Partial differentiation, total differentiation, Jacobian, Taylor's expansion, stationary points,maxima & minima (Extreme values),constraint maxima & minima ( Lagrangian multiplier), differentiation of implicit functions.

1. Functions of
severable
variables

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

6. Problem 1:
If 𝒇 𝒙, 𝒚 = 𝟑𝒙𝒚 𝟐 − 𝟐𝒙 𝟐 𝒚,then find
𝒇 𝒙, 𝒇 𝒚, 𝒇 𝒙𝒙, 𝒇 𝒚𝒚, 𝒇 𝒙𝒚
Solution:
Given 𝑓 𝑥, 𝑦 = 3𝑥𝑦2 − 2𝑥2 𝑦 ---(1)
Diff. (1) partially w.r.to x,
𝑓𝑥 = 3 1 𝑦2 − 2 2𝑥 𝑦
𝑓𝑥 = 3𝑦2
− 4𝑥𝑦 --------- (2)
Diff. (1) partially w.r.to y,
𝑓𝑦 = 3𝑥 (2𝑦) − 2𝑥2
(1)
𝑓𝑦 = 6𝑥𝑦 − 2𝑥2
-----------(3)
Diff. (2) p.w.r.to x,
𝑓𝑥𝑥 = 0 − 4 1 𝑦
𝑓𝑥𝑥 = −4𝑦
Diff. (3) p.w.r.to y,
𝑓𝑦𝑦 = 6𝑥 1 − 0
𝑓𝑦𝑦 = 6𝑥
Diff. (2) p.w.r.to y,
𝑓𝑥𝑦 = 3 2𝑦 − 4𝑥(1)
𝑓𝑥𝑦 = 6𝑦 − 4𝑥
Note :
Diff.(3) p.w.r.to x,
𝑓𝑦𝑥 = 6 1 𝑦 − 2 2𝑥 = 6𝑦 − 4𝑥
𝑓𝑥𝑦 = 𝑓𝑦𝑥

7. Problem 2:
Find
𝝏𝒖
𝝏𝒙
,
𝝏𝒖
𝝏𝒚
, if 𝒖 = 𝒙𝒆 𝒚
+ 𝒚 𝒆 𝒙
Solution:
Given 𝑢 = 𝑥𝑒 𝑦 + 𝑦 𝑒 𝑥 -------(1)
Diff. (1) p.w.r.to x,
𝜕𝑢
𝜕𝑥
= 1 𝑒 𝑦
+ 𝑦 𝑒 𝑥
= 𝑒 𝑦
+ 𝑦 𝑒 𝑥
Diff.(1) p.w.r.to y,
𝜕𝑢
𝜕𝑦
= 𝑥 𝑒 𝑦
+ 1 𝑒 𝑥
= 𝑥𝑒 𝑦
+ 𝑒 𝑥

8. Problem 3:
If 𝒖 = 𝒙 𝟐
+ 𝒚 𝟐
+ 𝒛 𝟐 − 𝟏
𝟐
then find the value of
𝝏 𝟐 𝒖
𝝏𝒙 𝟐 +
𝝏 𝟐 𝒖
𝝏𝒚 𝟐 +
𝝏 𝟐 𝒖
𝝏𝒛 𝟐
Solution:
Given 𝑢 = 𝑥2
+ 𝑦2
+ 𝑧2 −1
2 ------(1)
Diff. (1) p. w.r.to x,
𝜕𝑢
𝜕𝑥
= −
1
2
𝑥2 + 𝑦2 + 𝑧2 −
1
2
−1
2𝑥 + 0 + 0
= −
1
2
𝑥2
+ 𝑦2
+ 𝑧2 −
3
2(2𝑥)
𝜕𝑢
𝜕𝑥
= −𝑥 𝑥2 + 𝑦2 + 𝑧2 −
3
2 ------- (2)
Diff. (2) p. w.r.to x,
𝜕2 𝑢
𝜕𝑥2 = −𝑥 −
3
2
𝑥2 + 𝑦2 + 𝑧2 −
3
2
−1
(2𝑥) + 𝑥2 + 𝑦2 + 𝑧2 −
3
2(−1)
= 3𝑥2 𝑥2 + 𝑦2 + 𝑧2 −
3
2
−1
− 𝑥2 + 𝑦2 + 𝑧2 −
3
2
= 𝑥2
+ 𝑦2
+ 𝑧2 −
3
2 3𝑥2
𝑥2
+ 𝑦2
+ 𝑧2 −1
− 1

9. = 𝑥2
+ 𝑦2
+ 𝑧2 −
3
2
3𝑥2
𝑥2+𝑦2+𝑧2 1 − 1
=
𝑥2+𝑦2+𝑧2 −
3
2 3𝑥2− 𝑥2+𝑦2+𝑧2
𝑥2+𝑦2+𝑧2
= 𝑥2
+ 𝑦2
+ 𝑧2 −
3
2 𝑥2
+ 𝑦2
+ 𝑧2 −1
3𝑥2
− 𝑥2
− 𝑦2
− 𝑧2
= 𝑥2 + 𝑦2 + 𝑧2 −
3
2
−1
2𝑥2 − 𝑦2 − 𝑧2
𝜕2 𝑢
𝜕𝑥2 = 𝑥2 + 𝑦2 + 𝑧2 −
5
2 2𝑥2 − 𝑦2 − 𝑧2 ------ (3)
Similarly,
𝜕2 𝑢
𝜕𝑦2 = 𝑥2
+ 𝑦2
+ 𝑧2 −
5
2 2𝑦2
− 𝑧2
− 𝑥2
------- (4)
𝜕2 𝑢
𝜕𝑧2 = 𝑥2
+ 𝑦2
+ 𝑧2 −
5
2 2𝑧2
− 𝑥2
− 𝑦2
------ (5)
(3)+(4)+(5) ⇒
𝜕2 𝑢
𝜕𝑥2+
𝜕2 𝑢
𝜕𝑦2+
𝜕2 𝑢
𝜕𝑧2 = 𝑥2
+ 𝑦2
+ 𝑧2 −
5
2(2𝑥2
− 𝑦2
− 𝑧2
+2𝑦2
− 𝑧2
− 𝑥2
+2𝑧2
− 𝑥2
− 𝑦2
)
= 𝑥2
+ 𝑦2
+ 𝑧2 −
5
2 2𝑥2
+ 2𝑦2
+ 2𝑧2
− 2𝑥2
− 2𝑦2
− 2𝑧2
= 0

10. Problem 4
If 𝒖 = 𝒍𝒐𝒈 𝒕𝒂𝒏𝒙 + 𝒕𝒂𝒏𝒚 + 𝒕𝒂𝒏𝒛 𝒇𝒊𝒏𝒅 𝒔𝒊𝒏𝟐𝒙.
𝝏𝒖
𝝏𝒙
.
Solution:
Given 𝑢 = log 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧 ------- (1)
Diff. (1) p.w.r.to x,
𝜕𝑢
𝜕𝑥
=
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
(sec2 𝑥)
=
1
cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧)
, (∵ sec2
𝑥 =
1
cos2 𝑥
)
Now 𝑠𝑖𝑛2𝑥.
𝜕𝑢
𝜕𝑥
= 𝑠𝑖𝑛2𝑥
1
cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧)
=
2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
cos2 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧)
, (∵ 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥)
=
2𝑠𝑖𝑛𝑥
cos 𝑥(𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧)
= 2
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
(
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
)

11. ∴ 𝑠𝑖𝑛2𝑥.
𝜕𝑢
𝜕𝑥
= 2𝑡𝑎𝑛𝑥 (
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
)----(2)
Similarly,
𝑠𝑖𝑛2𝑦.
𝜕𝑢
𝜕𝑦
= 2𝑡𝑎𝑛𝑦 (
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
)----- (3)
𝑠𝑖𝑛2𝑧.
𝜕𝑢
𝜕𝑧
= 2𝑡𝑎𝑛𝑧 (
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
)------ (4)
(3)+(4)+(5)⟹
𝑠𝑖𝑛2𝑥.
𝜕𝑢
𝜕𝑥
+𝑠𝑖𝑛2𝑦.
𝜕𝑢
𝜕𝑦
+𝑠𝑖𝑛2𝑧.
𝜕𝑢
𝜕𝑧
=2𝑡𝑎𝑛𝑥
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
+ 2𝑡𝑎𝑛𝑦
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
+2𝑡𝑎𝑛𝑧
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
=2
1
𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧 = 2

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

16. Note : If 𝑢 = 𝑓 𝑝, 𝑞, 𝑟 and p = 𝑔 𝑥, 𝑦, 𝑧 , 𝑞 = ℎ(𝑥, 𝑦, 𝑧)
𝑟 = 𝑘(𝑥, 𝑦, 𝑧)
Total differential
𝑑𝑢
𝑑𝑥
= 𝑢 𝑝
𝑑𝑝
𝑑𝑥
+ 𝑢 𝑞
𝑑𝑞
𝑑𝑥
+ 𝑢 𝑟
𝑑𝑟
𝑑𝑥
𝑑𝑢
𝑑𝑦
= 𝑢 𝑝
𝑑𝑝
𝑑𝑦
+ 𝑢 𝑞
𝑑𝑞
𝑑𝑦
+ 𝑢 𝑟
𝑑𝑟
𝑑𝑦
𝑑𝑢
𝑑𝑧
= 𝑢 𝑝
𝑑𝑝
𝑑𝑧
+ 𝑢 𝑞
𝑑𝑞
𝑑𝑧
+ 𝑢 𝑟
𝑑𝑟
𝑑𝑧

17. Partial derivative:
If 𝑧 = 𝑓 𝑢, 𝑣 and 𝑢 = 𝑔 𝑥, 𝑦 , 𝑣 = ℎ(𝑥, 𝑦) then
𝜕𝑧
𝜕𝑥
=
𝜕𝑧
𝜕𝑢
𝜕𝑢
𝜕𝑥
+
𝜕𝑧
𝜕𝑣
𝜕𝑣
𝜕𝑥
= 𝑧 𝑢 𝑢 𝑥 + 𝑧 𝑣 𝑣 𝑥
𝜕𝑧
𝜕𝑦
=
𝜕𝑧
𝜕𝑢
𝜕𝑢
𝜕𝑦
+
𝜕𝑧
𝜕𝑣
𝜕𝑣
𝜕𝑦
= 𝑧 𝑢 𝑢 𝑦 + 𝑧 𝑣 𝑣 𝑦
Note : If 𝑢 = 𝑓 𝑝, 𝑞, 𝑟 and p = 𝑔 𝑥, 𝑦, 𝑧 , 𝑞 = ℎ(𝑥, 𝑦, 𝑧)
𝑟 = 𝑘(𝑥, 𝑦, 𝑧)
Partial differential
𝜕𝑢
𝜕𝑥
= 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 + 𝑢 𝑟 𝑟𝑥
𝜕𝑢
𝜕𝑦
= 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 + 𝑢 𝑟 𝑟𝑦
𝜕𝑢
𝜕𝑧
= 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 + 𝑢 𝑟 𝑟𝑧

18. Problem 1:
If 𝒘 = 𝒇(𝒚 − 𝒛, 𝒛 − 𝒙, 𝒙 − 𝒚) then show that
𝝏𝒘
𝝏𝒙
+
𝝏𝒘
𝝏𝒚
+
𝝏𝒘
𝝏𝒛
= 𝟎
Solution:
Given 𝑤 = 𝑓(𝑦 − 𝑧, 𝑧 − 𝑥, 𝑥 − 𝑦)
Take p = 𝑦 − 𝑧, 𝑞 = 𝑧 − 𝑥, 𝑟 = 𝑥 − 𝑦
⇒ 𝑤 = 𝑓(𝑝, 𝑞, 𝑟) and 𝑝 𝑦, 𝑧 , 𝑞 𝑧, 𝑥 , 𝑟(𝑥, 𝑦)
p = 𝑦 − 𝑧, 𝑞 = 𝑧 − 𝑥, 𝑟 = 𝑥 − 𝑦
𝑝 𝑥 = 0 𝑞 𝑥 = −1 𝑟𝑥 = 1
𝑝 𝑦 = 1 𝑞 𝑦 = 0 𝑟𝑦 = −1
𝑝 𝑧 = −1 𝑞 𝑧 = 1 𝑟𝑧 = 0
𝜕𝑤
𝜕𝑥
= 𝑤 𝑝 𝑝 𝑥 + 𝑤 𝑞 𝑞 𝑥 + 𝑤𝑟 𝑟𝑥 = 𝑤 𝑝 0 + 𝑤 𝑞 −1 + 𝑤𝑟 1
𝜕𝑤
𝜕𝑥
= −𝑤 𝑞 + 𝑤𝑟 ------ (1)
𝜕𝑤
𝜕𝑦
= 𝑤 𝑝 𝑝 𝑦 + 𝑤 𝑞 𝑞 𝑦 + 𝑤𝑟 𝑟𝑦 = 𝑤 𝑝 1 + 𝑤 𝑞 0 + 𝑤𝑟 −1
𝜕𝑤
𝜕𝑦
= 𝑤 𝑝 − 𝑤𝑟 ------ (2)
𝜕𝑤
𝜕𝑧
= 𝑤 𝑝 𝑝 𝑧 + 𝑤 𝑞 𝑞 𝑧 + 𝑤𝑟 𝑟𝑧 = 𝑤 𝑝 −1 + 𝑤 𝑞 1 + 𝑤𝑟 0
𝜕𝑤
𝜕𝑧
= −𝑤 𝑝 + 𝑤 𝑞 ------ (3)
(1)+(2)+(3) ⇒
𝜕𝑤
𝜕𝑥
+
𝜕𝑤
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
= −𝑤 𝑞 + 𝑤𝑟+𝑤 𝑝 − 𝑤𝑟 −𝑤 𝑝 +𝑤 𝑞 = 0

19. Problem2:
If 𝒖 = 𝒇(𝟐𝒙 − 𝟑𝒚, 𝟑𝒚 − 𝟒𝒛, 𝟒𝒛 − 𝟐𝒙) then find
𝟏
𝟐
𝝏𝒖
𝝏𝒙
+
𝟏
𝟑
𝝏𝒖
𝝏𝒚
+
𝟏
𝟒
𝝏𝒖
𝝏𝒛
Solution:
Given 𝑢 = 𝑓(2𝑥 − 3𝑦, 3𝑦 − 4𝑧, 4𝑧 − 2𝑥)
Take p = 2𝑥 − 3𝑦, 𝑞 = 3𝑦 − 4𝑧, 𝑟 = 4𝑧 − 2𝑥
⇒ 𝑢 = 𝑓(𝑝, 𝑞, 𝑟) and 𝑝 𝑥, 𝑦 , 𝑞 𝑦, 𝑧 , 𝑟(𝑧, 𝑥)
p = 2𝑥 − 3𝑦, 𝑞 = 3𝑦 − 4𝑧, 𝑟 = 4𝑧 − 2𝑥
𝑝 𝑥 = 2 𝑞 𝑥 = 0 𝑟𝑥 = −2
𝑝 𝑦 = −3 𝑞 𝑦 = 3 𝑟𝑦 = 0
𝑝 𝑧 = 0 𝑞 𝑧 = −4 𝑟𝑧 = 4
𝜕𝑢
𝜕𝑥
= 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 + 𝑢 𝑟 𝑟𝑥 = 𝑢 𝑝 2 + 𝑢 𝑞 0 + 𝑢 𝑟 −2
1
2
𝜕𝑢
𝜕𝑥
=
1
2
2𝑢 𝑝 − 2𝑢 𝑟 =
1
2
(2 𝑢 𝑝 − 𝑢 𝑟 ) = 𝑢 𝑝 − 𝑢 𝑟 --------- (1)
𝜕𝑢
𝜕𝑦
= 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 + 𝑢 𝑟 𝑟𝑦 = 𝑢 𝑝 −3 + 𝑢 𝑞 3 + 𝑢 𝑟 0
1
3
𝜕𝑢
𝜕𝑦
=
1
3
(−3𝑢 𝑝+3𝑢 𝑞) =
1
3
3 −up + uq = −𝑢 𝑝 + 𝑢 𝑞 ---------(2)
𝜕𝑢
𝜕𝑧
= 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 + 𝑢 𝑟 𝑟𝑧 = 𝑢 𝑝 0 + 𝑢 𝑞 −4 + 𝑢 𝑟 4
1
4
𝜕𝑢
𝜕𝑧
=
1
4
−4𝑢 𝑞 + 4𝑢 𝑟 =
1
4
4 −𝑢 𝑞 ∓ 𝑢 𝑟 = −𝑢 𝑞 + 𝑢 𝑟 ------- (3)
(1)+(2)+(3) ⇒
1
2
𝜕𝑢
𝜕𝑥
+
1
3
𝜕𝑢
𝜕𝑦
+
1
4
𝜕𝑢
𝜕𝑧
= 𝑢 𝑝 − 𝑢 𝑟 − 𝑢 𝑝 + 𝑢 𝑞 − 𝑢 𝑞 + 𝑢 𝑟 = 0

20. Problem3:
If 𝒖 = 𝒇(
𝒚−𝒙
𝒙𝒚
,
𝒛−𝒙
𝒙𝒛
) then find 𝒙 𝟐 𝝏𝒖
𝝏𝒙
+ 𝒚 𝟐 𝝏𝒖
𝝏𝒚
+ 𝒛 𝟐 𝝏𝒖
𝝏𝒛
Solution:
Given𝒖 = 𝒇(
𝒚−𝒙
𝒙𝒚
,
𝒛−𝒙
𝒙𝒛
)
Take p =
𝒚−𝒙
𝒙𝒚
, 𝑞 =
𝒛−𝒙
𝒙𝒛
⇒ 𝑢 = 𝑓(𝑝, 𝑞) and 𝑝 𝑥, 𝑦 , 𝑞 𝑥, 𝑧
𝐩 =
𝐲 − 𝐱
𝐱𝐲
𝐪 =
𝐳 − 𝐱
𝐱𝐳
𝐩 𝐱 =
𝐱𝐲 −𝟏 − 𝐲−𝐱 𝐲
𝐱𝐲 𝟐
=
−𝐱𝐲−𝐲 𝟐+𝐱𝐲
𝐱 𝟐 𝐲 𝟐 = −
𝐲 𝟐
𝐱 𝟐 𝐲 𝟐 = −
𝟏
𝐱 𝟐
𝐪 𝐱 =
𝐱𝐳 −𝟏 − 𝐳−𝐱 𝐳
𝐱𝐳 𝟐
=
−𝐱𝐳−𝐳 𝟐+𝐱𝐳
𝐱 𝟐 𝐳 𝟐 = −
𝐳 𝟐
𝐱 𝟐 𝐳 𝟐 = −
𝟏
𝐱 𝟐
𝐩 𝐲 =
𝐱𝐲 𝟏 − 𝐲−𝐱 𝐱
𝐱𝐲 𝟐
=
𝐱𝐲−𝐱𝐲+𝐱 𝟐
𝐱 𝟐 𝐲 𝟐 =
𝐱 𝟐
𝐱 𝟐 𝐲 𝟐 =
𝟏
𝐲 𝟐
𝐪 𝐲 = 𝟎
𝐩 𝐳 = 𝟎
𝐪 𝐳 =
𝐱𝐳 𝟏 − 𝐳−𝐱 𝐱
𝐱𝐳 𝟐
=
𝐱𝐳−𝐳𝐱+𝐱 𝟐
𝐱 𝟐 𝐳 𝟐 =
𝐱 𝟐
𝐱 𝟐 𝐳 𝟐 =
𝟏
𝐳 𝟐

21. 𝜕𝑢
𝜕𝑥
= 𝑢 𝑝 𝑝 𝑥 + 𝑢 𝑞 𝑞 𝑥 = 𝑢 𝑝 −
1
𝑥2 + 𝑢 𝑞 −
1
𝑥2 = −
1
𝑥2 (𝑢 𝑝 + 𝑢 𝑞)
𝑥2 𝜕𝑢
𝜕𝑥
= 𝑥2
−
1
𝑥2 𝑢 𝑝 + 𝑢 𝑞 = −𝑢 𝑝 − 𝑢 𝑞 --------- (1)
𝜕𝑢
𝜕𝑦
= 𝑢 𝑝 𝑝 𝑦 + 𝑢 𝑞 𝑞 𝑦 = 𝑢 𝑝
1
𝑦2 + 𝑢 𝑞 0
𝑦2 𝜕𝑢
𝜕𝑦
= 𝑦2
(
1
𝑦2 𝑢 𝑝) = 𝑢 𝑝 ---------(2)
𝜕𝑢
𝜕𝑧
= 𝑢 𝑝 𝑝 𝑧 + 𝑢 𝑞 𝑞 𝑧 = 𝑢 𝑝 0 + 𝑢 𝑞
1
𝑧2
𝑧2 𝜕𝑢
𝜕𝑧
= 𝑧2 1
𝑧2 (𝑢 𝑞) = 𝑢 𝑞 ------- (3)
(1)+(2)+(3) ⇒ 𝑥2 𝜕𝑢
𝜕𝑥
+ 𝑦2 𝜕𝑢
𝜕𝑦
+ 𝑧2 𝜕𝑢
𝜕𝑧
= −𝑢 𝑝 − 𝑢 𝑞 + 𝑢 𝑝 + 𝑢 𝑞 = 0

22. • If 𝑓(𝑥, 𝑦) is a function of two variables x and y, then Taylor’s expansion of 𝑓(𝑥, 𝑦)
about the point (𝑎, 𝑏) is given by
𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 +
1
2!
𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 +
1
3!
𝑥 − 𝑎 3 𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2 𝑦 − 𝑏 𝑓𝑥𝑥𝑦 + 3 𝑥 − 𝑎 𝑦 − 𝑏 2 𝑓𝑥𝑦𝑦
+ 𝑦 − 𝑏 3 𝑓𝑦𝑦𝑦 𝑎, 𝑏
+ ⋯
• Taylor’s expansion of 𝑓(𝑥, 𝑦) about the origin or (0,0) is given by
𝑓 𝑥, 𝑦 = 𝑓 0,0 + 𝑥𝑓𝑥 0,0 + 𝑦𝑓𝑦 0,0
+
1
2!
𝑥2 𝑓𝑥𝑥 0,0 + 2𝑥𝑦𝑓𝑥𝑦 0,0 + 𝑦2 𝑓𝑦𝑦 0,0 + ⋯
Taylor’s Theorem for functions of two variables

23. Problem 1:
Obtain the Taylor series of 𝒙 𝟑 + 𝒚 𝟑 + 𝒙𝒚 𝟐 in powers of 𝒙 − 𝟏 𝒂𝒏𝒅 𝒚 − 𝟐.
Solution:
W.kt, Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by
𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 +
1
2!
𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 +
1
3!
𝑥 − 𝑎 3 𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2 𝑦 − 𝑏 𝑓𝑥𝑥𝑦(𝑎, 𝑏) + 3 𝑥 − 𝑎 𝑦 − 𝑏 2 𝑓𝑥𝑦𝑦(𝑎, 𝑏)
+ 𝑦 − 𝑏 3 𝑓𝑦𝑦𝑦 𝑎, 𝑏
+ ⋯
Here 𝑓 𝑥, 𝑦 = 𝑥3
+ 𝑦3
+ 𝑥𝑦2
,
the points are 𝑎 = 1, 𝑏 = 2 , (∵ 𝑥 − 1 = 0 & 𝑦 − 2 = 0 ⇒ 𝑥 = 1, 𝑦 = 2 )

24. 𝑓 𝑥, 𝑦 = 𝑥3
+ 𝑦3
+ 𝑥𝑦2
𝑓 1,2 = 13
+ 23
+ 1 22
= 1 + 8 + 4 = 13
𝑓𝑥 = 3𝑥2
+ 𝑦2
𝑓𝑥 1,2 = 3 1 2
+ 22
= 3 + 4 = 7
𝑓𝑦 = 3𝑦2
+ 𝑥 2𝑦 = 3𝑦2
+ 2𝑥𝑦 𝑓𝑦 1,2 = 3 2 2
+ 2 1 2 = 12 + 4 = 16
𝑓𝑥𝑥 = 6𝑥 + 0 = 6𝑥 𝑓𝑥𝑥 1,2 = 6 1 = 6
𝑓𝑥𝑦 = 0 + 2𝑦 = 2𝑦 𝑓𝑥𝑦 1,2 = 2 2 = 4
𝑓𝑦𝑦 = 3 2𝑦 + 2𝑥 1 = 6𝑦 + 2𝑥 𝑓𝑦𝑦 1,2 = 6 2 + 2 1 = 12 + 2 = 14
𝑓𝑥𝑥𝑥 = 6 1 = 6 𝑓𝑥𝑥𝑥 1,2 = 6
𝑓𝑥𝑥𝑦 =0 𝑓𝑥𝑥𝑦(1,2) = 0
𝑓𝑥𝑦𝑦 = 2 1 = 2 𝑓𝑥𝑦𝑦(1,2) = 2
𝑓𝑦𝑦𝑦 = 6 1 + 0 = 6 𝑓𝑦𝑦𝑦(1,2) = 6

25. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (1,2) is given by
𝑓 𝑥, 𝑦 = 𝑓 1,2 + 𝑥 − 1 𝑓𝑥 1,2 + 𝑦 − 2 𝑓𝑦 1,2 +
1
2!
𝑥 − 1 2 𝑓𝑥𝑥 1,2 + 2 𝑥 − 1 𝑦 − 2 𝑓𝑥𝑦 1,2 + 𝑦 − 2 2 𝑓𝑦𝑦 1,2 +
1
3!
𝑥 − 1 3
𝑓𝑥𝑥𝑥 1,2 + 3 𝑥 − 1 2
𝑦 − 2 𝑓𝑥𝑥𝑦(1,2) + 3 𝑥 − 1 𝑦 − 2 2
𝑓𝑥𝑦𝑦 (1,2)
+ 𝑦 − 2 3 𝑓𝑦𝑦𝑦 1,2
+ ⋯
𝑓 𝑥, 𝑦 = 13 + 7 𝑥 − 1 + 16 𝑦 − 2
+
1
2!
[6 𝑥 − 1 2
+ 2 4 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2
]
+
1
3!
6 𝑥 − 1 3 + 3 0 𝑥 − 1 2 𝑦 − 2 + 3 2 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3
+ ⋯
= 13 + 7 𝑥 − 1 + 16 𝑦 − 2
+
1
2!
[6 𝑥 − 1 2
+ 8 𝑥 − 1 𝑦 − 2 + 14 𝑦 − 2 2
]
+
1
3!
6 𝑥 − 1 3 + 0 + 6 𝑥 − 1 𝑦 − 2 2 + 6 𝑦 − 2 3 + ⋯
= 13 + 7 𝑥 − 1 + 16 𝑦 − 2 +
6
2
𝑥 − 1 2
+
8
2
𝑥 − 1 𝑦 − 2 +
14
2
𝑦 − 2 2
+
6
6
𝑥 − 1 3
+
6
6
𝑥 − 1 𝑦 − 2 2
+
6
6
𝑦 − 2 3
+… , (2! = 1.2 = 2, 3! = 1.2.3 = 6)

26. = 13 + 7 𝑥 − 1 + 16 𝑦 − 2 +
6
2
𝑥 − 1 2
+
8
2
𝑥 − 1 𝑦 − 2 +
14
2
𝑦 − 2 2
+ 𝑥 − 1 3 + 𝑥 − 1 𝑦 − 2 2 + 𝑦 − 2 3 + ⋯
Problem 2
Expand 𝒆 𝒙
𝒄𝒐𝒔𝒚 at 𝟎,
𝝅
𝟐
upto the third term by using Taylor’s series.
Solution
W.kt, Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏) is given by
𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 +
1
2!
𝑥 − 𝑎 2 𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2 𝑓𝑦𝑦 𝑎, 𝑏 +
1
3!
𝑥 − 𝑎 3
𝑓𝑥𝑥𝑥 𝑎, 𝑏 + 3 𝑥 − 𝑎 2
𝑦 − 𝑏 𝑓𝑥𝑥𝑦(𝑎, 𝑏) + 3 𝑥 − 𝑎 𝑦 − 𝑏 2
𝑓𝑥𝑦𝑦(𝑎, 𝑏)
+ 𝑦 − 𝑏 3
𝑓𝑦𝑦𝑦 𝑎, 𝑏
+ ⋯
Here 𝑓 𝑥, 𝑦 = 𝑒 𝑥
𝑐𝑜𝑠𝑦
the points are 𝑎 = 0, 𝑏 =
𝜋
2
,

27. 𝑓 𝑥, 𝑦 = 𝑒 𝑥
𝑐𝑜𝑠𝑦 𝑓 0,
𝜋
2
= 𝑒0
cos
𝜋
2
= 1.0 = 0, (∵ 𝑒0
= 1, cos
𝜋
2
= 0
)
𝑓𝑥 = 𝑒 𝑥
𝑐𝑜𝑠𝑦 𝑓𝑥 0,
𝜋
2
= 𝑒0
cos
𝜋
2
= 0
𝑓𝑦 = 𝑒 𝑥
−𝑠𝑖𝑛𝑦 = −𝑒 𝑥
𝑠𝑖𝑛𝑦 𝑓𝑦 0,
𝜋
2
= −𝑒0
sin
𝜋
2
= −1.1 = −1 (∵ 𝑒0
=
1, sin
𝜋
2
= 1)
𝑓𝑥𝑥 = 𝑒 𝑥
𝑐𝑜𝑠𝑦 𝑓𝑥𝑥 0,
𝜋
2
= 𝑒0
cos
𝜋
2
= 0
𝑓𝑥𝑦 = −𝑒 𝑥
𝑠𝑖𝑛𝑦 𝑓𝑥𝑦 0,
𝜋
2
= −1
𝑓𝑦𝑦 = −𝑒 𝑥
(𝑐𝑜𝑠𝑦) 𝑓𝑦𝑦 0,
𝜋
2
= −𝑒0
cos
𝜋
2
= 0
𝑓𝑥𝑥𝑥 = 𝑒 𝑥
𝑐𝑜𝑠𝑦 𝑓𝑥𝑥𝑥 0,
𝜋
2
= 0
𝑓𝑥𝑥𝑦 = −𝑒 𝑥
𝑠𝑖𝑛𝑦 𝑓𝑥𝑥𝑦 0,
𝜋
2
= −1
𝑓𝑥𝑦𝑦 = −𝑒 𝑥
𝑐𝑜𝑠𝑦 𝑓𝑥𝑦𝑦(0,
𝜋
2
) = 0
𝑓𝑦𝑦𝑦 = −𝑒 𝑥
−𝑠𝑖𝑛𝑦 = 𝑒 𝑥
𝑠𝑖𝑛𝑦 𝑓𝑦𝑦𝑦 0,
𝜋
2
= 1.1 = 1

28. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (0,
𝜋
2
) is given by
𝑓 𝑥, 𝑦 = 𝑓 0,
𝜋
2
+ 𝑥 − 0 𝑓𝑥 0,
𝜋
2
+ 𝑦 −
𝜋
2
𝑓𝑦 0,
𝜋
2
+
1
2!
𝑥 − 0 2 𝑓𝑥𝑥 0,
𝜋
2
+ 2 𝑥 − 0 𝑦 −
𝜋
2
𝑓𝑥𝑦 0,
𝜋
2
+ 𝑦 −
𝜋
2
2
𝑓𝑦𝑦 0,
𝜋
2
+
1
3!
𝑥 − 0 3 𝑓𝑥𝑥𝑥 0,
𝜋
2
+ 3 𝑥 − 0 2 𝑦 −
𝜋
2
𝑓𝑥𝑥𝑦(0,
𝜋
2
) + 3 𝑥 − 0 𝑦 −
𝜋
2
2
𝑓𝑥𝑦𝑦(0,
𝜋
2
)
+ 𝑦 −
𝜋
2
3
𝑓𝑦𝑦𝑦 0,
𝜋
2
+ ⋯
= 0 + 𝑥 0 + 𝑦 −
𝜋
2
−1 +
1
2!
𝑥 2 0 + 2 𝑥 𝑦 −
𝜋
2
−1 + 𝑦 −
𝜋
2
2
0 +
1
3!
𝑥 3
0 + 3 𝑥 2
𝑦 −
𝜋
2
−1 + 3 𝑥 𝑦 −
𝜋
2
2
(0)
+ 𝑦 −
𝜋
2
3
(1)
+ ⋯

29. = − 𝑦 −
𝜋
2
+ −
2
2
𝑥 𝑦 −
𝜋
2
+ −
3
6
𝑥 2
𝑦 −
𝜋
2
+
1
6
𝑦 −
𝜋
2
3
+ ⋯
(∵ 2! = 1.2 = 2, 3! = 1.2.3 = 6)
= − 𝑦 −
𝜋
2
− 𝑥 𝑦 −
𝜋
2
−
1
2
𝑥 2 𝑦 −
𝜋
2
+
1
6
𝑦 −
𝜋
2
3
+ ⋯
Problem 3
Find the Taylor’s series expansion of 𝟏 + 𝒙 + 𝒚 𝟐 in powers of 𝒙 − 𝟏 and 𝒚 upto
second degree term.
Solution:
W.kt, Taylor’s expansion (upto second degree) of 𝑓(𝑥, 𝑦) about the point (𝑎, 𝑏)
is given by
𝑓 𝑥, 𝑦 = 𝑓 𝑎, 𝑏 + 𝑥 − 𝑎 𝑓𝑥 𝑎, 𝑏 + 𝑦 − 𝑏 𝑓𝑦 𝑎, 𝑏 +
1
2!
𝑥 − 𝑎 2
𝑓𝑥𝑥 𝑎, 𝑏 + 2 𝑥 − 𝑎 𝑦 − 𝑏 𝑓𝑥𝑦 𝑎, 𝑏 + 𝑦 − 𝑏 2
𝑓𝑦𝑦 𝑎, 𝑏 + ⋯
Here 𝑓 𝑥, 𝑦 = 1 + 𝑥 + 𝑦2
the points are 𝑎 = 1, 𝑏 = 0,

30. 𝑓 𝑥, 𝑦 = 1 + 𝑥 + 𝑦2 = 1 + 𝑥 + 𝑦2
1
2 𝑓 1,0 = 1 + 1 + 0 = 2
1
2
𝑓𝑥 =
1
2
1 + 𝑥 + 𝑦2
1
2
−1
1 =
1
2
(1 + 𝑥 + 𝑓𝑥 1,0 =
1
2
1 + 1 + 0 −
1
2 =
1
2
2 −
1
2 =
2−
1
2
−1
= 2−
3
2
𝑓𝑦 =
1
2
1 + 𝑥 + 𝑦2
1
2
−1
2𝑦 =
𝑦 1 + 𝑥 + 𝑦2 −
1
2
𝑓𝑦 1,0 = 0
𝑓𝑥𝑥 =
1
2
−
1
2
1 + 𝑥 + 𝑦2 −
1
2
−1
1 =
−1
4
1 + 𝑥 + 𝑦2 −
3
2
𝑓𝑥𝑥 1,0 = −
1
4
1 + 1 + 0 −
3
2 =
1
4
2 −
3
2 =
−
1
22 2 −
3
2 = −2−
3
2
−2
= −2−
7
2
𝑓𝑥𝑦 =
1
2
−
1
2
1 + 𝑥 + 𝑦2 −
1
2
−1
2𝑦
=
−𝑦
2
1 + 𝑥 + 𝑦2 −
3
2
𝑓𝑥𝑦 1,0 = 0
𝑓𝑦𝑦 = 𝑦 −
1
2
1 + 𝑥 + 𝑦2 −
1
2
−1
2𝑦
+ 1 + 𝑥 + 𝑦2 −
1
2 1
= −𝑦 1 + 𝑥 + 𝑦2 −
3
2 +(1 + 𝑥 +
𝑓𝑦𝑦 1,0 = 0 + 1 + 1 + 0 −
1
2 = 2−
1
2

31. Taylor’s expansion of 𝑓(𝑥, 𝑦) about the point (1,0) is given by
𝑓 𝑥, 𝑦 = 𝑓 1,0 + 𝑥 − 1 𝑓𝑥 1,0 + 𝑦 𝑓𝑦 1,0 +
1
2!
𝑥 − 1 2 𝑓𝑥𝑥 1,0 + 2 𝑥 − 1 𝑦 𝑓𝑥𝑦 1,0 + 𝑦 2 𝑓𝑦𝑦 1,0 + ⋯
= 2
1
2 + 𝑥 − 1 2−
3
2 + 𝑦 0 +
1
2
𝑥 − 1 2 −2 −
7
2 +
2
2
𝑥 − 1 𝑦 0 + 𝑦22−
1
2
= 2
1
2 + 2−
3
2 𝑥 − 1 − −2 −
7
2
−1
𝑥 − 1 2
+ 2−
1
2 𝑦2
= 2
1
2 + 2−
3
2 𝑥 − 1 − −2 −
9
2 𝑥 − 1 2
+ 2−
1
2 𝑦2

32. Maximum and Minimum
(Extreme Values)
of two variable function

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

35. Problem 1:
Discuss the maxima and minima of the function 𝒇 𝒙, 𝒚 = 𝒙 𝟒 + 𝒚 𝟒 − 𝟐𝒙 𝟐 + 𝟒𝒙𝒚 − 𝟐𝒚 𝟐
Solution:
Given 𝑓 𝑥, 𝑦 = 𝑥4
+ 𝑦4
− 2𝑥2
+ 4𝑥𝑦 − 2𝑦2
--------- (1)
To find the stationary point:
Put
𝝏𝒇
𝝏𝒙
= 𝑓𝑥 = 0 &
𝝏𝒇
𝝏𝒚
= 𝒇 𝒚 = 0
𝝏𝒇
𝝏𝒙
= 0 ⟹ 4𝑥3 − 4𝑥 + 4𝑦 = 0 ⟹ 4(𝑥3 − 𝑥 + 𝑦) = 0
⟹ 𝑥3 − 𝑥 + 𝑦 = 0 ---------(2)
𝝏𝒇
𝝏𝒚
= 0 ⟹ 4𝑦3 + 4𝑥 − 4𝑦 = 0 ⟹ 4 𝑦3 + 𝑥 − 𝑦 = 0
⟹ 𝑦3 + 𝑥 − 𝑦 = 0 ------(3)
𝐩 =
𝛛𝐟
𝛛𝐱
= 𝒇 𝒙
= 𝟒𝐱 𝟑
− 𝟒𝐱 + 𝟒𝐲
𝐫 =
𝛛 𝟐
𝐟
𝛛𝐱 𝟐
= 𝐟 𝐱𝐱 =
= 𝟏𝟐𝐱 𝟐
− 𝟒
𝐬 =
𝛛 𝟐
𝐟
𝛛𝐱𝛛𝐲
= 𝒇 𝒙𝒚 = 𝟒
𝐪 =
𝛛𝐟
𝛛𝐲
= 𝐟 𝐲
= 𝟒𝐲 𝟑
+ 𝟒𝐱 − 𝟒𝐲
𝐭 =
𝛛 𝟐𝐟
𝛛𝐲 𝟐
= 𝐟 𝐲𝐲
= 𝟏𝟐𝐲 𝟐
− 𝟒

36. 2 ⟹ 𝑥3
− 𝑥 + 𝑦 = 0
3 ⟹ 𝑦3
+ 𝑥 − 𝑦 = 0
Add ------------------------
𝑥3 + 𝑦3 = 0
⟹ 𝑥 + 𝑦 𝑥2
− 𝑥𝑦 + 𝑦2
= 0 , (∵ 𝑎3
+ 𝑏3
= (𝑎 + 𝑏)(𝑎2
− 𝑎𝑏 + 𝑏2
)
⟹ 𝑥 + 𝑦 = 0 𝑜𝑟 (𝑥2 − 𝑥𝑦 + 𝑦2) = 0
⟹ 𝑥 = −𝑦 𝑜𝑟 𝑥2
− 𝑥𝑦 + 𝑦2
= 0
Put 𝑥 = −𝑦 in (2)
⟹ −𝑦 3
− −𝑦 + 𝑦 = 0
⟹ −𝑦3
+ 𝑦 + 𝑦 = 0
⟹ −𝑦3
+ 2𝑦 = 0 ⟹ 𝑦(−𝑦2
+ 2) = 0
⟹ 𝑦 = 0 𝑜𝑟 − 𝑦2
+ 2 = 0
⟹ 𝑦 = 0 𝑜𝑟 𝑦2
= 2
⟹ 𝑦 = 0 𝑜𝑟 𝑦 = ± 2
When 𝑦 = 0 , from (3) 0 + 𝑥 − 0 = 0 ⟹ 𝑥 = 0
The point is ( 𝑥, 𝑦) = (0,0)
When 𝑦 = 2 , from (3) 2
3
+ 𝑥 − 2 = 0
⟹ 𝑥 = − 2
3
+ 2 ⟹ 𝑥 = 2 (− 2
2
+ 1)
⟹ 𝑥 = 2 −2 + 1 = 2 −1 = − 2
The point is ( 𝑥, 𝑦) = (− 2, 2)

37. When 𝑦 = − 2 , from (3), − 2
3
+ 𝑥 − (− 2) = 0
⟹ − 2
3
+ 𝑥 + 2 = 0
⟹ 𝑥 = 2
3
− 2 = 2 2
2
− 1 = 2 2 − 1 = 2 1 = 2
The point is ( 𝑥, 𝑦) = ( 2, − 2)
The stationary / critical points are (0,0), − 2, 2 , ( 2, − 2)
To find the point at which 𝒇(𝒙, 𝒚) has maximum/minimum:
𝑟𝑡 − 𝑠2 = 12𝑥2 − 4 12𝑦2 − 4 − 4 2
= 12𝑥2 − 4 12𝑦2 − 4 − 16
At the Point (0,0):
𝑟𝑡 − 𝑠2
𝑎𝑡 0,0 = 12 0 2
− 4 12 0 2
− 4 − 16
= −4 −4 − 16 = 16 − 16 = 𝟎
Hence there is no maximum / minimum for 𝑓(𝑥, 𝑦) at (0,0) (we can’t judge)
At the Point − 𝟐, 𝟐 :
𝑟𝑡 − 𝑠2
𝑎𝑡 − 2, 2 = 12 − 2
2
− 4 12 2
2
− 4 − 16
= 12 2 − 4 12 2 − 4 − 16 = 24 − 4 24 − 4 − 16
= 20 20 − 16 = 400 − 16 = 394 > 𝟎
𝑟 𝑎𝑡 − 2, 2 = 12 − 2
2
− 4 = 12 2 − 4 = 24 − 4 = 20 > 𝟎
Hence 𝑓(𝑥, 𝑦) has minimum at − 2, 2

38. To find the minimum value :
Put x, y = − 2, 2 in (1)
The minimum value is 𝑓(− 2, 2)
= − 2
4
+ 2
4
− 2 − 2
2
+ 4 − 2 2 − 2 2
2
= 4 + 4 − 4 − 8 − 4 = −8
Hence the minimum value is −8
At the Point 𝟐, − 𝟐 :
𝑟𝑡 − 𝑠2
𝑎𝑡 2,− 2 = 12 2
2
− 4 12 − 2
2
− 4 − 16
= 12 2 − 4 12 2 − 4 − 16 = 24 − 4 24 − 4 − 16
= 20 20 − 16 = 400 − 16 = 394 > 𝟎
𝑟 𝑎𝑡 − 2, 2 = 12 − 2
2
− 4 = 12 2 − 4 = 24 − 4 = 20 > 𝟎
Again 𝑓(𝑥, 𝑦) has minimum at 2, − 2

39. Problem 2:
Find the maximum or minimum values of 𝒇 𝒙, 𝒚 = 𝟑𝒙 𝟐
− 𝒚 𝟐
+ 𝒙 𝟑
Solution:
Given 𝑓 𝑥, 𝑦 = 3𝑥2 − 𝑦2 + 𝑥3 ------- (1)
To find the stationary point:
Put
𝜕𝑓
𝜕𝑥
= 𝑓𝑥 = 0 &
𝜕𝑓
𝜕𝑦
= 𝑓𝑦 = 0
𝜕𝑓
𝜕𝑥
= 0 ⇒ 6𝑥 + 3𝑥2 = 0 ⟹ 3𝑥 2 + 𝑥 = 0
⟹ 𝑥 = 0, 𝑥 + 2 = 0 ⟹ 𝑥 = 0, 𝑥 = −2
𝜕𝑓
𝜕𝑦
= 0 ⟹ −2𝑦 = 0 ⟹ 𝑦 = 0
The stationary points are 0,0 , (−2,0)
𝑝 = 𝐟 𝐱 =
𝜕𝑓
𝜕𝑥
= 6𝑥 + 3𝑥2
𝑟 = 𝐟 𝐱𝐱 =
𝜕2
𝑓
𝜕𝑥2
= 6 + 6𝑥
𝑠 = 𝐟 𝐱𝐲 =
𝜕2
𝑓
𝜕𝑥𝜕𝑦
= 0
𝑞 = 𝐟 𝐲 =
𝜕𝑓
𝜕𝑦
= −2𝑦 𝑡 = 𝑓𝑦𝑦 =
𝜕2𝑓
𝜕𝑦2
= −2

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

44. To find the stationary points:
Put
𝜕F
𝜕𝑥
= 0,
𝜕F
𝜕𝑦
= 0 &
𝜕F
𝜕𝑧
= 0
𝜕𝐹
𝜕𝑥
= 0 ⟹ 2𝑥 − 𝜆𝑦 = 0 ⟹ 𝜆𝑦 = 2𝑥 ⟹ 𝜆 =
2𝑥
𝑦
------- (3)
𝜕F
𝜕𝑦
= 0 ⟹ 2𝑦 − 𝜆x = 0 ⟹ 𝜆𝑥 = 2𝑦 ⟹ 𝜆 =
2𝑦
𝑥
------ (4)
𝜕F
𝜕𝑧
= 0 ⟹ 2 z −
25
9
+ λ = 0 ⟹ λ = −2 z −
25
9
----- (5)
From (3), (4), (5),
2𝑥
𝑦
=
2𝑦
𝑥
= −2 z −
25
9
⟹
𝑥
𝑦
=
𝑦
𝑥
= −z +
25
9
Consider
𝑥
𝑦
=
𝑦
𝑥
⟹ 𝑥2
= 𝑦2
⟹ 𝑥 = ±𝑦
Consider
𝑦
𝑥
= −z +
25
9
,
If 𝑥 = 𝑦
−𝑧 +
25
9
=
𝑦
𝑦
= 1 ⟹ 𝑧 =
25
9
− 1 =
25−9
9
=
16
9
Given 𝑧 = 𝑥𝑦 = 𝑦. 𝑦 = 𝑦2
(∵ 𝑥 = 𝑦)
⟹ 𝑦2
=
16
9
∵ 𝑧 =
16
9
⟹ 𝑦 = ±
16
9
= ±
4
3
If 𝑥 = −𝑦
−𝑧 +
25
9
=
𝑦
−𝑦
= −1 ⟹ 𝑧 =
25
9
+ 1 =
25+9
9
=
34
9
Given 𝑧 = 𝑥𝑦 = −𝑦. 𝑦 = −𝑦2 (∵ 𝑥 = −𝑦)
⟹ −𝑦2
=
34
9
⟹ 𝑦2
= −
34
9
∵ 𝑧 =
34
9
⟹ 𝑦 = ±
−34
9
(Imaginary , this is not possible)
Hence 𝑥 = −𝑦 is ruled out.

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)

47. Note:

49. Hence the total surface area of a open rectangular box is 𝒙𝒚 + 𝟐𝒙𝒛 + 𝟐𝒚𝒛

50. Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙
𝐹 = 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 + 𝜆 𝑥𝑦𝑧 − 32 ( by using (1) & (2))
To find the stationary points:
Put
𝜕F
𝜕𝑥
= 0,
𝜕F
𝜕𝑦
= 0 &
𝜕F
𝜕𝑧
= 0
𝜕𝐹
𝜕𝑥
= 0 ⟹ y + 2z + 𝜆𝑦𝑧 = 0 ⟹ 𝜆𝑦𝑧 = − 𝑦 + 2𝑧 ⟹ 𝜆 =
−(𝑦+2𝑧)
𝑦𝑧
------ (3)
𝜕F
𝜕𝑦
= 0 ⟹ 𝑥 + 2𝑧 + 𝜆xz = 0 ⟹ 𝜆𝑥𝑧 = −(𝑥 + 2𝑧) ⟹ 𝜆 =
−(𝑥+2𝑧)
𝑥𝑧
------ (4)
𝜕F
𝜕𝑧
= 0 ⟹ 2x + 2y + λxy = 0 ⟹ λxy = −2x − 2y = −2 x + y
𝜆 =
−2 𝑥+𝑦
𝑥𝑦
----- (5)
𝜕𝐹
𝜕𝑥
= 𝑦 + 2𝑧 + 𝜆𝑦𝑧
𝜕𝐹
𝜕𝑦
= 𝑥 + 2𝑧 + 𝜆𝑥𝑧
𝜕𝐹
𝜕𝑧
= 2𝑥 + 2𝑦 + 𝜆𝑥𝑦

51. From (3), (4), (5),
−(𝑦+2𝑧)
𝑦𝑧
=
−(𝑥+2𝑧)
𝑥𝑧
=
−2 𝑥+𝑦
𝑥𝑦
Consider
−(𝑦+2𝑧)
𝑦𝑧
=
−(𝑥+2𝑧)
𝑥𝑧
⇒
−𝑥(𝑦+2𝑧)
𝑥𝑦𝑧
=
−𝑦(𝑥+2𝑧)
𝑦𝑥𝑧
⇒ 𝑥𝑦 + 2𝑧𝑥 = (𝑦𝑥 + 2𝑧𝑦)
⇒ 2𝑧𝑥 = 2𝑧𝑦
⇒ 𝑥 = 𝑦 ------ (6)
Consider
−(𝑥+2𝑧)
𝑥𝑧
=
−2 𝑥+𝑦
𝑥𝑦
⇒
𝑥+2𝑧
𝑥𝑧
=
2 𝑥+𝑦
𝑥𝑦
Using (6),
𝑥+2𝑧
𝑥𝑧
=
2 𝑥+𝑥
𝑥𝑥
⇒
𝑥+2𝑧
𝑥𝑧
=
4𝑥
𝑥2
⇒
𝑥+2𝑧
𝑥𝑧
=
4
𝑥
⇒
𝑥+2𝑧
𝑧
= 4

52. ⇒ 𝑥 + 2𝑧 = 4𝑧
⇒ 4𝑧 − 2𝑧 = 𝑥
⇒ 2𝑧 = 𝑥
⇒ 𝑧 =
𝑥
2
-------- (7)
Given volume 𝑥𝑦𝑧 = 32
From (6) & (7)
𝑥 . 𝑥 .
𝑥
2
= 32 , (∵ 𝑦 = 𝑥 & 𝑧 =
𝑥
2
)
𝑥3
= 2 32 = 64
𝑥 = 64
1
3
𝑥 = 43
1
3 = 4
3
3 = 4
Hence 𝑥 = 4 , 𝑦 = 𝑥 = 4 and 𝑧 =
𝑥
2
=
4
2
= 2
The required dimension is 𝑥 = 4, 𝑦 = 4 𝑎𝑛𝑑 𝑧 = 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)

54. Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙
𝐹 = 𝑥𝑦𝑧 + 𝜆 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 − 432 ( by using (1) & (2))
To find the stationary points:
Put
𝜕F
𝜕𝑥
= 0,
𝜕F
𝜕𝑦
= 0 &
𝜕F
𝜕𝑧
= 0
𝜕𝐹
𝜕𝑥
= 0 ⟹ yz + 𝜆 𝑦 + 2z = 0 ⟹ 𝜆 𝑦 + 2𝑧 = −𝑦𝑧 ⟹ 𝜆 =
−𝑦𝑧
𝑦+2𝑧
------ (3)
𝜕F
𝜕𝑦
= 0 ⟹ 𝑥𝑧 + 𝜆(x + 2z) = 0 ⟹ 𝜆(𝑥 + 2𝑧) = −𝑥𝑧 ⟹ 𝜆 =
−𝑥𝑧
𝑥+2𝑧
------ (4)
𝜕F
𝜕𝑧
= 0 ⟹ xy + λ 2x + 2y = 0 ⟹ λ 2x + 2y = −xy ⇒ 𝜆 =
−𝑥𝑦
2𝑥+2𝑦
----- (5)
𝜕𝐹
𝜕𝑥
= 𝑦𝑧 + 𝜆(𝑦 + 2𝑧)
𝜕𝐹
𝜕𝑦
= 𝑥𝑧 + 𝜆(𝑥 + 2𝑧)
𝜕𝐹
𝜕𝑧
= 𝑥𝑦 + 𝜆(2𝑥 + 2𝑦)

55. From (3), (4), (5),
−𝑦𝑧
𝑦+2𝑧
=
−𝑥𝑧
𝑥+2𝑧
=
−𝑥𝑦
2𝑥+2𝑦
Consider
−𝑦𝑧
𝑦+2𝑧
=
−𝑥𝑧
𝑥+2𝑧
⇒
𝑦𝑧
𝑦+2𝑧
=
𝑥𝑧
𝑥+2𝑧
⇒
𝑦
𝑦+2𝑧
=
𝑥
𝑥+2𝑧
⇒ 𝑦 𝑥 + 2𝑧 = 𝑥(𝑦 + 2𝑧)
⇒ 𝑥𝑦 + 2𝑧𝑦 = 𝑥𝑦 + 2𝑧𝑥
⇒ 2𝑧𝑦 = 2𝑧𝑥
⇒ 𝑦 = 𝑥 -------- (6)
Consider
−𝑥𝑧
𝑥+2𝑧
=
−𝑥𝑦
2𝑥+2𝑦
⇒
𝑧
𝑥+2𝑧
=
𝑦
2 𝑥+𝑦
⇒
𝑧
𝑥+2𝑧
=
𝑥
2 𝑥+𝑥
(by using (6))
⇒
𝑧
𝑥+2𝑧
=
𝑥
2 2𝑥
⇒
𝑧
𝑥+2𝑧
=
1
4

56. ⇒
𝑧
𝑥+2𝑧
=
1
4
⇒ 4𝑧 = 𝑥 + 2𝑧
⇒ 4𝑧 − 2𝑧 = 𝑥
⇒ 2𝑧 = 𝑥
⇒ 𝑧 =
𝑥
2
------- (7)
Given surface area 𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 432
Using (6) & (7),
𝑥. 𝑥 + 2𝑥.
𝑥
2
+ 2𝑥.
𝑥
2
= 432
⇒ 𝑥2
+ 𝑥2
+ 𝑥2
= 432
⇒ 3𝑥2
= 432
⇒ 𝑥2
=
432
3
⇒ 𝑥2
= 144
⇒ 𝑥 = ± 144
⇒ 𝑥 = ±12
(since 𝑥 is the length , 𝑥 = −12 is not possible )
Hence 𝑥 = 12 , 𝑦 = 𝑥 = 12 , & 𝑧 =
𝑥
2
=
12
2
= 6
The required dimension for maximum capacity is 𝑥 = 12 , 𝑦 = 12 & 𝑧 = 6

57. Problem 4:
Find the minimum distance from the point (𝟏, 𝟐, 𝟎) to the cone 𝒛 𝟐 = 𝒙 𝟐 + 𝒚 𝟐
Solution:
Wkt, the square of the distance (𝑑2
) from the point (1, 2, 0) to (𝑥, 𝑦, 𝑧) is
𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2
Let 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 --------- (1)
Subject to 𝑧2
= 𝑥2
+ 𝑦2
⇒ 𝑧2
− 𝑥2
− 𝑦2
= 0
Let ∅ 𝑥, 𝑦, 𝑧 = 𝑧2
− 𝑥2
− 𝑦2
---------- (2)
Consider the Lagrangian function 𝐹 = 𝑓 + 𝜆𝜙
𝐹 = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2 + 𝜆(𝑧2 − 𝑥2 − 𝑦2)
𝜕F
𝜕x
= 2 x − 1 − λ(2x)
𝜕F
𝜕y
= 2 y − 2 − λ(2y)
𝜕F
𝜕z
= 2z + λ 2z
= 2z(1 + λ)

58. To find the stationary points:
Put
𝜕F
𝜕𝑥
= 0,
𝜕F
𝜕𝑦
= 0 &
𝜕F
𝜕𝑧
= 0
𝜕𝐹
𝜕𝑥
= 0 ⟹ 2 𝑥 − 1 − 𝜆 2𝑥 = 0 ⟹ 2𝜆𝑥 = 2 𝑥 − 1
⟹ 𝜆 =
(𝑥−1)
𝑥
------ (3)
𝜕F
𝜕𝑦
= 0 ⟹ 2 𝑦 − 2 − 𝜆 2𝑦 = 0 ⟹ 2𝜆𝑦 = 2 𝑦 − 2
⟹ 𝜆 =
(𝑦−2)
𝑦
------ (4)
𝜕F
𝜕𝑧
= 0 ⟹ 2𝑧(1 + 𝜆) = 0 ----- (5)
5 ⟹ 𝑧 = 0 𝑜𝑟 1 + 𝜆 = 0
⟹ 𝑧 = 0 𝑜𝑟 𝜆 = −1
Suppose 𝑧 = 0, Since 𝑥2
+ 𝑦2
= 𝑧2
⇒ 𝑥2
+ 𝑦2
= 0
⇒ 𝑥 = 0 & 𝑦 = 0 , this is not possible.
Hence 𝜆 = −1

59. From (3) , -1=
(𝑥−1)
𝑥
⇒ −𝑥 = 𝑥 − 1 ⇒ −𝑥 − 𝑥 = −1 ⇒ −2𝑥 = −1
⇒ 2𝑥 = 1 ⇒ 𝑥 =
1
2
From(4), −1 =
𝑦−2
𝑦
⇒ −𝑦 = 𝑦 − 2 ⇒ −𝑦 − 𝑦 = −2 ⇒ −2𝑦 = −2
⇒ 𝑦 = 1
Since 𝑥2 + 𝑦2 = 𝑧2
Substitute 𝑥 =
1
2
, 𝑦 = 1
𝑧2 =
1
2
2
+ 1 2 =
1
4
+ 1 =
1+4
4
=
5
4
⇒ 𝑧 = ±
5
4
= ±
5
2
Hence , 𝑥 =
1
2
, 𝑦 = 1 , 𝑧 = ±
5
2
The square distance (𝑑2) is 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧2
=
1
2
− 1
2
+ 1 − 2 2
+
5
2
2
= −
1
2
2
+ −1 2
+
5
4
=
1
4
+ 1 +
5
4
=
1+4+5
4
=
10
4
=
5
2
The minimum distance (𝑑) =
5
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

64. 𝜕 𝑢,𝑣,𝑤
𝜕 𝑥,𝑦,𝑧
=
𝑢 𝑥 𝑢 𝑦 𝑢 𝑧
𝑣 𝑥 𝑣 𝑦 𝑣𝑧
𝑤 𝑥 𝑤 𝑦 𝑤𝑧
=
−
𝑦𝑧
𝑥2
𝑧
𝑥
𝑦
𝑥
𝑧
𝑦
−
𝑧𝑥
𝑦2
𝑥
𝑦
𝑦
𝑧
𝑥
𝑧
−
𝑥𝑦
𝑧2
=
1
𝑥
1
𝑦
1
𝑧
−
𝑦𝑧
𝑥
𝑧 𝑦
𝑧 −
𝑧𝑥
𝑦
𝑥
𝑦 𝑥 −
𝑥𝑦
𝑧
=
1
𝑥𝑦𝑧
−
𝑦𝑧
𝑥
−
𝑧𝑥
𝑦
−
𝑥𝑦
𝑧
− 𝑥2 − 𝑧 𝑧 −
𝑥𝑦
𝑧
− 𝑦𝑥 + 𝑦 𝑧𝑥 − −
𝑧𝑥
𝑦
𝑦
=
1
𝑥𝑦𝑧
[ −
𝑦𝑧
𝑥
𝑥2 − 𝑥2 − 𝑧 −𝑥𝑦 − 𝑥𝑦 + 𝑦 𝑧𝑥 + 𝑧𝑥 ]
=
1
𝑥𝑦𝑧
−
𝑦𝑧
𝑥
0 − 𝑧 −2𝑥𝑦 + 𝑦 2𝑧𝑥 =
1
𝑥𝑦𝑧
2𝑥𝑦𝑧 + 2𝑥𝑦𝑧 =
4𝑥𝑦𝑧
𝑥𝑦𝑧
= 4

65. Problem 2:
If 𝒙 = 𝒖 𝟐 – 𝒗 𝟐 , 𝒚 = 𝟐𝒖𝒗 find the Jacobian of 𝒙, 𝒚 with respect to 𝒖 𝒂𝒏𝒅 𝒗
Solution:
Given 𝑥 = 𝑢2 – 𝑣2 , 𝑦 = 2𝑢𝑣
To find the jacobian of 𝑥, 𝑦 w . r. to 𝑢 & 𝑣
To find
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
𝑥 = 𝑢2 – 𝑣2 𝑦 = 2𝑢𝑣
𝑥 𝑢 = 2𝑢 𝑦𝑢 = 2𝑣
𝑥 𝑣 = −2𝑣 𝑦𝑣 = 2𝑢

66. 𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
=
2𝑢 −2𝑣
−2𝑣 2𝑢
= 4𝑢2
− 4𝑣2
= 4 𝑢2
− 𝑣2
Problem 3:
If 𝒙 = 𝒖𝒗 𝒂𝒏𝒅 𝒚 =
𝒖
𝒗
, then find
𝝏 𝒙,𝒚
𝝏 𝒖,𝒗
Solution:
Given
𝑥 = 𝑢𝑣 𝑎𝑛𝑑 𝑦 =
𝑢
𝑣
𝑥 = 𝑢𝑣
𝑦 =
𝑢
𝑣
𝑥 𝑢 = 𝑣
𝑦𝑢 =
1
𝑣
𝑥 𝑣 = 𝑢
𝑦𝑣 = −
𝑢
𝑣2

67. 𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
=
𝑣 𝑢
1
𝑣
−
𝑢
𝑣2
= 𝑣
𝑢
𝑣2 −
1
𝑣
𝑢 =
𝑢
𝑣
−
𝑢
𝑣
= 0
Problem 4:
If 𝒙 = 𝒓 𝐜𝐨𝐬 𝜽 and 𝒚 = 𝒓 𝐬𝐢𝐧 𝜽 then find
𝝏𝒓
𝝏𝒙
Solution:
Given 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃
𝑥2 + 𝑦2 = 𝑟2 cos2 𝜃 + 𝑟2 sin2 𝜃 = 𝑟2(cos2 𝜃 + sin2 𝜃) = 𝑟2
∴ 𝑟2 = 𝑥2 + 𝑦2
2𝑟
𝜕𝑟
𝜕𝑥
= 2𝑥 ⟹
𝜕𝑟
𝜕𝑥
=
2𝑥
2𝑟
=
𝑥
𝑟
=
𝑥
𝑥2+𝑦2
, ∵ 𝑟 = 𝑥2 + 𝑦2

68. Problem 5:
If 𝒙 = 𝒖 (𝟏 – 𝒗) and 𝒚 = 𝒖𝒗, find
𝝏 𝒖,𝒗
𝝏 𝒙,𝒚
Solution:
Given 𝑥 = 𝑢 (1 – 𝑣) and 𝑦 = 𝑢𝑣
𝑖. 𝑒 𝑥 = 𝑢 − 𝑢𝑣 and 𝑦 = 𝑢𝑣
Wkt
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
=
1
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
Now
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
𝑥 = 𝑢 − 𝑢𝑣 𝑦 = 𝑢𝑣
𝑥 𝑢 = 1 − 𝑣 𝑦𝑢 = 𝑣
𝑥 𝑣 = −𝑢 𝑦𝑣 = 𝑢

69. 𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
=
1 − 𝑣 −𝑢
𝑣 𝑢
= 1 − 𝑣 𝑢 + 𝑢𝑣 = 𝑢 − 𝑢𝑣 + 𝑢𝑣 = 𝑢
Hence
𝜕 𝑢,𝑣
𝜕 𝑥,𝑦
=
1
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
1
𝑢
Problem 6:
If 𝒙 = 𝒓 𝒄𝒐𝒔𝜽 𝒚 = 𝒓 𝐬𝐢𝐧 𝜽 then find
𝝏 𝒓,𝜽
𝝏 𝒙,𝒚
Solution:
Given 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 sin 𝜃
Wkt,
𝜕 𝑥,𝑦
𝜕 𝑟,𝜃
=
𝑥 𝑟 𝑥 𝜃
𝑦𝑟 𝑦 𝜃

70. 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑦 = 𝑟 𝑠𝑖𝑛𝜃
𝑥 𝑟 = 𝑐𝑜𝑠𝜃 𝑦𝑟 = 𝑠𝑖𝑛𝜃
𝑥 𝜃 = −𝑟𝑠𝑖𝑛𝜃 𝑦 𝜃 = 𝑟 𝑐𝑜𝑠𝜃
𝜕 𝑥,𝑦
𝜕 𝑟,𝜃
=
𝑥 𝑟 𝑥 𝜃
𝑦𝑟 𝑦 𝜃
=
𝑐𝑜𝑠𝜃 −𝑟𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃 𝑟 𝑐𝑜𝑠𝜃
= 𝑟 cos2 𝜃 + 𝑟𝑠𝑖𝑛2 𝜃 = 𝑟 (cos2 𝜃 + sin2 𝜃) = 𝑟
𝜕 𝑟,𝜃
𝜕 𝑥,𝑦
=
1
𝜕 𝑥,𝑦
𝜕 𝑟,𝜃
=
1
𝑟

71. Problem 7
If 𝒖 = 𝒙 + 𝒚 & 𝒚 = 𝒖𝒗, then find the jacobian
𝝏 𝒙,𝒚
𝝏 𝒖,𝒗
Solution:
Given 𝑢 = 𝑥 + 𝑦 & 𝑦 = 𝑢𝑣
⟹ 𝑢 = 𝑥 + 𝑢𝑣 ∵ 𝑦 = 𝑢𝑣
⟹ 𝑥 = 𝑢 − 𝑢𝑣 & 𝑦 = 𝑢𝑣
Wkt,
𝜕 𝑥,𝑦
𝜕 𝑢,𝑣
=
𝑥 𝑢 𝑥 𝑣
𝑦 𝑢 𝑦𝑣
𝑥 = 𝑢 − 𝑢𝑣 𝑦 = 𝑢𝑣
𝑥 𝑢 = 1 − 𝑣 𝑦𝑢 = 𝑣
𝑥 𝑣 = −𝑢 𝑦𝑣 = 𝑢

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

74. Now
𝜕𝑓
𝜕𝑥
=
𝜕
𝜕𝑥
𝑥 𝑦 + 𝑦 𝑥 − 𝑐 =
𝜕
𝜕𝑥
𝑥 𝑦 +
𝜕
𝜕𝑥
𝑦 𝑥 −
𝜕
𝜕𝑥
(𝑐)
= 𝑦𝑥 𝑦−1
+ 𝑦 𝑥
log 𝑦 − 0 , (∵
𝑑
𝑑𝑥
𝑥 𝑛
= 𝑛𝑥 𝑛−1
&
𝑑
𝑑𝑥
𝑎 𝑥
= 𝑎 𝑥
log 𝑎 )
𝜕𝑓
𝜕𝑥
= 𝑦𝑥 𝑦−1
+ 𝑦 𝑥
log 𝑦
Now
𝜕𝑓
𝜕𝑦
=
𝜕
𝜕𝑦
𝑥 𝑦 + 𝑦 𝑥 − 𝑐 =
𝜕
𝜕𝑦
𝑥 𝑦 +
𝜕
𝜕𝑦
𝑦 𝑥 −
𝜕
𝜕𝑦
(𝑐)
= 𝑥 𝑦
log 𝑥 + 𝑥 𝑦 𝑥−1
− 0
𝜕𝑓
𝜕𝑦
= 𝑥 𝑦
log 𝑥 + 𝑥 𝑦 𝑥−1
Wkt
𝑑𝑦
𝑑𝑥
= −
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑦
= −
𝑦𝑥 𝑦−1 +𝑦 𝑥 log 𝑦
𝑥 𝑦 log 𝑥+𝑥 𝑦 𝑥−1