Partial differentiation, total differentiation, Jacobian, Taylor's expansion, stationary points,maxima & minima (Extreme values),constraint maxima & minima ( Lagrangian multiplier), differentiation of implicit functions.
2. A function of several variables is a function where the domain is a
subset of 𝑅 𝑛 and range is 𝑅.
A real valued function of 𝑛–variables is a function
𝑓 ∶ 𝐷 → R, where the domain D is a subset of 𝑅 𝑛.
So: for each (𝑥1, 𝑥2, . . . , 𝑥 𝑛) in D, the value of f is a real number
𝑓(𝑥1, 𝑥2, . . . , 𝑥 𝑛 ).
For example,
1.𝑓(𝑥, 𝑦) = 𝑥 − 𝑦
(a function of 2 variables defined for all (𝑥, 𝑦) ∈ 𝑅2)
2. If 𝑓 is a function defined by
𝑓(𝑥, 𝑦) = 9 − cos(𝑥) + sin(𝑥2 + 𝑦2),
(a function of 2 variables defined for all (𝑥, 𝑦) ∈ 𝑅2
)
3. 2. 𝑓 𝑥, 𝑦, 𝑧 =
1
𝑥2+𝑦2+𝑧2
Then f is a function of 3 variables, defined whenever 𝑥2 + 𝑦2 + 𝑧2 ≠
0
This is all (𝑥, 𝑦, 𝑧) ∈ 𝑅3 except for (𝑥, 𝑦, 𝑧) = (0, 0, 0).
4. Partial Derivative
If 𝑓(𝑥, 𝑦) is a function of two variables 𝑥 and 𝑦,
the partial derivative of 𝑓 with respect to 𝑥, is given by 𝑓𝑥 𝑥, 𝑦 =
lim
ℎ→0
𝑓 𝑥+ℎ,𝑦 −𝑓 𝑥,𝑦
ℎ
The partial derivative of 𝑓 with respect to 𝑦, is given by 𝑓𝑦 𝑥, 𝑦 =
lim
ℎ→0
𝑓 𝑥,𝑦+ℎ −𝑓 𝑥,𝑦
ℎ
5. Note
• If 𝑓(𝑥, 𝑦) is a function of two variables 𝑥 and 𝑦, then
Partial derivative of 𝑓(𝑥, 𝑦) with respect to 𝑥 is
𝜕𝑓
𝜕𝑥
, it is
denoted by 𝑓𝑥
• Partial derivative of 𝑓(𝑥, 𝑦) with respect to 𝑦 is
𝜕𝑓
𝜕𝑦
, it is denoted by
𝑓𝑦
• Partial second derivative of 𝑓(𝑥, 𝑦) with respect to 𝑥 is
𝜕2 𝑓
𝜕𝑥2 , it is
denoted by 𝑓𝑥𝑥
• Partial second derivative of 𝑓(𝑥, 𝑦) with respect to 𝑦 is
𝜕2 𝑓
𝜕𝑦2 , it is
denoted by 𝑓𝑦𝑦
• Partial derivative of
𝜕𝑓
𝜕𝑥
with respect to 𝑦 is
𝜕2 𝑓
𝜕𝑦𝜕𝑥
, it is denoted by 𝑓𝑥𝑦
and so on
12. Homogeneous Function:
A function 𝑓(𝑥, 𝑦) is called a homogeneous function of the degree ′𝑛′ if
the following relationship is valid for all 𝑡 > 0: 𝑓 𝑡𝑥, 𝑡𝑦 = 𝑡 𝑛
𝑓 𝑥, 𝑦 .
Example:
Consider 𝑓 𝑥, 𝑦 =
𝑥2+𝑦2
𝑥𝑦
𝑓 𝑡𝑥, 𝑡𝑦 =
𝑡𝑥 2+ 𝑡𝑦 2
𝑡𝑥 𝑡𝑦
=
𝑡2 𝑥2+𝑡2 𝑦2
𝑡2 𝑥𝑦
=
𝑡2 𝑥2+𝑦2
𝑡2 𝑥𝑦
=
𝑥2+𝑦2
𝑥𝑦
= 𝑡0 𝑓(𝑥, 𝑦)
Hence f(x, y ) is homogeneous of order zero
13. Euler’s Theorem:
If 𝑢 = 𝑓 𝑥, 𝑦 is a homogeneous function of degree ‘n’, then 𝑥
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
= 𝑛𝑢
(OR) 𝑥
𝜕𝑓
𝜕𝑥
+ 𝑦
𝜕𝑓
𝜕𝑦
= 𝑛𝑓
Problem 1:
If 𝒖 = 𝒇
𝒙
𝒚
, prove that 𝒙
𝝏𝒖
𝝏𝒙
+ 𝒚
𝝏𝒖
𝝏𝒚
= 𝟎
Solution:
Given 𝑢 𝑥, 𝑦 = 𝑓
𝑥
𝑦
For any 𝑡 > 0,
𝑢 𝑡𝑥, 𝑡𝑦 = 𝑓
𝑡𝑥
𝑡𝑦
= 𝑓
𝑥
𝑦
= 𝑡0
𝑓
𝑥
𝑦
= 𝑡0
𝑢(𝑥, 𝑦)
Hence 𝑢 = 𝑓
𝑥
𝑦
is a homogeneous function of order zero
𝑖. 𝑒 𝑛 = 0
By Euler’s Theorem , 𝑥
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
= 𝑛𝑢
⟹ 𝑥
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
= 0 u = 0
Hence proved.
.
14. Problem 2:
If 𝒖 = 𝐬𝐢𝐧−𝟏 𝒙
𝒚
+ 𝐭𝐚𝐧−𝟏 𝒚
𝒙
, prove that 𝒙
𝝏𝒖
𝝏𝒙
+ 𝒚
𝝏𝒖
𝝏𝒚
= 𝟎.
Solution:
Given 𝑢(𝑥, 𝑦) = sin−1 𝑥
𝑦
+ tan−1 𝑦
𝑥
For any 𝑡 > 0,
𝑢(𝑡𝑥, 𝑡𝑦) = sin−1 𝑡𝑥
𝑡𝑦
+ tan−1 𝑡𝑦
𝑡𝑥
= sin−1 𝑥
𝑦
+ tan−1 𝑦
𝑥
= 𝑡0
( sin−1 𝑥
𝑦
+ tan−1 𝑦
𝑥
)
Hence 𝑢(𝑥, 𝑦) is a homogeneous function of order zero
𝑖. 𝑒 𝑛 = 0
By Euler’s Theorem,
𝑥
𝜕𝑢
𝜕𝑥
+ 𝑦
𝜕𝑢
𝜕𝑦
= 𝑛𝑢 = 0 𝑢 = 0
15. Total differential coefficient
If 𝑢 = 𝑓 𝑥, 𝑦 , the total differential of 𝑢 is given by
𝑑𝑢 = 𝑢 𝑥 𝑑𝑥 + 𝑢 𝑦 𝑑𝑦 (OR) 𝑑𝑓 = 𝑓𝑥 𝑑𝑥 + 𝑓𝑦 𝑑𝑦
If 𝑢 = 𝑓 𝑥, 𝑦 , and x = g(t) , y = h(t) , the total differential of 𝑢 is given
by
𝑑𝑢
𝑑𝑡
=
𝜕𝑢
𝜕𝑥
𝑑𝑥
𝑑𝑡
+
𝜕𝑢
𝜕𝑦
𝑑𝑦
𝑑𝑡
= 𝑢 𝑥
𝑑𝑥
𝑑𝑡
+ 𝑢 𝑦
𝑑𝑦
𝑑𝑡
(OR)
𝑑𝑓
𝑑𝑡
=
𝜕𝑓
𝜕𝑥
𝑑𝑥
𝑑𝑡
+
𝜕𝑓
𝜕𝑦
𝑑𝑦
𝑑𝑡
= 𝑓𝑥
𝑑𝑥
𝑑𝑡
+ 𝑓𝑦
𝑑𝑦
𝑑𝑡
33. Necessary Conditions
The necessary conditions for 𝑓(𝑥, 𝑦) to have a maximum or minimum at a point
(𝑎, 𝑏) are that 𝑓𝑥 𝑎, 𝑏 = 0 𝑎𝑛𝑑 𝑓𝑦(𝑎, 𝑏) = 0 𝑎𝑡 (𝑎, 𝑏)
Sufficient Conditions
The sufficient conditions for 𝑓(𝑥, 𝑦) to have a maximum or minimum at a point
(𝑎, 𝑏) are as follows
Let 𝑟 = 𝑓𝑥𝑥 𝑎, 𝑏 , 𝑠 = 𝑓𝑥𝑦 𝑎, 𝑏 𝑎𝑛𝑑 𝑡 = 𝑓𝑦𝑦 𝑎, 𝑏
The function 𝑓(𝑥, 𝑦) has maximum or minimum (extreme values) at ( a, b) if
1. 𝑓𝑥 𝑎, 𝑏 = 0 𝑎𝑛𝑑 𝑓𝑦(𝑎, 𝑏) = 0
2. 𝑟𝑡 − 𝑠2
> 0
3. 𝑓(𝑥, 𝑦) has maximum or minimum at (𝑎, 𝑏) according as 𝑟 < 0 or 𝑟 > 0
34. Procedure to find maximum or minimum (extreme values):
Step 1. Find 𝑓𝑥 𝑎𝑛𝑑 𝑓𝑦
Step 2. Put 𝑓𝑥 = 0 𝑎𝑛𝑑 𝑓𝑦 = 0
Step 3. Solve the simultaneous equations in a and y, find the values of x and y
Say (𝑎, 𝑏), (𝑐, 𝑑) … . ( these points are called stationary / critical points)
Step 4.Find 𝑟 = 𝑓𝑥𝑥 , 𝑡 = 𝑓𝑦𝑦 𝑎𝑛𝑑 𝑠 = 𝑓𝑥𝑦 (at each pair (𝑎, 𝑏) , (𝑐, 𝑑) … . )
Step 5. Find 𝑟𝑡 − 𝑠2 (at each pair (𝑎, 𝑏) , (𝑐, 𝑑) … . )
Step 6.
(i) If r𝑡 − 𝑠2
> 0 𝑎𝑛𝑑 𝑟 < 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has maximum at (𝑎, 𝑏)
And the maximum value is 𝑓(𝑎, 𝑏)
(ii) If r𝑡 − 𝑠2 > 0 𝑎𝑛𝑑 𝑟 > 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has minimum at (𝑎, 𝑏)
And the minimum value is 𝑓(𝑎, 𝑏)
(iii) If r𝑡 − 𝑠2 < 0 𝑎𝑡 (𝑎, 𝑏) , then 𝑓(𝑥, 𝑦) has neither maximum nor minimum at
(𝑎, 𝑏) (no extreme values at (𝑎, 𝑏)) and the point (𝑎, 𝑏) is called saddle point.
(iv) If 𝑟𝑡 − 𝑠2
= 0 𝑎𝑡 (𝑎, 𝑏), the case is doubtful and need further investigation.
Step 7. Step 6 to be repeated for other pair of values (c, d) … to examine extreme values
40. To find the point at which 𝒇(𝒙, 𝒚) has maximum/minimum:
𝑟𝑡 − 𝑠2 = 6 + 6𝑥 −2 − 0 = −12 − 12𝑥
At the Point (0,0):
𝑟𝑡 − 𝑠2
𝑎𝑡 0,0 = −12 − 12 0 = −12 < 𝟎
Hence the point (0,0) is the saddle point of 𝑓(𝑥, 𝑦)
At the Point (-2,0):
𝑟𝑡 − 𝑠2
𝑎𝑡 −2,0 = −12 − 12 −2 = −12 + 24 = 12 > 𝟎
𝑟at −2,0 = 6 + 6 −2 = 6 − 12 = −6 < 𝟎
Hence at the point −2,0 the function 𝑓 𝑥, 𝑦 has maximum.
To find the maximum value:
Put 𝑥, 𝑦 = (−2,0) in (1), 𝑓 −2,0 = 3(−2)2 − 0 2 + −2 3 = 12 − 8 = 4
41. Constrained Maxima and Minima – Lagrangian Multiplier
If 𝑓(𝑥, 𝑦, 𝑧) is a function of three variables 𝑥, 𝑦, 𝑧 , we will find the extreme values
(maximum or minimum) of 𝑓(𝑥, 𝑦, 𝑧) with respect to a constraint ∅ 𝑥, 𝑦, 𝑧 = 0
Procedure
Step 1. Identify the constraint equation ∅ 𝑥, 𝑦, 𝑧 = 0
Step 2. Identify the main function for which we have to find the extreme value, let it be
𝑓(𝑥, 𝑦, 𝑧)
Step 3. Form the equation 𝐹 = 𝑓 + 𝜆∅
Step 4. Find 𝐹𝑥 , 𝐹𝑦, 𝐹𝑧
Step 5. Put 𝐹𝑥 = 0 , 𝐹𝑦 = 0, 𝐹𝑧 = 0 and solve all the equations including ∅ 𝑥, 𝑦, 𝑧 = 0
Step 6. Find the values of 𝑥, 𝑦, 𝑧 and 𝜆
Step 7. The values of 𝑥, 𝑦, 𝑧 gives the extreme values of 𝑓(𝑥, 𝑦, 𝑧)
42. Note :
Distance of a point 𝑥1, 𝑦1, 𝑧1 𝑓𝑟𝑜𝑚 𝑥, 𝑦, 𝑧 is given by
𝑑 = 𝑥 − 𝑥1
2 + 𝑦 − 𝑦1
2 + 𝑧 − 𝑧1
2
Square of the distance is 𝑑2
= 𝑥 − 𝑥1
2
+ 𝑦 − 𝑦1
2
+ 𝑧 − 𝑧1
2
43. Problem 1:
Find the length of the shortest line form the point (𝟎, 𝟎,
𝟐𝟓
𝟗
) to the surface
𝒛 = 𝒙𝒚
Solution:
The square of the distance from the point (0,0,
25
9
) to (𝑥, 𝑦, 𝑧) is (𝑑2)
𝑓 𝑥, 𝑦, 𝑧 = 𝑥 − 0 2 + 𝑦 − 0 2 + 𝑧 −
25
9
2
𝑓 𝑥, 𝑦, 𝑧 = 𝑥2 + 𝑦2 + 𝑧 −
25
9
2
------- (1)
Subject to ( to the surface) ∅ 𝑥, 𝑦𝑧 = 𝑧 − 𝑥𝑦 = 0 ------------(2)
(∵ 𝑧 = 𝑥𝑦 ⟹ 𝑧 − 𝑥𝑦 = 0)
Consider the Lagrangian function F = 𝑓 𝑥, 𝑦, 𝑧 + 𝜆𝜙 𝑥, 𝑦, 𝑧
𝐹 = 𝑥2
+ 𝑦2
+ 𝑧 −
25
9
2
+ 𝜆(𝑧 − 𝑥𝑦)
𝜕F
𝜕𝐱
= 2x − λy
𝜕F
𝜕y
= 2y − λx
𝜕F
𝜕z
= 2 z −
25
9
+ λ
45. Hence 𝑥 = 𝑦 = ±
4
3
& 𝑧 = 𝑥𝑦 = 𝑦2 =
4
3
2
=
16
9
From (1) square distance is 𝑑2 = 𝑥2 + 𝑦2 + 𝑧 −
25
9
2
=
4
3
2
+
4
3
2
+
16
9
−
25
9
2
=
16
9
+
16
9
+ −
9
9
2
= 2
16
9
+ −1 2
=
32
9
+ 1 =
32+9
9
=
41
9
The required minimum distance is (d) =
41
9
=
41
3
units
Problem 2:
A rectangular box open at the top is to have a volume of 32 cc .Find the dimensions of the
box that requires the least material for its construction
Solution:
Given a rectangular open box with volume 32cc
Let us take the length, width and height of the box be x, y and z respectively.
46. Hence the volume of the box is 𝑥𝑦𝑧 = 32 , which is the given constrain
(condition).
Let ∅ 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 − 32 ------ (1)
Requirement of least material to construct the open top box is the total least surface
area of the box.
Total surface area of the open rectangular box is 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧
Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 ------ (2)
53. Problem 3:
Find the dimensions of the rectangular box, open at the top, of maximum capacity
whose surface area is 432 square meter.
Solution:
Given a rectangular open box with surface area 432 sq.m
Let us take the length, width and height of the box be x, y and z respectively.
Hence the surface area 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 432, which is the given constrain(condition)
Let ∅ 𝑥, 𝑦, 𝑧 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 − 432 ------ (1)
Requirement of a open rectangular box with maximum capacity (volume)
Total volume of the open rectangular box is 𝑥𝑦𝑧
Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦𝑧 ------ (2)
60. Jacobian
• If 𝑢 = 𝑓(𝑥, 𝑦) and 𝑣 = 𝑔(𝑥, 𝑦) are two continuous functions of two
independent variables x and y then the functional determinant
𝐽 =
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑣
𝜕𝑥
𝜕𝑣
𝜕𝑦
=
𝑢 𝑥 𝑢 𝑦
𝑣 𝑥 𝑣 𝑦
is called Jacobian of 𝑢 , 𝑣 with respect to 𝑥, 𝑦 and it is
denoted by
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
• If u , v, w are functions of x ,y , z then jacobian of u , v , w with respect to x , y , z
is given by
𝜕 𝑢,𝑣,𝑤
𝜕 𝑥,𝑦,𝑧
=
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑢
𝜕𝑧
𝜕𝑣
𝜕𝑥
𝜕𝑣
𝜕𝑦
𝜕𝑣
𝜕𝑧
𝜕𝑤
𝜕𝑥
𝜕𝑤
𝜕𝑦
𝜕𝑤
𝜕𝑧
=
𝑢 𝑥 𝑢 𝑦 𝑢 𝑧
𝑣 𝑥 𝑣 𝑦 𝑣𝑧
𝑤 𝑥 𝑤 𝑦 𝑤𝑧
61. Two important Properties of Jacobian
1. If 𝑢, 𝑣 are functions of 𝑥, 𝑦 and 𝑥, 𝑦 are the function of 𝑟, 𝑠 then
𝜕 𝑢,𝑣
𝜕 𝑟,𝑠
=
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
𝜕 𝑥,𝑦
𝜕 𝑟,𝑠
2. If 𝑢, 𝑣 are the functions of 𝑥, 𝑦 then
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
1
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
Note :
Two functions 𝑢(𝑥, 𝑦) & 𝑣(𝑥, 𝑦) are functionally depended if
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
=0
62. Problem 1:
If 𝒖 =
𝒚𝒛
𝒙
, 𝒗 =
𝒛𝒙
𝒚
, 𝒘 =
𝒙𝒚
𝒛
find
𝝏 𝒖,𝒗,𝒘
𝝏 𝒙,𝒚,𝒛
Solution:
Wkt, the jacobian of 𝑢, 𝑣, 𝑤 with respect to 𝑥, 𝑦, 𝑧 is given by
𝜕 𝑢,𝑣,𝑤
𝜕 𝑥,𝑦,𝑧
=
𝑢 𝑥 𝑢 𝑦 𝑢 𝑧
𝑣 𝑥 𝑣 𝑦 𝑣𝑧
𝑤 𝑥 𝑤 𝑦 𝑤𝑧
Given
𝑢 =
𝑦𝑧
𝑥
, 𝑣 =
𝑧𝑥
𝑦
, 𝑤 =
𝑥𝑦
𝑧
72. 𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦𝑢 𝑦𝑣
=
1 − 𝑣 −𝑢
𝑣 𝑢
= 1 − 𝑣 𝑢 + 𝑢𝑣 = 𝑢 − 𝑢𝑣 + 𝑢𝑣 = 𝑢
Note :
Implicit vs Explicit
Explicit: "y = some function of x". When we know x we can calculate y directly.
An explicit function is one which is given in terms of the independent variable.
Example , consider 𝑦 = 𝑥2 + 3𝑥 – 8
here y is the dependent variable and is given in terms of the independent variable x.
More Examples : 𝑦 = 𝑥 + 3 , 𝑦 = 𝑥2
− 𝑟2
𝑒𝑡𝑐.,
Implicit: "some function of y and x equals something else".
Implicit functions, on the other hand, are usually given in terms of both dependent and
independent variables.
Example, consider 𝑦 + 𝑥2 − 3𝑥 + 8 = 0
More Examples: 𝑥2
+ 𝑦2
= 𝑎2
, 𝑥3
+ 𝑥𝑦2
+ 4𝑥 = 5, 𝑒𝑡𝑐. ,
73. Differentiation of implicit functions
If 𝑓(𝑥, 𝑦 ) is the given implicit function , then
𝑑𝑦
𝑑𝑥
= −
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑦
Producer to find the differentiation:
(i) Take 𝑓(𝑥, 𝑦)
(ii) Find
𝜕𝑓
𝜕𝑥
&
𝜕𝑓
𝜕𝑦
(iii) Find
𝑑𝑦
𝑑𝑥
= −
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑦
Problem 1:
If 𝒙 𝒚
+ 𝒚 𝒙
= 𝒄, then find
𝒅𝒚
𝒅𝒙
Solution:
Given 𝑥 𝑦
+ 𝑦 𝑥
= 𝑐
⟹ 𝑥 𝑦
+ 𝑦 𝑥
− 𝑐 = 0
Take 𝑓 𝑥, 𝑦 = 𝑥 𝑦
+ 𝑦 𝑥
− 𝑐