function
(d) If 𝑓 ′′ 𝑥0 = 0 = 𝑓 ′′′ 𝑥0 , then find 𝑓 𝑖𝑣 𝑥0 .
(i) If 𝑓 𝑖𝑣 𝑥0 < 0, then 𝑓 𝑥 has maximum at 𝑥0 .
(ii) If 𝑓 𝑖𝑣 𝑥0 > 0, then 𝑓 𝑥 has minimum at 𝑥0 .
Examples
 Find the relative extrema of the function
𝒇 𝒙 = 𝟑𝒙𝟓 − 𝟓𝒙𝟑
So, the relative maximum value 𝑓 −1 = 3 −1 5 − 5 −1 3 = −3 + 5 = 2
Also, the relative minimum value 𝑓 1 = 3 1 5 − 5 1 3 = 3 − 5 = −2
Problems
Thank
you