University of Plymouth

213451

Intermediate Mathematics

Introduction to Partial

Differentiation

R Horan & M LavelleTheaim of this document is to provide a short, self assessment programme for students who wish to acquire a basic understanding of partial differentiation.Copyright c?2004rhoran@plymouth.ac.uk,mlavelle@plymouth.ac.uk

Last Revision Date: May 25, 2005Version 1.0

Table of Contents

1.Partial Differentiation (Introduction)

2.The Rules of Partial Differentiation

3.Higher Order Partial Derivatives

4.Quiz on Partial Derivatives

Solutions to Exercises

Solutions to Quizzes

The full range of these packages and some instructions, should they be required, can be obtained from our web pageMathematics Support Materials. Section 1: Partial Differentiation (Introduction) 3

1. Partial Differentiation (Introduction)

In the package onintroductory differentiation, rates of change of functions were shown to be measured by thederivative. Many applications require functions with more than one variable: the ideal gas law, for example, ispV=kTwherepis the pressure,Vthe volume,Tthe absolute temperature of the gas, andkis a constant. Rearranging this equation as p=kTV shows thatpis a function ofTandV. If one of the variables, sayT, is kept fixed andVchanges, then the derivative ofpwith respect to Vmeasures therate of changeofpressurewith respect tovolume. In this case, it is calledthe partial derivative ofpwith respect toVand written as∂p ∂V Section 1: Partial Differentiation (Introduction) 4 Example 1Ifp=kTV, find the partial derivatives ofp: (a)with respect toT,(b)with respect toV.

Solution

(a)This part of the example proceeds as follows: p=kT V, ?∂p ∂T=k V, whereVis treated as a constant for this calculation. (b)For this part,Tis treated as a constant. Thus p=kT

1V=kTV

-1, ∂p∂V=-kTV-2=- kTV 2. Section 1: Partial Differentiation (Introduction) 5 The symbol∂is used whenever a function with more than one variable is being differentiated but the techniques ofpartialdifferentiation are exactly the same as for (ordinary) differentiation.

Example 2Find∂z

∂xand∂z ∂yfor the functionz=x2y3.

Solutionz=x

2y3?∂z

∂x=2xy3, and∂z ∂y=x 23y2,
=3x2y2.For the first party

3is treated as

a constant and the derivative ofx

2with respect toxis2x.

For the second partx

2is treated

as a constant and the derivative ofy

3with respect toyis3y2.

Exercise 1.Find∂z

∂xand∂z ∂yfor each of the following functions. (Click on thegreenletters for solutions.) (a)z=x2y4,(b)z= (x4+x2)y3,(c)z=y12 sin(x).

Section 2: The Rules of Partial Differentiation 6

2. The Rules of Partial Differentiation

Sincepartial differentiationis essentially the same asordinary differ- entiation, theproduct,quotientandchainrules may be applied.

Example 3Find∂z

∂xfor each of the following functions. (a)z=xycos(xy),(b)z=x-yx+y,(c)z= (3x+y)2.

Solution

(a)Herez=uv, whereu=xyandv= cos(xy)so theproduct rule applies (see the package on theProduct and Quotient Rules). u=xyandv=cos(xy)?∂u ∂x=yand∂v ∂x=-ysin(xy).

Thus∂z

∂x=∂u ∂x v+u∂v∂x=ycos(xy)-xy2sin(xy).

Section 2: The Rules of Partial Differentiation 7

(b)Herez=u/v, whereu=x-yandv=x+yso thequotient rule applies (see the package on theProduct and Quotient Rules). u=x-yandv=x+y?∂u ∂x=1and∂v ∂x=1.

Thus∂z

∂x=v ∂u∂x -u∂v∂x v

2=(x+y)-(x-y)(x+y)2=2y(x+y)2.

(c)In this casez= (3x+y)2soz=u2whereu= 3x+y, and the chain ruleapplies (see the package on theChain Rule). z=u

2andu=3x+y?∂z

∂u=2uand∂u ∂x=3.

Thus∂z

∂x=∂z ∂u ∂u∂x=2(3x+y)3=6(3x+y).

Section 2: The Rules of Partial Differentiation 8

Exercise 2.Find∂z

∂xand∂z ∂yfor each of the following functions. (Click on thegreenletters for solutions.) (a)z= (x2+3x)sin(y),(b)z=cos(x)y

5,(c)z= ln(xy),

(d)z= sin(x)cos(xy),(e)z= e(x2+y2),(f)z= sin(x2+y). QuizIfz= cos(xy), which of the following statements is true? (a)∂z ∂x=∂z ∂y,(b)∂z ∂x=1 x ∂z∂y, (c)1 y ∂z∂x=∂z

Intermediate Mathematics

Introduction to Partial

Differentiation

R Horan & M LavelleTheaim of this document is to provide a short, self assessment programme for students who wish to acquire a basic understanding of partial differentiation.Copyright c?2004rhoran@plymouth.ac.uk,mlavelle@plymouth.ac.uk

Last Revision Date: May 25, 2005Version 1.0

Table of Contents

1.Partial Differentiation (Introduction)

2.The Rules of Partial Differentiation

3.Higher Order Partial Derivatives

4.Quiz on Partial Derivatives

Solutions to Exercises

Solutions to Quizzes

The full range of these packages and some instructions, should they be required, can be obtained from our web pageMathematics Support Materials. Section 1: Partial Differentiation (Introduction) 3

1. Partial Differentiation (Introduction)

In the package onintroductory differentiation, rates of change of functions were shown to be measured by thederivative. Many applications require functions with more than one variable: the ideal gas law, for example, ispV=kTwherepis the pressure,Vthe volume,Tthe absolute temperature of the gas, andkis a constant. Rearranging this equation as p=kTV shows thatpis a function ofTandV. If one of the variables, sayT, is kept fixed andVchanges, then the derivative ofpwith respect to Vmeasures therate of changeofpressurewith respect tovolume. In this case, it is calledthe partial derivative ofpwith respect toVand written as∂p ∂V Section 1: Partial Differentiation (Introduction) 4 Example 1Ifp=kTV, find the partial derivatives ofp: (a)with respect toT,(b)with respect toV.

Solution

(a)This part of the example proceeds as follows: p=kT V, ?∂p ∂T=k V, whereVis treated as a constant for this calculation. (b)For this part,Tis treated as a constant. Thus p=kT

1V=kTV

-1, ∂p∂V=-kTV-2=- kTV 2. Section 1: Partial Differentiation (Introduction) 5 The symbol∂is used whenever a function with more than one variable is being differentiated but the techniques ofpartialdifferentiation are exactly the same as for (ordinary) differentiation.

Example 2Find∂z

∂xand∂z ∂yfor the functionz=x2y3.

Solutionz=x

2y3?∂z

∂x=2xy3, and∂z ∂y=x 23y2,
=3x2y2.For the first party

3is treated as

a constant and the derivative ofx

2with respect toxis2x.

For the second partx

2is treated

as a constant and the derivative ofy

3with respect toyis3y2.

Exercise 1.Find∂z

∂xand∂z ∂yfor each of the following functions. (Click on thegreenletters for solutions.) (a)z=x2y4,(b)z= (x4+x2)y3,(c)z=y12 sin(x).

Section 2: The Rules of Partial Differentiation 6

2. The Rules of Partial Differentiation

Sincepartial differentiationis essentially the same asordinary differ- entiation, theproduct,quotientandchainrules may be applied.

Example 3Find∂z

∂xfor each of the following functions. (a)z=xycos(xy),(b)z=x-yx+y,(c)z= (3x+y)2.

Solution

(a)Herez=uv, whereu=xyandv= cos(xy)so theproduct rule applies (see the package on theProduct and Quotient Rules). u=xyandv=cos(xy)?∂u ∂x=yand∂v ∂x=-ysin(xy).

Thus∂z

∂x=∂u ∂x v+u∂v∂x=ycos(xy)-xy2sin(xy).

Section 2: The Rules of Partial Differentiation 7

(b)Herez=u/v, whereu=x-yandv=x+yso thequotient rule applies (see the package on theProduct and Quotient Rules). u=x-yandv=x+y?∂u ∂x=1and∂v ∂x=1.

Thus∂z

∂x=v ∂u∂x -u∂v∂x v

2=(x+y)-(x-y)(x+y)2=2y(x+y)2.

(c)In this casez= (3x+y)2soz=u2whereu= 3x+y, and the chain ruleapplies (see the package on theChain Rule). z=u

2andu=3x+y?∂z

∂u=2uand∂u ∂x=3.

Thus∂z

∂x=∂z ∂u ∂u∂x=2(3x+y)3=6(3x+y).

Section 2: The Rules of Partial Differentiation 8

Exercise 2.Find∂z

∂xand∂z ∂yfor each of the following functions. (Click on thegreenletters for solutions.) (a)z= (x2+3x)sin(y),(b)z=cos(x)y

5,(c)z= ln(xy),

(d)z= sin(x)cos(xy),(e)z= e(x2+y2),(f)z= sin(x2+y). QuizIfz= cos(xy), which of the following statements is true? (a)∂z ∂x=∂z ∂y,(b)∂z ∂x=1 x ∂z∂y, (c)1 y ∂z∂x=∂z