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
Analysis of Function
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
Analysis of Function
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
Example 2 : Page (133)
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.
Example 3 : Page (133)
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
Example 4 : Page (134)
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.
Example 1 : Page (143)
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
Example 2 : Page (145)
(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 .
Example 4 : Page (145)
(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
Example 6 : Page (148)
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
EXERCISES SET 2.2: (Home Work) Page (152)
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
Example 1 : Page (155)
* 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
Example 2 : Page (156)
* 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
Example 3 : Page (157)
* 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
Example 4 : Page (158)
* 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
Example 5 : Page (158)
* 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
Example 6 : Page (159)
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
Example 9 : Page (160)
* 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
EXERCISES SET 2.3: (Home Work) Page (161)
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 . .
Example 1 : Page (164)
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.
Example 2 : Page (165)
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
f x .g x f x . g x g x . f xdx dx dx
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
.
Example 3 : Page (166)
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
EXERCISES SET 2.4: (Home Work) Page (168)
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
Example 1 : Page (170)
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
f x .g x f x . g x g x . f xdx dx dx
d d
x . xsin si x xt
n .d dt
Remember that :
* d
sin x cos xdx
x xos xc sin .
38
Example 2 : Page (170)
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 .
Example 3 : Page (170)
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
f x .g x f x . g x g x . f xdx dx dx
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 .
EXERCISES SET 2.5: (Home Work) Page (172)
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
Example 1 : Page (175)
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.
Example 4 : Page (176)
* 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
Example 5 : Page (177)
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
f x .g x f x . g x g x . f xdx dx dx
* 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
.
EXERCISES SET 2.6: (Home Work) Page (178)
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
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