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KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABADB. Tech. (I SEM), 2020-21
MATHEMATICS-I (KAS-103T)
MODULE 3(Differential Calculus II)
Syllabus: Partial derivatives, Total derivative, Euler’s Theorem for homogeneous functions,
Taylor, and Maclaurin’s Theorem for function of one and two variables, Maxima and Minima of
functions of several variables, Lagrange’s Method of Multipliers, Jacobians, Approximation of
errors
Course Outcomes:
S.NO. Course Outcome BL
CO 3 Associate the concept of partial differentiation for determining 2,3
maxima, minima, expansion of series and Jacobians
CONTENT
S.NO. TOPIC PAGE NO.
3.1 Partial Derivatives and its applications 2
3.1.1 First order partial derivatives 2
3.1.2 Second and higher order partial derivatives 2
3.2 Total Derivative 6
3.2.1 Composite function and its Derivative 6
3.2.2 Differentiation of an implicit function 12
3.3 Euler’s Theorem for Homogeneous function 15
3.3.1 Homogeneous Function 15
3.3.2 Euler’s Theorem for Homogeneous Function 15
3.3.3 Deduction on Euler’s Theorem for homogeneous function 19
3.4 Taylor and Maclaurin’s Theorem for functions of one and two variables 23
3.4.1 Taylor and Maclaurin’s Theorem for functions of one variable 23
3.4.1.1 Taylor’s theorem for function of one variable 23
3.4.1.2 Maclaurin’s theorem for function of one variable 23
3.4.2.1 Taylor’s Theorem for function of two variables 26
3.4.2.2 Maclaurin’s Theorem for function of two variables 26
3.5 Maxima and minima of function of two variables 32
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, KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
3.5.1 Rule to find the extreme values of a function z = f(x, y) 32
3.5.2 Condition or f(x, y, z) to be maximum or minimum 32
3.6 Lagrange’s method of undetermined multipliers 38
3.6.1 Working procedure of Lagrange’s method of undetermined multipliers 38
3.7 Jacobian 44
3.7.1 Properties of Jacobians 44
3.7.2 Functional Relationship 45
3.8 Approximation of Errors 51
3.9 E- Link for more understanding 57
3.1. Partial Derivatives and its Applications: The concept of partial differentiation is applied for function of two
or more independent variables. Partial differentiation is the process of finding various orders partial derivatives of
given function. Partial derivatives are very useful in determining Maxima and Minima, Jacobian, Series expansion
of function of several variables, Error and Approximation, In solution of wave and heat equations etc.
3.1.1. First order partial derivatives:
If𝑢 = 𝑓(𝑥, 𝑦) be any function of two independent variables 𝑥, 𝑦hen first order partial derivative of u with respect to ′𝑥′
𝜕𝑢
is obtained by differentiating u with respect to ‘x’ treating y and function of ‘y’ as constant and is denoted by
𝜕𝑥
Similarly fist order partial derivative of u with respect to ‘y’ is obtained by differentiating u with respect to ‘y’ treating
𝜕𝑢
‘x’ and functions of ‘x’ as constant and is denoted by𝜕𝑦 .
𝜕𝑢 𝜕𝑢
Example-1:If 𝑢 = 𝑥 2 + 2𝑥𝑦 − 𝑦 2 + 2𝑥 − 3𝑦 + 5 then find 𝑎𝑛𝑑 .
𝜕𝑥 𝜕𝑦
Solution: Here we have given 𝑢 = 𝑥 2 + 2𝑥𝑦 − 𝑦 2 + 2𝑥 − 3𝑦 + 5
𝜕𝑢
Therefore, to get value of 𝜕𝑥 we shall differentiate u with respect to ‘x’ treating ‘y’ as constant
𝝏𝒖
= 𝟐𝒙 + 𝟐𝒚 + 𝟐
𝝏𝒙
𝜕𝑢
Also, to find we shall differentiate u with respect to ‘y’ treating ‘x’ as constant
𝜕𝑦
𝝏𝒖
𝝏𝒚
= 𝟐𝒙 − 𝟐𝒚 − 𝟑.
3.1.2. Second and higher order partial derivatives:
If u f ( x, y ) be any function of two independent variables x, y then second order partial derivative of u with respect
to ′𝑥′ is obtained by differentiating u with respect to ‘x’ twice in succession treating ‘y’ and function of ‘y’ as constant
𝜕2 𝑢
each time and is denoted by 𝜕𝑥 2.
2
, KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Similarly second order partial derivative of u with respect to ‘y’ is obtained by differentiating u with respect to ‘y’
𝜕2 𝑢
twice treating ‘x’ and functions of ‘x’ as constant each time and is denoted by 𝜕𝑦2 .
𝜕𝑢
Here if we differentiate 𝜕𝑦 with respect to ‘x’ treating ‘y’ as constant then we get second order derivative of u denoted
𝜕2 𝑢 𝜕𝑢
as𝜕𝑥𝜕𝑦 . Similarly differentiating 𝜕𝑥 with respect to ‘y’ treating ‘x’ and function of ‘x’ as constant we get second order
𝜕2 𝑢 𝜕2 𝑢 𝜕2 𝑢
derivative of u denoted as 𝜕𝑦𝜕𝑥 . It is interesting to see that 𝜕𝑥𝜕𝑦
= 𝜕𝑦𝜕𝑥 . In the similar way third and other higher
order partial derivatives can be evaluated.
𝜕2 𝑢 𝜕2 𝑢
Example-2: Find all first and second order partial derivatives of𝑢(𝑥, 𝑦) = 𝑥 𝑦 . Hence show that = .
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
Solution: Here we have 𝑢(𝑥, 𝑦) = 𝑥 𝑦 … … … … . (1)
Differentiating equation (1) partially with respect to ‘x’ and ‘y’ we get
𝜕𝑢
= 𝑦𝑥 𝑦−1 … … … … … … … … … …. (2)
𝜕𝑥
𝜕𝑢
= 𝑥 𝑦 𝑙𝑜𝑔𝑥 … … … … … … … … … … … . . (3)
𝜕𝑦
Differentiating equation (2) partially with respect to ‘x’ and ‘y’ we get
𝜕2𝑢
= 𝑦(𝑦 − 1)𝑥 𝑦−2 … … … … … … … … … …. (4)
𝜕𝑥 2
𝝏𝟐 𝒖
= 𝒚𝒙𝒚−𝟏 𝒍𝒐𝒈𝒙 + 𝒙𝒚−𝟏 … … … … … … … … … …. (5)
𝝏𝒚𝝏𝒙
Now differentiating equation (3) partially with respect to ‘y’ and ‘x’ we get
𝜕2𝑢
= 𝑥 𝑦 (𝑙𝑜𝑔𝑥)2 … … … … … … … … … …. (6)
𝜕𝑦 2
𝜕2𝑢 1
= 𝑦𝑥 𝑦−1 𝑙𝑜𝑔𝑥 + 𝑥 𝑦 ( )
𝜕𝑥𝜕𝑦 𝑥
𝝏𝟐 𝒖
= 𝒚𝒙𝒚−𝟏 𝒍𝒐𝒈𝒙 + 𝒙𝒚−𝟏 … … … … … … … . (7)
𝝏𝒙𝝏𝒚
𝜕2 𝑢 𝜕2 𝑢
From equations (5) & (7), it is clear that = .
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
𝜕𝑢 𝜕𝑢 𝜕𝑢
Example-3: If 𝑢 = log(𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧), then prove that 𝑠𝑖𝑛2𝑥 𝜕𝑥 + 𝑠𝑖𝑛2𝑦 𝜕𝑦
+ 𝑠𝑖𝑛2𝑧 𝜕𝑧
=2.
Solution: In this problem, we have 𝑢 = log(𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧) … … … … … … … (1)
Differentiating equation (1) partially with respect to ‘x’ we get
𝜕𝑢 1
= (𝑠𝑒𝑐 2 𝑥) … … … … … … … … . . (2)
𝜕𝑥 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧
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, KIET GROUP OF INSTITUTIONS, DELHI – NCR, GHAZIABAD
Similarly differentiating (1) partially with respect to ‘y’ and ‘z’ we get
𝜕𝑢 1
= (𝑠𝑒𝑐 2 𝑦) … … … … … … … … . . (3)
𝜕𝑦 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧
𝜕𝑢 1
= (𝑠𝑒𝑐 2 𝑧) … … … … … … … … . . (4)
𝜕𝑧 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑦 + 𝑡𝑎𝑛𝑧
Multiplying equations (2), (3) and (4) with 𝑠𝑖𝑛2𝑥, 𝑠𝑖𝑛2𝑦 𝑎𝑛𝑑 𝑠𝑖𝑛2𝑧 respectively and then adding, we get
𝜕𝑢 𝜕𝑢 𝜕𝑢 1
𝑠𝑖𝑛2𝑥 𝜕𝑥
+sin2y𝜕𝑦 + 𝑠𝑖𝑛2𝑧 𝜕𝑧
= 𝑠𝑖𝑛2𝑥 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑥)} +
1 1
𝑠𝑖𝑛2𝑦 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑦)}+ 𝑠𝑖𝑛2𝑧 {𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 (𝑠𝑒𝑐 2 𝑧)}
𝜕𝑢 𝜕𝑢 𝜕𝑢 2𝑡𝑎𝑛𝑥 2𝑡𝑎𝑛𝑦 2𝑡𝑎𝑛𝑧
𝑠𝑖𝑛2𝑥 +sin2y + 𝑠𝑖𝑛2𝑧 = + + +
𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 𝑡𝑎𝑛𝑥+𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧
𝝏𝒖 𝝏𝒖 𝝏𝒖
𝒔𝒊𝒏𝟐𝒙 + 𝐬𝐢𝐧𝟐𝐲 + 𝒔𝒊𝒏𝟐𝒛 = 𝟐.
𝝏𝒙 𝝏𝒚 𝝏𝒛
𝜕𝑥 𝜕𝑦 𝜕𝑟 𝜕𝜃
Example-4: If 𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃, then find (a) ( 𝜕𝑟 ) 𝑎𝑛𝑑 (𝜕𝜃 ) , and (b) (𝜕𝑥 ) 𝑎𝑛𝑑 (𝜕𝑦 ) .
𝜃 𝑟 𝑦 𝑥
Solution: Here we have
𝑥 = 𝑟𝑐𝑜𝑠𝜃 … … … … … … … … . . (1),
and 𝑦 = 𝑟𝑠𝑖𝑛𝜃 … … … … … … … … . . (2)
Differentiating equation (1) partially with respect to ‘r’ treating ‘𝜃′ as constant, we get
𝝏𝒙
( ) = 𝒄𝒐𝒔𝜽
𝝏𝒓 𝜽
Similarly differentiating (2) partially with respect to ‘𝜃′ treating ‘r’ as constant we get
𝝏𝒚
( ) = 𝒓𝒄𝒐𝒔𝜽
𝝏𝜽 𝒓
Again, from equations (1) and (2), we get
𝑟 2 = 𝑥 2 + 𝑦 2 … … … … … … … … … … … … … … (3)
𝑦
and, 𝜃 = tan−1 (𝑥 ) … … … … … … … … … … … … (4)
Differentiating equation (3) partially with respect to ‘x’ treating ‘y’ as constant, we get
𝜕𝑟 𝝏𝒓 𝒙
2𝑟 (𝜕𝑥 ) = 2𝑥 𝒐𝒓 (𝝏𝒙) = 𝒓
𝑦 𝒚
Similarly Differentiating equation (3) partially with respect to ‘y’ treating ‘x’ as constant, we get
4
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