Chapter 2 - Calculus I - MATH 203

1

Calculus I

MATH 203

Course Description :

Chapter 2 : The Derivative

Chapter 3 : Topics in Differentiation

Chapter 4 : The Derivative in Graphing and Applications

Chapter 5 : Integration

Text Book :

Howard Anton , Irl Bivens , Stephen Davis , '' Calculus '' Early

Transcendentals , John Wiley & Sons , Inc., 10 th

Edition , 2012 .

Grade Distribution :

Home Works 7 %

Participation 3 %

Quizzes 10 %

First Periodic Exam 20 %

Second Periodic Exam 20 %

Final Exam 40 %

2

Kingdom of Saudi Arabia

Ministry of Higher Education

Taibah University

Department of Mathematics

Syllabus of Calculus I

MATH 203

First Semester

1435/1436 H

Week Section Topics Items Examples Exercises

1 Review of Introduction to Mathematics MATH 101

2 2.1

2.2

Tangent Lines and

Rates of Change

The Derivative

Function

2.1.1 Def., For.(2)

2.2.1 Def. , For.(3),

2.2.2 Def., 2.2.3 Th.,

For.(10), For.(12)

1,2,3,4

1,2(a),4,6(a)

15,16

9,15

3 2.3

2.4

Introduction to

Techniques of

Differentiation

The Product and

Quotient Rules

2.3.1 Th., For.(2),

2.3.2 Th., 2.3.3 Th.,

2.3.4 Th., 2.3.5 Th.,

For.(11), For.(12)

2.4.1 Th., 2.4.2 Th.,

Tab. 2.4.1

1,2,3,4,5,6,9

1,2,3

2,10,41(a)

4,9,11

4 2.5

2.6

Derivatives of

Trigonometric

Functions

The Chain Rule

For.(3), For.(4),

For.(5-6), For.(7-8)

2.6.1 Th., For.(2),

For.(3), Tab.2.6.1

1,2,3

1,3,4,5

1,8,21

8,19,21

5 3.1

3.2

Implicit

Differentiation

Derivative of

Logarithmic

Functions

3.11 Def.

For.(1), For.(2),

For.(3), For.(4-5),

For.(6), For.(8)

2,3,4,5(a,b)

2,3,4,5

4,13,17

13,27,35

6 3.3

Derivatives of

Exponential and

Inverse

Trigonometric

For.(2), For.(3),

3.3.1 Th., For.(5),

For.(6), For.(7-8),

For.(9-10), For.(11-

3,4,5

21,32,47

3

Functions 12), For.(13-14),

7 3.6

L'Hopital's Rule;

Indeterminate Forms

3.6.1 Th., Applying

L'Hopital's Rule

P.g.(220), 3.6.2 Th.,

For.(5-6)

1,2,3,4,5 7,13, 21,36

8 4.1

4.2

Analysis of Function

I: Increase,

Decrease, and

Concavity


II: Relative Extrema

; Graphing

Polynomials

4.1.1 Def., 4.1.2 Th.,

4.1.3 Def., 4.1.4 Th.,

4.1.5 Def.

4.2.1 Def., 4.2.2 Th.,

4.2.3 Th., 4.2.4 Th.

1,2,4,5

1,2,4,5

15,19

7,34

9 4.3


III: Rational

Functions, Cusps

and Vertical

Tangents

Graphing a Rational

Function

f(x)=P(x)/Q(x) if

Q(x) and Q(x) have

no Common Factors.

P.g.(255)

1,3

1,13

10 4.4

4.5

Absolute Maxima

and Minima

Applied Maximum

and Minimum

Problems

4.4.1 Def., 4.4.2 Th.,

4.4.3 Th., Procedure.

P.g. (258), 4.4.4 Th.

A procedure for

Solving Applied

Maximum and

Minimum Problems.

P.g.(276)

1,4

1,2

7,24

3

11 4.8

Roll's Theorem;

Mean-Value

Theorem

4.8.1 Th., 4.8.2 Th.,

4.1.2 Th., 4.8.3 Th.

1,4 1,5

12 5.1

5.2

An Overview of the

Area Problem

The Indefinite

Integral

5.1.1 The Area

Problem

5.2.1 Def., 5.2.2 Th.,

For.(3) , Tab.5.2.1,

5.2.3 Th., For.(4),

1(a,b,c)

1,2,3,4

15

11,21,23

4

For.(5), For.(6),

For.(7), For.(10)

13 5.3

5.4

Integration by

Substitution

The Definition of

Area as a Limit;

Sigma Notation

For.(2), For.(3),

Guidelines for

u-substitution

P.g.(334), For.(5),

For.(6), For.(7)

5.4.1 Th., 5.4.2 Th.,

5.4.3 Def.

1,2,3,4,5,6,7,

8,9,10,14

4

17,31,35

39

14 5.5

5.6

The Definite Integral

The Fundamental

Theorem of Calculus

5.5.1 Def., 5.5.2 Th.,

5.5.3 Def., 5.5.4 Th.,

5.5.5 Def., 5.5.6 Th.,

5.5.7 Def., 5.5.8 Th.

5.6.1 Th., 5.6.2 Th.,

5.6.3 Th.

1,4

1,2,3,4,5,6

13(a),25

13,19,25

15 5.8

5.9

5.10

Average Value of

a Function and its

Applications

Evaluating Definite

Integrals by

Substitution

Logarithmic and

Other Functions

Defined by Integrals

5.8.1 Def.

5.9.1 Th.

5.10.1 Def.,

5.10.2Th., 5.10.3 Th.,

5.10.4Def.,5.10.5Th.,

5.10.6Def.,5.10.7Th.,

5.10.8Th.,5.10.9 Def.

2

1,2,3

-

5

5,11,16

-

Symbols :

Def. Definition , Th. Theorem , For. Formula , Tab. Table .

5

Notes :

1. First periodic exam through week No. 7.

2. Second periodic exam through week No. 12.

3. Simple calculators are allowed (e.g. Casio fx-82 MS or less). Advanced scientific

calculators which graph functions or have programs are not allowed (e.g. Casio fx-991

ES+ , TI 83 or more).

6

CHAPTER (2)

THE DERIVATIVE

2.1 TANGENT LINES AND RATES OF CHANGE :

Page (131)

Tangent Lines :

* Referring to Figure 2.1.1 the slope PQm of the secant line

through 0 0P x ,f x and Q x ,f x on the curve of y xf is

0PQ

0

f f

xm

xx

x

.

Figure 2.1.1

* If we let x approach 0x , then the point Q will move along the curve and approach the point P. If the secant line through P

and Q approaches a limiting position as 0x x , then we will

regard that position to be the position of the tangent line at P.

7

2.1.1 Definition : Page (132)

Suppose that 0x is in the domain of the function f. The tangent

line to the curve y xf at the point 0 0P x ,f x is the line with equation

ta 0n0y mf x xx where

0xan

0

0t

xm

f f xxl

xim

x

1

provided the limit exists. For simplicity , we will also call this

the tangent line to y xf at 0x .

Example 1 : Page (132)

Use definition 2.1.1 to find an equation for the tangent line to the

parabola 2y x at the point P 1 ,1 . Solution

* Applying Formula 1 with 2f x x and 0x 1 , we have

0xan

0

0t

xm

f f xxl

xim

x

x 1

f 1

1

xlim

x

f

x

2

1

x

x

1i

1l m

Remember that :

* 2 2a b a b a b

1x

1 1xl m

x 1

xi

8

1x

lim x 1

2 .

* Thus , the tangent line to 2y x at 1 ,1 has equation ta 0n0y mf x xx

y x1 12

y x1 22

or equivalently y 2 x 1 .

* There is an alternative way of expressing Formula 1 that is commonly used. If we let h denote the difference

0h xx

then the statement that 0x x is equivalent to the statement

h 0 , so we can rewrite 1 in terms of 0x and h as

tan

0h

0 0xlim

f

h

xm

h f

2

* Figure 2.1.2 shows how Formula 2 express the slope of the tangent line as a limit of slopes of secant lines.

Figure 2.1.2

9


Compute the slope in Example 1 using Formula 2 . Solution

* Applying Formula 2 with 2f x x and 0x 1 , we obtain

tan0h

0 0xlim

f

h

xm

h f

h 0

f h

h

1l

f1im

h

2 2

0

1li

h

1hm

Remember that :

* 2 2 2

a b a 2ab b

0h

21 h h

hl m

2i

1

0h

22 h h

hlim

h 0 0h

h hh

hlim lim

22

2

which agrees with the slope found in Example 1.


Find an equation for the tangent line to the curve y 2 / x at

the point 2 ,1 on this curve. Solution

* First , we will find the slope of the tangent line by Formula 2 with f x x2 / and 0x 2 . This yields

tan0h

0 0xlim

f

h

xm

h f

10

h 0

f h

h

2l

f2im

h 0

21

2lih

m h

Remember that :

* a a b

1b b

h 0

h2 2

2 hl

him

0h him

2 hl

h

h 0

1

2 hlim

1

2 .

* Thus , an equation of the tangent line to y x2 / at 2 ,1 is ta 0n0y mf x xx

1

y 12

x 2

1

y 12

x 1

or equivalently 1

y x 22

.

11

Figure 2.1.3


Find the slopes of the tangent lines to the curve y x at

0x 1 , 0x 4 , and 0x 9 .

Solution

* We could compute each of these slopes separately , but it will be more efficient to find the slope for a general value of 0x

and substitute the specific numerical values. Proceeding in

this way we obtain

tan0h

0 0xlim

f

h

xm

h f

h

0 0

0

x h x

hlim

0 0 0 0

0 00h

lim .x x x x

x

h h

h xh

12

Remember that :

* a b a b a b

0

00 0

h

0h

h hl

x x

xim

x

0 0 0h

h

h xl m

xi

h

0

0 0h

1

x hlim

x

0

1

2 x .

* The slopes at 0x ,1 , 4 and 9 can now be obtained by substituting these values into our general formula . Thus ,

0

1slop t x

1e 1a :

2

1

2 .

0

1slop t x

4e 4a :

2

1

4 .

0

1slop t x

9e 9a :

2

1

6 .

Figure 2.1.4

13

EXERCISES SET 2.1: (Home Work) Page (140)

15 , 16 A function y xf and an x-value 0x are given. (a) Find a formula for the slope of the tangent line to the graph

of f at a general point 0x x .

(b) Use the formula obtained in part (a) to find the slope of the

tangent line for the given value of 0x .

15. 2 01 xx xf ; 1

16. 2 0x x x ; xf 3 2 2

14

2.2 THE DERIVATIVE FUNCTION : Page (143)

Definition of the Derivative Function :

* In the last section we showed that if the limit

00

0

hl

hx xim

h

f f

exists , then it can be interpreted as the slope of the tangent line

tanm to the curve y xf at 0x x .

* This limit is so important that it has a special notation :

0h

0 00

fli

h

hf '

x xm

fx

.

1

2.2.1 Definition : Page (143)

The function f ' defined by the formula

0h

f ff

x h

h

x' x lim

2

is called the derivative of f with respect to x . The domain of

f ' consists of all x in the domain of f for which the limit exists.


Find the derivative with respect to x of 2f x x , and use it to

find the equation of the tangent line to 2y x at x 2 .

Solution

* It follows from 2 that

0h

f ff

x h

h

x' x lim

15

h

2 2

0

x xli

h

hm

Remember that :

* 2 2 2

a b a 2ab b

0

2 2 2

h

2x xhl

h

xim

h

h

2

0

2 hxlim

h

h

h 0 0h

h hh

h

xlim lim

22 x

2 x .

* Then the slope of the tangent line to 2y x at x 2 is f ' 2 42 2 .

* Since 2

y 42 if x 2 , thus an equation of the tangent

line to 2y x at 2 ,4 is 0 0 0y f f 'x x xx

y x4 24

y x4 84

or equivalently y 4 x 4 .

Figure 2.2.1

16

Finding an Equation for the Tangent Line to y xf at

0x x : Page (144)

Step 1. Evaluate 0f x ; the point of tangency is 0 0x ,f x .

Step 2. Find f ' x and evaluate 0f ' x , which is the slope m of the line.

Step 3. Substitute the value of the slope m and the point

0 0x ,f x into the point-slope form of the line

0 0 0y f f 'x x xx

or , equivalently ,

0 0 0y f f 'x x xx 3


(a) Find the derivative with respect to x of 3f x x x . Solution

* It follows from 2 that

0h

f ff

x h

h

x' x lim

0

3 3

h

h hx x x xi

hl m

Remember that :

* 3 3 2 2 3

a b a 3a b 3ab b

h

3 2 2 3 3

0

h h h h3x x x x x xlim

h

3

17

2 2 3

0h

h3 3x xm

h

hl

hi

h

0

2 2

h

3 3x xh h h 1

hlim

2

h

2

0lim x x h3 3 1h

23 x 1 .


(a) Find the derivative with respect to x of f x x .

(b) Find the slope of the tangent line to f x x at x 9 .

(c) Find the limits of f ' x as x 0 and as x , and

explain what those limits say about the graph of f .

Solution

(a) From Example 4 of section 2.1

0h

f ff

x h

h

x' x lim

0h

x xi

h

hl m

h 0

h h

h

x x x xlim .

x xh

Remember that :

* a b a b a b

h 0h

h h

x xlim

x x

18

h 0

h

hlim

x h x

h 0

1lim

x h x

1

2 x .

(b) The slope of the tangent line at x 9 is

f ' 99

1

2

1

6 .

(c) The graphs of f x x and f ' x x are shown in Figure 2.2.5. Observe that f ' x 0 if x 0 , which

means that all tangent lines to the graph of y x have

positive slope at all points in this interval.

Figure 2.2.5

* Since 0x

1lim

2 x and

x

1lim

2 x 0

the graph of f becomes more and more vertical as x 0 and more and more horizontal as x .

19

Differentiability : Page (146)

2.2.2 Definition :

A function f is said to be differentiable at 0x if the limit

0h

0 00

fli

h

hf '

x xm

fx

5

exists. If f is differentiable at each point of the open interval

a ,b , then we say that it is differentiable on a ,b , and similarly for open intervals of the form a , , ,b , and , . In the last case we say that f is differentiable everywhere.

* Figure 2.2.6 illustrates two common ways in which a function that is continuous at 0x can fail to be differentiable at 0x .

Figure 2.2.6

* At a corner point , the slopes of the secant lines have different limits from the left and from the right , and hence the two-

sided limit that defines the derivative does not exist (Figure

2.2.7).

20

Figure 2.2.7

* At a point of vertical tangency the slopes of the secant lines approach or from the left and from the right (Figure

2.2.8) , so again the limit that defines the derivative does not

exist.

Figure 2.2.8

21


The graph of y x in Figure 2.2.10 has a corner at x 0 ,

which implies that f x x is not differentiable at x 0 . (a) Prove that f x x is not differentiable at x 0 by

showing that the limit in Definition 2.2.2 does not exist at

x 0 .

Figure 2.2.10

Solution

Remember that :

* x , x 0

xx , x 0

(a) From Formula (5) with 0x 0 , the value of f ' 0 , if it were exist , would be given by

0h

0 00

fli

h

hf '

x xm

fx

0h

f ff

0 h

h

0' 0 lim

0h 0h

h h

hlim li

h

0m

* But h, 0

,

1

h h 0

h

1

22

so that 0h

l mh

ih

1 and 0h

l mh

ih

1 .

* Since these one-sided limits are not equal , the two-sided limit in 5 does not exist , and hence

f is not differentiable at x 0 .

The Relationship Between Differentiability and Continuity :

Page (148)

2.2.3 Theorem :

If a function f is differentiable at 0x , then f is continuous at

0x .

Other Derivative Notations : Page (150)

* When the independent variable is x , the differentiation operation is also commonly denoted by

d

x xdx

f ' f or

xf ' fx D x .

* In the case where there is a dependent variable y xf , the derivative is also commonly denoted by

f ' x xy ' or

f 'd

xx

y

d .

* With the above notations , the value of the derivative at a point

0x can be expressed as

0 0

xx x0 0x x

dx , D x ,

df ' f f ' f

xx x

0

0x

0 0x

f ' y ' f 'dy

,d

x xx

x

23

* It is common to regard the variable h in the derivative formula

0h

f ff

x h

h

x' x lim

9

as an increment x in x and write 9 as

0x

f x ff '

xx l m

x

xi

10

* Moreover , if y xf , then the numerator in 10 can be regarded as the increment

xy f fx x 11 in which case

x 0 x 0

x xdlim l

xim

dx

f

x x

fy y

12

* The geometric interpretation of x and y are shown in Figure 2.2.14.

Figure 2.2.14

24


9. Use Definition 2.2.1 to find f ' x , and then find the tangent

line to the graph of y xf at x a if 2f 2 1x x ; a .

15. Use Formula (12) to find dy / dx if yx

1 .

25

2.3 INTRODUCTION TO TECHNIQUES OF

DIFFERENTIATION : Page (155)

Derivative of a Constant :

2.3.1 Theorem :

The derivative of a constant function is 0 ; if c is any real

number , then

d

dxc 0 . 1


* d

dx1 0 .

* d

dx3 0 .

* d

dx 0 .

* 2d

dx 0 .

Remember that :

* d

c 0dx

Derivative of Power Functions : Page (156)

2.3.2 Theorem (The Power Rule):

If n is a positive integer , then

n n 1dx xn

dx

. 5

26


* 4d

xdx

3

4 x .

* 5d

xdx

4

5 x .

* 12d

tdt

11

12 t .

Remember that :

* n n 1d

x n xdx

2.3.3 Theorem (Extended Power Rule): Page (157)

If r is any real number , then

r r 1dx xr

dx

. 7


* d

xdx

1

x .

* 1 1 1 2d d

x x xdx x d

1x

1 2

1

x .

* 100

100 101d dw w

dw dw0

w

11 0

101

100

w .

* 4 / 5 1 1 / 51 / 5

4 / 5 4 4 4

5

dx

5x

x5x

dx

5

4

5 x .

27

* 1 / 2 1 / 2d d

x x xdx dx

1

2

1

2 x .

* 3 1 / 3 2 / 32 / 3

d 1dx x x

dx d

1

x3x 3

3 2

1

3 x .

Remember that :

* r r 1d

x r xdx

* d 1

xdx 2 x

Derivative of a Constant Times a Function : Page (157)

2.3.4 Theorem (Constant Multiple Rule):

If f is differentiable at x and c is any real number , then c f is

also differentiable at x and

d d

x xfcx

fd dx

c . 8


* 8 8 7d d

x x xdx d

4 84x

4 7

32 x .

* 12 12d d

x xdx dx

1 11

12 x .

* 1 2d d

x xdx x dx

2x

.

28

Remember that :

* d d

c f x c f xdx dx

Derivatives of Sums and Difference : Page (158)

2.3.2 Theorem (Sum and Difference Rules):

If f and g are differentiable at x , then so are f g and f g

and

f fd d d

x x x xdx dx d

g gx

. 9

f fd d d

x x x xdx dx d

g gx

. 10


* 6 9 6 92 2d d d

x x x xdx dx dx

5 10

2 6 x x9

5 1012 9x x

.

* d x x d

x2

1 2dx dxx

d d

xdx d

11

x2 2

x0

2

x

1 .

29

Remember that :

* d d d

f x g x f x g xdx dx dx

* a b a b

c c c

* d 1

xdx 2 x


Find dy / dx if 8 5

y 3 x 2 x 6 x 1 .

Solution 8 5

x x 1x2y 3 6

8 5

3 2y

6 1d d

x x xdx dx

Remember that :

* d d d

f x g x f x g xdx dx dx

8 5d d d d

x x xdx dx dx d

3 2 6 1x

Remember that :

* d d

c f x c f xdx dx

* r r 1d

x r xdx

* d

c 0dx

7 424 10 6x x .

30

Higher Derivatives : Page (159)

* The derivative f ' of a function f is itself a function and hence may have a derivative of its own. If f ' is differentiable , then

its derivative is denoted by f '' and is called second derivative

of f . As long as we have differentiability , we can continue

the process of differentiating to obtain third , fourth , fifth ,

and even higher derivatives of f . These successive

derivatives are denoted by

4 5 4, , ,f ' f '' f ' ' f ''' f '' ' f f , , ..''' ' f f .' * If y xf , then successive derivatives can also be denoted by

4 5y ' y '' y ''', , , ,y y , ...

* Other common notations are

d d

xdx d

yy ' f

x

2 2

2 2

d d d dx x

dx dxdx df

x

yy '' f

3 2 3

3 2 3

d d d dx x

dxdx d

yy ''' f f

x dx

* The number of times that f is differentiated is called the order of the derivative. A general nth order derivative can be

denoted by

n n

n

n n

d dx x

dx dx

yf f 11

31

and the value of a general nth order derivative at a specific

point 0x x can be denoted by

0 0x

n nn

n

x x

n0

x

xd d

xdx dx

yf f

12


* If 4 3 2x x x2 2xf 4x3 , then

3 2x x x x2f ' 1 6 2 4

2

x xf ' 36 x 2' 12

7f 2'' 1x' x 2

4f x 72

5f x 0

nf x 50 n


2. Find dy / dx if 12

y 3 x .

10. Find f ' x if

f x1

xx

.

41.(a) Find 2 2

d y / dx if 3 2

7 x xy 5 x .

32

2.4 THE PRODUCT AND QUOTIENT RULES :

Page (163)

Derivative of a Product :

2.4.1 Theorem (The Product Rule):

If f and g are differentiable at x , then so is the product f . g ,

and

g gd d d

x . x x . x x . xdx

f f fdx d

gx

1

* Formula 1 can also be expressed as

'

g gf f '. f. 'g . .


Find dy / dx if 2 3y 4 x 1 7 x x . Solution

2 34 1x xy 7 x Method 1.: (Using the Product Rule)

2 3d d

x x xd

4 1 7x d

y

x

Remember that :

* d d d

f x .g x f x . g x g x . f xdx dx dx

2 3 3 24 1 7 7 4d d

x . x x x x . xdx dx

1

2 2 3x x4 1 21 1 x 8 x7 x

4 240 x1 9 1x .

33

Method 2.: (Multiplying First)

2 35 3 3 5 3

x x x

x x x x x x

4 1 7

28 4 7 28 3

y

x

* Thus , 5 328d d

x x xdx d

y3

x

4 240 x1 9 1x .

which agrees with the result obtained using the product rule.


Find ds / dt if s 1 t t . Solution

s t1 t

* Applying the product rule yields

1d d

t tdt dt

s

Remember that :

* d d d


d d

t . t t . tdt dt

1 1

Remember that :

* d 1

xdx 2 x

1

1 12

t tt

t t

tt2 t

1 2

2

1

t

3t

2

.

34

Derivative of a Quotient : Page (165)

2.4.1 Theorem (The Product Rule):

If f and g are differentiable at x and if g x 0 , then f / g is differentiable at x and

2

d dx . x x . x

xd dx dx

dx

f ff

g x

g g

g x

. 2

* Formula 2 can also be expressed as '

2

gf f ' f '.

g g

. g

.


Find y ' x if 3 2

x 2 x 1y

x 5

.

Solution 3 2

x x

x

2y

1

5

3 2d d x x2 1

dx dx x 5

y

Remember that :

*

2

d dg x . f x f x . g x

f xd dx dx

dx g x g x

35

3 2 3

2

2d dx . x x x x . x

dx dx

x

5 2 1 2 1 5

5

2 3

2

2x x x x5 3 4 2

x

1

5

x 1

2

2

3 3 23 19 2x x x x0 1

x

2

5

x

2

2

32 17 2x x x0

5x

1

.

Table 2.4.1 Rules for Differentiation: Page (167)

d

dxc 0 c f ' 'c f

f ' fg g'' f ' fg g''

g g'.f ' .f f 'g . '

2

gf f '. g'

g

f .

g

2

'g'

g g

1

r r 1dx xr

dx

36


4. Compute the derivative of the given function

2x x x x1f 1 by (a) multiplying and then differentiating and

(b) using the product rule.

Verify that (a) and (b) yield the same result.

9. Find f ' x if 2x x x2 4f 2 x .

11. Find f ' x if 2x3

1

xf

4

x

.

37

2.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS :

Page (169)

dx x

dsin

xcos

dx x

dos n

xc si 1 2

2dxtan s x

xec

d

2dx x

dt

xco csc 3 4

sec secd

x x xdx

tan d

x x xcsc csd

c tx

co 5 6


Find dy / dx if y x sin x .

Solution y x xsin

* Using Formula (3) and the product rule we obtain

d d

x xdx dx

ysin

Remember that :

* d d d


d d

x . xsin si x xt

n .d dt

Remember that :

* d

sin x cos xdx

x xos xc sin .

38


Find dy / dx if sin x

y1 cos x

.

Solution

x

1

siny

xcos

* Using the quotient rule together with Formula (3) and (4) we obtain

d d x

dx dx 1

y sin

cos x

Remember that :

*

2

d dg x . f x f x . g x

f xd dx dx

dx g x g x

2

cos sin sind d

x . x x . xdx dx

cos1

co

1

x1 s

Remember that :

* d d

sin x cos x , cos x sin xdx dx

2

cos cos sin sinx x x x

os xc

1

1

2 2

2

cos cos s

1

x x i

xo

xn

c s

39

Remember that :

* 2 2cos x sin x 1

* 2 21 tan x sec x

* 2 2cot x 1 csc x

2

1

1

cos

s

x

xco

1

1 xcos .


Find f '' / 4 if f x sec x . Solution

f x xsec

* d

x xf x x' sec s nx

ecd

ta

Remember that :

* d

sec x sec x tan xdx

* f '' secd

x x xdx

tan

Remember that :

* d d d


d d

x . xsec tan ta x . xd

nt

cdt

se

40

Remember that :

* 2d

tan x sec xdx

2

sec sec tx . x x .an sec xnx ta

3 2

sec sec tx x xan

*

3 2/ 4 / 4f '' sec se / 4c tan / 4

Remember that :

* 180

45 , tan 1 , sec 24 4 4 4

3 22 2 1 3 2 .


1. Find f ' x if x xf xcos n4 2 si .

8. Find f ' x if 2f cx x 1 xse . 21. Find

2 2d y / dx if x x xf sin s x3 co .

41

2.6 THE CHAIN RULE : Page (174)

Derivatives of Compositions :

2.6.1 Theorem (The Chain Rule):

If g is differentiable and f is differentiable at g x , then the composition f g is differentiable at x . Moreover , if

y xf g and u xg then y uf and

d d d.

dx d

y

u x

y u

d . 1


Find dy / dx if 3y cos x . Solution

3y cos x * Let 3u x and express y as y ucos . Applying Formula

(1) yields

d d d

.dx d

y

u x

y u

d

3d d

. xd d

cosu

u xu

Remember that :

* r r 1d

x r xdx

* d

cos x sin xdx

42

2usin . x3

3 2xin xs . 3

2 3sx3 in x .

An Alternative Version of the Chain Rule : Page (175)

* Formula (1) for the chain rule can be expressed in the form

d

x ' x x . xg gf f f ' 'x

g gd

2

* A convenient way to remember this formula is to call f the "outside function" and g the "inside function" in the

composition f g x and the express (2) in words as : "The derivative of f g x is the derivative of the outside

function evaluated at the inside function times the derivative of

the inside function".

Example 3 : (Example 1 revisited) Page (176)

Find h' x if 3h x cos x . Solution

3h x cos x * We can think of h as a composition f g x in which

3g x x is the inside function and f x xcos is the outside function. Thus , Formula (2) yields

gh x x' f ' . xg'

3 2f ' x .3 x

43

3 2xsin x .3

2 3sx3 in x . which agrees with the result obtained in Example 1.


* 2 2tan tand d

x xdx dx

1 dx . x

dxa2 t n tan

Remember that :

* 2d

tan x sec xdx

2x .tan s xec2

2tan s2 ecx x .

* 2 22

1d dx . x

dx dxx1 1

2 1

Remember that :

* d 1

xdx 2 x

2

1

2x

x2

1.

2

x

x 1 .

44

Generalized Derivative Formulas : Page (176)

* There is a useful third variation of the chain rule that strikes a middle ground between Formulas (1) and (2) . If we let

u xg in (2) , then we can rewrite that formula as

d d

.dx dx

uuf f ' u 3

Table 2.6.1 Generalized Derivative Formula: Page (176)

If u is a differentiable function of x , then

r r 1r

d d.

dx

u

xu u

d

d d

.dx d

uu

x

1

2 u

sin cosu

u ud d

.dx dx

cos sinu

u ud d

.dx dx

2tan seu

u ud d

.dx

cdx

2cot cd d

.d d

scx

ux

uu

sec sec tanu

u u ud d

.dx dx

csc csc cotd d

.dx

ux

ud

uu


Find

(a) d

sin 2 xdx

(b) 2d tan x 1

dx

(c) 3d

x csc xdx

(d)

3 / 42dx x 2

dx

(e) 8

5d1 x cot x

dx

Solution

45

(a)

2d

xdx

sin

Taking u x2 in the generalized derivative formula for

sinu yields

d d d

x .dx dx

uusin sin2

dxucos

Remember that :

* d du

sinu cos u .dx dx

2 2d

x . xcos cos 2.dx

2 x

2 cos 2 x .

(b)

2d xtan 1dx

Taking 2

u 1x in the generalized derivative formula for

tanu yields

22d d dx .dx d

uutan ta

x dxn1 usec

Remember that :

* 2d du

tanu sec u .dx dx

2 2 2dx . xdx

se 1c 1

22 xec 2s x1 .

2 22 ecx xs 1 .

46

(c) 3

cscd

x xdx

Taking 3

xcxu cs in the generalized derivative formula

for u yields

3d d d

x x .dx dx

uu

ucs

1c

2 dx

Remember that :

* d 1 du

u .dx dx2 u

3

3

d. x xdxx

1

2csc

csc x

Remember that :

* d

csc x csc x cot xdx

23

31

. x x xx

csc o2

tc x

ccs

2

3

csc cotx x

c

x

2 x x

3

cs

.

(d) 2 3 / 4

2d

x xdx

Taking 2

u x x 2 in the generalized derivative formula

for 3 / 4u yields

3 / 4 3 / 4 12 / 4 u

ud d d

x x .d

u3

24x dx dx

47

Remember that :

* r r 1d du

u r u .dx dx

1

2/ 4

2 dx x .

3x 2x2

4 dx

1 / 4

2x x x

32 2 1

4

.

(e) 58

cotd

x xdx

1

Taking 5

u 1 x xcot in the generalized derivative

formula for 8u yields

58

8 9d d uu

dx x .u8

dx dx dc1

xot

Remember that :

* r r 1d du

u r u .dx dx

5 59 d

x x . x xdx

cot cot1 18

Remember that :

* d d d


* d

c 0dx

48

* 2d

cot x csc xdx

5 4259

cot cx x . x x xsc co18 t x5

49

5 52csc c8 40 1x x x x xt xo cot

.


8. Find f ' x if 26

3 2 1xf x x .

19. Find f ' x if 2f co xsx 3 . 21. Find f ' x if 2 7f se2 cx x .

49

Revision of MATH 101

The Trigonometric Functions :

* Of an Acute Angle :

sinb

c

cosa

c

tanb

a

cscc

b

secc

a

cota

b

Definition :

* 1 1

csc , sec , cot ,sin c

1

os tan

sin cos

tan , cot , tan .cos sin cot

1

* 2 2sin co 1s ,

2 21tan sec ,

2 2cot csc1 .

50

Notes (9) :

* 2 2

cos cos s2 in

2

cos2 1

2

s n1 2 i .

* sin sin o2 c s2 .

* sin sin , cos cos , tan tan ,

cot cot , sec sec , csc csc .

Special Values of the Trigonometric Functions :

(Rad.)

(Deg.) sin cos tan cot sec csc

6

30

1

2 3

2

3

3 3

2 3

3 2

4

45

2

2

2

2 1 1 2 2

3

60

3

2

1

2 3

3

3 2

2 3

3

Remember that :

* 180 180 180

30 , 45 , 606 6 4 4 3 3

51

y xsin

Remember that :

* 3

sin0 0 sin 1 sin 0 sin, , 1 si2

, n, 2 02

y xcos

Remember that :

* 3

cos 0 1 cos 0 cos 1 cos, , 0 cos, 2 12

,2

52

y xtan

Remember that :

* x x

2 2

, lim , lim ,tan0 0 tan x tan x tan 0

y xcot

Remember that :

* x 0 x x

lim , , limcot limx cot 0 cot x , cot x2

53

y xsec

Remember that :

* x x

2 2

, lim lsec 0 1 sec , sec , s2

im ec 12

y xcsc

Remember that :

* x 0 x x

lim , , limcsc x csc 1 tan x , tanim2

l x

Chapter 2 - Calculus I - MATH 203

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