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Chapter 1 Overview: Review of Derivatives The purpose of this chapter is to review the “how” of differentiation. We will review all the derivative rules learned last year in PreCalculus. In the next several chapters, we will be exploring the “why” of differentiation as well. As a quick reference, here are those rules:
The Power Rule: - n nd duu nu
dx dx 1
The Product Rule: d dv du
u v u vdx dx dx
The Quotient Rule:
du dv
v uu xd dx dxdx vv x
2
The Chain Rule: ( ( )) '( ( )) '( )
df g x f g x g x
dx
sin cos
cos sin
tan sec
d duu u
dx dx
d duu u
dx dx
d duu u
dx dx
2
csc csc cot
sec sec tan
cot csc
d duu u u
dx dx
d duu u u
dx dx
d duu u
dx dx
2
ln
u u
u u
d due e
dx dx
d dua a a
dx dx
ln
log
ln a
d duu
dx u dx
d duu
dx u a dx
1
1
-
-
-
sin
cos
tan
d duu
dx dxu
d duu
dx dxu
d duu
dx dxu
1
2
1
2
12
1
1
1
1
1
1
-
-
-
csc
sec
cot
d duu
dx dxu u
d duu
dx dxu u
d duu
dx dxu
1
2
1
2
12
1
1
1
1
1
1
2
Here is a quick review of from last year:
Identities While all will eventually be used somewhere in Calculus, the ones that occur most often early are the Reciprocals and Quotients, the Pythagoreans, and the Double Angle Identities.
sin
tan cos
xx
x ;
cos cot
sin
xx
x ;
1sec
cos x
x ;
1csc
sin x
x
2 2sin cos 1x x ; 2 2tan 1 sec x x ; 2 2cot 1 csc x x
sin 2 2sin cosx x x ; 2 2cos2 cos sinx x x
Inverses Because of the quadrants, taking an inverse yields two answers, only one of which your calculator can show. How the second answer is found depends on the kind of inverse:
-1 2cos
2
calculator nx
calculator n
-1 2
sin 2
calculator nx
calculator n
-1 2tan
2
calculator nx calculator n
calculator n
log log log ( )
log log log
log log
a a a
a a a
na a
x y xy
xx y
y
x n x
3
1.1: The Power Rule and the Exponential Rules In PreCalculus, we developed the idea of the Derivative geometrically. That is, the derivative initially arose from our need to find the slope of the tangent line. In Chapter 2 and 3, that meaning, its link to limits, and other conceptualizations of the Derivative will be Explored. In this Chapter, we are primarily interested in how to find the Derivative and what it is used for.
Derivative—Def’n: ( ) ( )
'( ) limh
f x h f xf x
h
0
—Means: The function that yields the slope of the tangent line.
Numerical Derivative—Def’n: ( ) ( )
'( ) lim x a
f x f af a
x a
—Means: The numerical value of the slope of the tangent line at x = a.
Symbols for the Derivative
dy
dx= "d-y-d-x" f '(x) = "f prime of x" y' = "y prime"
d
dx
= "d-d-x" Dx = "d sub x"
OBJECTIVES Use the Power Rule and Exponential Rules to find Derivatives. Find the equations of tangent and normal lines. Use the equation of a tangent line to approximate a function value.
Key Idea from PreCalc: The derivative yields the slope of the tangent line. [But there is more to it than that.]
Key Concept: The derivative of a function is the slope of the function. The definitions of the derivative, above, both come from the expression for the slope of a line, combined with the idea of a limit. The numerical derivative of a function is the slope of that function at a point.
4
Ex. 1 Find the values of ' 1f , ' 3f , ' 1g , and ' 4g
x
y
' 1 3f , 2
' 33
f , ' 1 2g , 2
' 45
g
Note that at x = 2 that the derivative of both f and g does not exist because there is not a single tangent line at this point. We will discuss this further in the chapter on limits and differentiability.
Ex. 2 Find the values of x where the function whose graph is below has a
derivative value of 0.
x
y
f(x)
g(x)
5
x
y
A slope of 0 is a horizontal line, so this function has tangent lines with slope of 0 at x = –2, x = 0, and x = 2.
The first and most basic derivative rule is the Power Rule. Among the last rules we learned in PreCalculus were the Exponential Rules. They look similar to one another, therefore it would be a good idea to view them together.
The Power Rule: The Exponential Rules:
- n nd
x n xdx
1
ln
x x
x x
de e
dx
da a a
dx
The difference between these is where the variable is. The Power Rule applies when the variable is in the base, while the Exponential Rules apply when the variable is in the Exponent. The difference between the two Exponential rules is what the base is. e = 2.718281828459…, while a is any positive number other than 1.
Ex 3 Find a) d
xdx
5 and b)
xd
dx 5
6
The first is a case of the Power Rule while the second is a case of the second Exponential Rule. Therefore,
a)
d
xdx
5 = x45 b)
xd
dx 5 = lnx5 5
There were a few other basic rules that we need to remember.
Dx [constant] is always 0
1n nxD cx cn x
x x xx g x x g xD f D f D
These rules allow us to easily differentiate a polynomial--term by term.
Ex 4 y x x 23 5 1; find dy
dx
[ ]
( ) ( ) ( )
dy dx x
dx dx
x x
2
2 1 1 1
3 5 1
3 2 5 1 1 0
x 6 5
Ex 5 ( ) ; find '( ).xf x x x e f x 2 4 3
'( ) xf x x e 2 4
7
Ex 6 ; find .dy
y x x edxx
43 3 44
y x x ex
y x x x e
43 3 4
33 1442 2
4
4
dy
x x xdx
3 1142 23 3
22 4
Note in Ex 6 that 4e is a constant, therefore, its derivative is 0. Ex 7 Find the equations of the lines tangent and normal to
( )f x x x x 4 3 22 1 at 1x
The slope of the tangent line will be '( )f 1
'( )
'( )
f x x x x
f
3 24 3 4
1 3
[Note that we could have gotten this more easily with the nDeriv function on our calculator.]
( )f 1 1, so the tangent line will be
( )
or
y x
y x
1 3 1
3 2
The normal line is perpendicular to the tangent line and, therefore, has the
negative reciprocal slope =1
3 . The normal line is
( )y x 1
1 13
One of the uses of the tangent line is based on the idea of Local Linearity. This means that in small areas, algebraic curves act like lines—namely their
8
tangent lines. Therefore, one can get an approximate y-value for points near the point of tangency by plugging x-values into the equation of the tangent line. We will explore this more in a later section.
As we have seen, when the variable is in the Exponent, we use the Exponential Rules. When the variable was in the base, we used the Power Rule. But what if
the variable is in both places, such as xd
xdx
2
2 1 ? It is definitely an
Exponential problem, but the base is not a constant as the rules above have. The Change of Base Rule allows us to clarify the problem:
ln x xxd d
x edx dx
22 2 12 1
but we will need the Product Rule for this derivative. Therefore, we will save this for section 1-4.
9
1.1 Homework Set A Differentiate.
1. ( )f x x x 2 3 4 2. ( )f t t 418
4
3. y x
2
3 4. xy e 35
5. ( )v r r 34
3 6. ( )g x x
x 2
2
1
7. x x
yx
2 4 3 8. u t t 3 2 32
9. yAz y Be
y 10
10. xy e 1 1
10
11. Find an equation of the tangent to the curve xy x e 4 2 at the point ( , )0 2 .
12. Find the points on the curve y x x x 3 22 3 12 1 where the tangent is
horizontal.
13. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x .
11
14. Find the tangent line equation to 2
5( ) at 1F x x x
x .
15. Find the equation of all tangent lines that have a slope of 5 for the function
2
42
xy x
16. Find the equation of the tangent line for 2 5 6f x x x that has a slope
of 4.
12
17. Find the equations of the lines tangent and normal to the function
35f x x x at the point (1,5)
13
Answers: 1.1 Homework Set A
1. ( )f x x x 2 3 4 2. ( )f t t 418
4
'( )f x x 2 3 '( )f t t 3
3. y x
2
3 4. xy e 35
dyx
dx
532
3 xdy
edx
5
5. ( )v r r 34
3 6. ( )g x x
x 2
2
1
'( )v r r 24 '( )g x x x 32 2
7. x x
yx
2 4 3 8. u t t 3 2 32
dyx x x
dx
31 12 2 2
3 32
2 2
dut t
dt
1 13 2
2
33
9. yAz y Be
y 10
10. xy e 1 1
' yyz y A Be 1110 xdy
edx
1
11. Find an equation of the tangent to the curve xy x e 4 2 at the point ( , )0 2 .
y x 2 2
12. Find the points on the curve y x x x 3 22 3 12 1 where the tangent is
horizontal. (1, –6) (–2, 21)
13. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x .
y x 21 75 2
14. Find the tangent line equation to 2
5( ) at 1F x x x
x .
y x
21
4 12
14
15. Find the equation of all tangent lines that have a slope of 5 for the function 2
42
xy x
y x 9 5 12
16. Find the equation of the tangent line for 2 5 6f x x x that has a slope
of 4.
y x 15 94 24
17. Find the equations of the lines tangent and normal to the function
35f x x x at the point (1,5)
y x 55 12
15
1.2: Composite Functions and the Chain Rule
Composite Function--A function made of two other functions, one within the
other. For Example, y x x 316 , sin y x 3 , cosy x 3 , and ( )y x x 2 32 5 .
The general symbols are ( ( ))f g x or [ ]( )f g x .
Ex 1 Given cos xf x 1 , xg x 2 1, and xh x 21 , find (a) f g 2 ,
(b) h g 1 , and (c) f h g 1 .
(a) g 2
2 2 1 1, so cosf g f 12 1 1 0 .
(b) g 01 , so h g h 21 0 1 0 1.
(c) g 01 and h g h 21 0 1 0 1, so
cosf h g f 11 1 1 0
So how do we take the derivative of a composite function? There are two (or more) functions that must be differentiated, but, since one is inside the other, the derivatives cannot be taken at the same time. Just as a radical cannot be distributed over addition, a derivative cannot be distributed concentrically. The composite function is like a matryoshka (Russian doll) that has a doll inside a doll. The derivative is akin to opening them. They cannot both be opened at the same time and, when one is opened, there is an unopened one within. You end up with two open dolls next to each other.
The Chain Rule: ( ( )) '( ( )) '( )
df g x f g x g x
dx
If you think of the inside function (the g x ) as equaling u, we could write The
Chain Rule like this: ( ) '
d duf u f u
dx dx . This is the way that most of the
derivatives are written with The Chain Rule.
16
The Chain Rule is one of the cornerstones of Calculus. It can be embedded within each of the other Rules, as it was in the introduction to this chapter. So the Power Rule and Exponential Rules in the last section really should have been stated as:
The Power Rule: The Exponential Rules:
- n nd du
u n udx dx
1
ln
u u
u u
d due e
dx dx
d dua a a
dx dx
where u is a function of x.
OBJECTIVE Find the Derivative of Composite Functions.
Ex 2 If y x x 316 , find dy
dx.
y x x x x 123 316 16
dyx x x
dx
123 21
16 16 32
x
x x
2
123
16 3
2 16
In this case, the is the f function and the polynomial x x 316 is the g. Each
derivative is found by the Power Rule, but, as x x 316 is inside the , it is inside
the derivative of the .
17
Ex 3
dx x
dx
10
24 2 1
dx x x x x
dx
10
2 2 94 2 1 10 4 2 1 8 2
x x x 29
20 4 2 1 4 1
Ex 4 xd
dxe
24
xd
dxe
24 x xe x xe
2 24 48 8
Ex 5 Given this table of values, find df g x
dx
and dg f x
dx
at 1x .
x f x g x 'f x 'g x
1 3 2 4 6
2 1 8 5 7
3 7 2 7 9
' 'd
f g x f g x g xdx
and ' 'd
g f x g f x f xdx
.
At 1x , At 1x ,
' 1 ' 1
' 2 6
5 6
30
df g x f g g
dx
f
' 1 ' 1
' 3 4
9 4
36
dg f x g f f
dx
g
18
Ex 6 Using the table from example 5, find 2x
fd
f xdx
.
2x
fd
f xdx
= 20
Of course, this kind of problem can extend to where we don’t have a table of values, but where we still only have a function that we can derive symbolically.
Ex 7 If 2 5g and ' 2 4g , find ' 2f if 3
3 6g xf x e g x g x
Notice that while we do not actually know the function that g represents, we still can take its derivative, because we know the derivative of g is g’. Of course, the Chain Rule is still essential in this process.
3
3 6g xf x e g x g x
2
3 2' ' ' 6 3 3 'g xf x e g x g x x g x g x
2
5
223
2
' 2 ' 2 ' 2 6 3 2 3 2 ' 2
' 2 4 ' 2 12 3 5 4
gf e g g g g
f e g
5 5
4 4' 2 4 12 300 348f
e e
19
1.2 Homework Set A
1. 7
3 4d
x xdx
2. xy e , find dy
dx.
3. 4 3( ) 1 2f x x x , find '( ).f x
4. Given the following table of values, find the indicated derivatives.
( ) ( )x f x f x
2 1 7
8 5 3
a. ( )g 2 , where ( ) ( )g x f x
3 b. '( )h 2 , where ( ) ( )h x f x 3
20
5. Given this graph and u f g x , v g f x , and w f f x , find
(a) ' 4.5u (b) ' 1v (c) ' 1w
x
y
6. If ( )f x x x 37
3 2 , find '( )f x .
f(x)
g(x)
21
7. If ( )g 2 3 and ( )g 2 4 , find ( )f 2 if ( )
( )g x
f x e .
8. 244 ; find ' 5
9f x x f
9. 2 3 4 9
dx x
dx
10. 7 3 2 ; find dy
y x xdx
22
1.2 Homework Set B
1. Given the table of values below, find ' 2g if f h x
g x e
x f x 'f x h x 'h x
2 ln4 1 3 -7 3 ln9 2 -9 11 -7 ln 49 3 8 -1
2. If 2g and ' 2 2g , find ' 2f if 2 2 2
g xf x e g x x
3. If 1 5h and ' 1 3h , find ' 1f if 3 h xf x h x e
23
4. Find 3df g h x
dx
5. If 0 5g and ' 0 3g , find ' 0f if 2
3 6f x g x g x g x
24
1.2 Homework Set A
1. 7
3 4d
x xdx
8
3 27 4 3 4x x x
2. xy e , find dy
dx.
2
x
x
dy
dx
e
3. 4 3( ) 1 2f x x x , find '( ).f x
3
43 211 2 2 3
4'( ) x x xf x
4. Given the following table of values, find the indicated derivatives.
( ) ( )x f x f x
2 1 7
8 5 3
a. ( )g 2 , where ( ) ( )g x f x
3 b. '( )h 2 , where ( ) ( )h x f x 3
( )g 2 21 '( )h 2 36
5. Given this graph and u f g x , v g f x , and w f f x , find
(a) ' 4.5u (b) ' 1v (c) ' 1w
x
y
(a) 4
' 4.515
u
(b) 2
' 115
v (c) 1
' 19
w
f(x)
g(x)
25
6. If ( )f x x x 37
3 2 , find '( )f x .
'( )f x x x x 36
3 237 2 3 2
7. If ( )g 2 3 and ( )g 2 4 , find ( )f 2 if ( )
( )g x
f x e .
'( )f e 32 4
8. 244 ; find ' 5
9f x x f
16' 5 5
27f
9. 2 3 4 9
dx x
dx
1
2213 4 9 6 4
2x x x
10. 7 3 2 ; find dy
y x xdx
6
73 212 3 2
7
dyx x x
dx
1.2 Homework Set B
1. Given the table of values below, find ' 2g if f h x
g x e
x f x 'f x h x 'h x
2 ln4 1 3 -7
3 ln9 2 -9 11
-7 ln 49 3 8 -1
' 2 126g
2. If 2g and ' 2 2g , find ' 2f if 2 2 2
g xf x e g x x
' 2 2 4f e
26
3. If 1 5h and ' 1 3h , find ' 1f if 3 h xf x h x e
5' 1 225 3f e
4. Find 3df g h x
dx
2 3 3 33 ' ' 'x f g h x g h x h x
5. If 0 5g and ' 0 3g , find ' 0f if 2
3 6f x g x g x g x
' 0 33f
27
1.3: Introduction to Implicit Differentiation, an Advanced Use of the Chain
Rule
One of the most common applications of the Chain Rule in Calculus is called implicit differentiation. Recall that differentiation just means “to take the derivative”. All of what we did in precalculus was explicit differentiation – that is, the functions were explicitly solved for the dependent variable (usually y). But we can actually use the Chain Rule to find the derivative of functions and relations where we do not have the dependent variable isolated.
Ex. 1 Find the dy
dx for the function 45 22y x x
45 22d
y x xdx
320 22dy
xdx
Now, this is an obvious and easy example, but notice what we have on the left side of the equation – in the process of taking the derivative, the y
became a dy
dx. That is because the derivative of y is
dy
dx!
OBJECTIVES Find Derivatives of functions and relations that are not explicitly solved for y.
As you saw in example 1, the derivative of y is dy
dx. This applies wherever the y is
located in a problem. And if the y is located within another function – like y4 –
which is actually just (y)4, then the Chain Rule applies, as it has in every other
aspect of the derivative we have ever worked with.
28
Ex. 2 Find dy
dx for 2 416 x x y
2 416d
dxx x y
30 1 2 4dy
dxx y
34 2 1dy
dxy x
3
2 1
4
dy
dx
x
y
So the derivative of 2 416 x x y is 3
2 1
4
dy
dx
x
y
.
Note that just like the original equation, there is still a y in the solution of the problem. This is a very common occurrence in implicit differentiation. In the case above, we could’ve solved it explicitly for y by using algebra, and done the derivative that way. But in many cases, this is not a possibility.
Ex. 3 Find dy
dx for 2 2 17yx y e
2 2 17yd
dxx y e
2 2 0ydy dy
dx dxx y e
2 2
2 2
y
y
dy dy
dx dx
dy
dx
y e x
y e x
Simply take the derivative, using the Chain
Rule on the 4y
Then just use basic algebra to solve for dy
dx
29
2
2 y
dy
dx
x
y e
Ex. 4 Find the equations of the lines tangent to the relation 2 2 17yx y e
at y = 0. Note that we were given a y-value, so we can plug that in and find an
x-value. We need a point to use to find the equation of the tangent line.
2 2 00 17x e 2 16x
4x So we have two tangent lines, one passing through (4,0) and the other passing through (–4,0).
From example 3, we know that 2
2 y
dy
dx
x
y e
, so we find the values for
dy
dx at
the two points.
4,00
4
0
28
2
dy
dx e
4,00
4
0
28
2
dy
dx e
Now we use the points and slopes and get the tangent lines:
8 4y x and 8 4y x
30
1.3 Homework Set A
Find dy
dx for each of the following relations.
1. 2 24 16x y 2. 3 2 25xe y y
3. 25 16x y y 4. 2 2
20y xe e y
5. Find the equations for the lines tangent and normal to the relation
2 22 4 20x x y y passing through the point (5,1).
6. Find the equations for the lines tangent to the relation 3 2 8 21 0x y y
when x = 1
31
1.3 Homework Set A
Find dy
dx for each of the following relations.
1. 2 24 16x y 2. 3 2 25xe y y
4
dy x
dx y
23 2
xdy e
dx y
3. 25 16x y y 4. 2 2
20y xe e y
121
2
1
16 10
dy
dx y y
2
2
2
20 2
x
y
dy xe
dx ye
or
2 16
1 20 16
ydy
dx y y
5. Find the equations for the lines tangent and normal to the relation
2 22 4 20x x y y passing through the point (5,1).
Tangent: 4
1 53
y x Normal: 3
1 54
y x
6. Find the equations for the lines tangent to the relation 3 2 8 21 0x y y
when x = 1
Tangent to (1,10): 1
10 14
y x
Tangent to (1,–2): 1
2 14
y x
32
1.4: Trig and Log Rules Trigonometric--Defn: "A function (sin, cos, tan, sec, csc, or cot) whose independent variable represents an angle measure." Means: an equation with sine, cosine, tangent, secant, cosecant, or cotangent in it. Logarithmic--Defn: "The inverse of an Exponential function." Means: there is a Log or Ln in the equation.
sin cos
cos sin
tan sec
d duu u
dx dx
d duu u
dx dx
d duu u
dx dx
2
csc csc cot
sec sec tan
cot csc
d duu u u
dx dx
d duu u u
dx dx
d duu u
dx dx
2
ln
log
ln a
d duu
dx u dx
d duu
dx u a dx
1
1
Note that all these Rules are expressed in terms of the Chain Rule.
OBJECTIVES Find Derivatives involving Trig and Logarithmic Functions.
33
Ex 1 sin
dx
dx3
sin sin cos
dx x x
dx3 23
Ex 2 sin
dx
dx
3
sin cos
dx x x
dx
3 3 23
cos x x 2 33
Ex 3
lnd
xdx
34
lnd
x xdx x
3 23
14 12
4
x
3
We could have also simplified algebraically before taking the derivative:
ln ln lnd d
x xdx dx
34 4 3
x
3
Oftentimes it is significantly easier to take a derivative if we simplify the function first – in the above case, it eliminated the need for The Chain Rule.
34
Of course, composites can involve more than two functions. The Chain Rule has as many derivatives in the chain as there are functions.
Ex 4 sec
dx
dx5 43
sec sec sec tan
dx x x x x
dx5 4 4 4 4 4 33 5 3 3 3 12
sec tan x x x 3 5 4 460 3 3
Ex 5 ln cos
dx
dx
ln cos sin
cos
tan
tan
dx x x
dx x
x x
x
x
1 12 2
12
1 12 2
12
12
1 1
2
1
2
2
35
1.4 Homework Set A For #1-12, find the derivatives of the given functions.
1. sin4y x 2. 54secy x
3. 3 3cosa xy 4. 2cot (sin )y
5. 3( ) 1 tanf t t 6. ( ) ln cosf
7. 3 3cosy a x 8. 2tan 3y
36
9. ( ) cos lnf x x 10. 5( ) lnf x x
11. 10( ) log 2 sinf x x 12. 2( ) log 1 3f x x
13. Find the equation of the tangent line to cosy x x at the point 0,1 .
14. Find the equation of the tangent line to sec 2cosy x x at the point ,13
.
37
15. Find all points on the graph of 2( ) 2sin sinf x x x at which the tangent line
is horizontal. 16. Find the equations of the lines tangent and normal to the function
2 2ln 4 5xf x e x x at the point (0,1+ln5)
17. Find the equation of the line tangent to 2
cos 4y x x
when 2
x
.
38
18. Find the equation of the line tangent to sec 2 cot 2y x x through the
point ,1 28
. Use exact values in your answers.
1.4 Homework Set B
1. Given the table of values below, find ' 3g if g x f h x
x f x 'f x h x 'h x
2 2
1
2
-7
3 4
2
3
1
3
2 3
4
-1
39
2. csc 2ln cot secd
ed
3. If 32
g
, ' 34
g
, and sin
3cos3
g xf x g x g x e
, find
' 3f
4. If ln cot sec lnz , find dz
d
40
5. 3
3ln sec 5ln 7d
x xdx
6. If 2ln cos sec 7tz t e , find dz
dt
7. If 2ln tan sin 7tz t e , find dz
dt
8. If ln cos sin lnz , find dz
d
41
9. Find '6
f
when 3cos 3f x x
10. Find ' 2g when 2ln 3g x x
11. 2ln 4 5d
x xdx
12. 5sin ln 7 3d
tdt
13. 2csc ln 7d
x xdx
14. 24 6ln td
edt
42
15. sec 5 cot 10lnxdx e x
dx
For problems 16 to 18, find the first derivative for the following functions
16. 10yA
z y Bey
17. sinrAf r Be
r 18.
3 2
2
3 12x xy
x
19. 23
59 27
d dx x
dx dx x
20. 23
59 27
d dx x
dx dx x
21. 3
2ln tan 5 7xdx e
dx
22.
2
2
cos ln 5
sin ln 5
xd
dx x
43
1.4 Homework Set A
1. sin4y x 2. 54secy x
4cos4dy
xdx
4 5 5' 20 sec tany x x x
3. 3 3cosa xy 4. 2cot (sin )y
23cos sindy
x xdx 22cos cot(sin )csc (sin )
dy
dx
5. 3( ) 1 tanf t t 6. ( ) ln cosf
2
23
sec'( )
3 1 tan
tf t
t
'( ) tanf
7. 3 3cosy a x 8. 2tan 3y
2 3 33 siny x a x 26tan 3 sec 3dy
dx
9. ( ) cos lnf x x 10. 5( ) lnf x x
sin ln
'( )x
f xx
45
1
5 ln'( )
x xf x
11. 10( ) log 2 sinf x x 12. 2( ) log 1 3f x x
cos
'( )ln10 2 sin
xf x
x
3
'( )ln 2 1 3
f xx
13. Find the equation of the tangent line to cosy x x at the point 0,1 .
1 1 0y x
14. Find the equation of the tangent line to sec 2cosy x x at the point ,13
.
1 3 33
y x
44
15. Find all points on the graph of 2( ) 2sin sinf x x x at which the tangent line
is horizontal.
2 ,3 , 2 ,12 2
n n
16. Find the equations of the lines tangent and normal to the function
2 2ln 4 5xf x e x x at the point (0,1+ln5)
1 ln 5 04
5y x
17. Find the equation of the line tangent to 2
cos 4y x x
when 2
x
.
22
2y x
18. Find the equation of the line tangent to sec 2 cot 2y x x through the
point ,1 28
. Use exact values in your answers.
1 2 2 2 48
y x
1.4 Homework Set B
1. Given the table of values below, find ' 3g if g x f h x
x f x 'f x h x 'h x
2
2
1
2
-7
3
4
2
3
1
3
2 3
4
-1
' 3 3g
45
2. csc 2ln cot secd
ed
2 2
csc2
2 csccsc cot sec tan
cote
3. If 32
g
, ' 34
g
, and sin3cos
3
g xf x g x g x e
, find ' 3f
' 34
f
4. If ln cot sec lnz , find dz
d
2 sec ln tan lncsc
cot
dz
d
5. 3
3ln sec 5ln 7d
x xdx
2 3
3 2 353 5ln 7 3 tan 5ln 7x x x x x
x
6. If 2ln cos sec 7tz t e , find dz
dt
tan sec tant t tdzt e e e
dt
7. If 2ln tan sin 7tz t e , find dz
dt
2sec
costan
t tdz te e
dt t
8. If ln cos sin lnz , find dz
d
cos ln
tandz
d
9. Find '6
f
when 3cos 3f x x
' 06
f
46
10. Find ' 2g when 2ln 3g x x
' 2 4g
11. 2ln 4 5d
x xdx
12. 5sin ln 7 3d
tdt
2
2
4 5
x
x x
435sin ln 7 3 cos ln 7 3
7 3
t t
t
13. 2csc ln 7d
x xdx
14. 24 6ln td
edt
2 2
2
14 1 csc ln 7 cot ln 7
7
x x x x x
x x
6t
15. sec 5 cot 10lnxdx e x
dx
2 105sec 5 tan 5 cscx xx x e e
x
16. 10yA
z y Bey
17. sinrAf r Be
r 18.
3 2
2
3 12x xy
x
11' 10 yz y Ay Be 2 sincos rf r Ar B r e 31 24dy
dxx
19. 23
59 27
d dx x
dx dx x
2 3 5 4 4
22 3
18 54 10 60 54 9 54 15 18 108 30
18 54 10
x x x x x x x x
x x x
20. 23
59 27
d dx x
dx dx x
554 60x
47
21. 3
2ln tan 5 7xdx e
dx
22.
2
2
cos ln 5
sin ln 5
xd
dx x
2 32 2 2
32
3 5 7 sec 5 7
tan 5 7
x x
x
x e x e
x e
2 22csc ln 5x
x
48
1.5: Product and Quotient Rules Remember:
The Product Rule: d dv du
dx dx dxu v u v
The Quotient Rule:
du dvd dx dxdx
uv
v
v
u
2
OBJECTIVE Find the Derivative of a product or quotient of two functions.
Ex 1 sin
dx x
dx2
sin cos sin
cos sin
dx x x x x x
dx
x x x x
2 2
2
2
2
Ex 2
d x x
dx x
2 2 3
4
, so
, so
( ) ( ) ( )’( )
( )
( )
( )
duU x x x
dx
dvV x
dx
du dvV U x x x xdx dxf x
V x
x x x x
x
x x
x
2
2
2 2
2 2
2
2
2 3 2 2
4 1
4 2 2 2 3 1
4
2 6 8 2 3
4
8 5
4 2
49
Ex 3 d x x
dx x x
2
2
4 3
2 5 3
Notice that this problem becomes much easier if we simplify before applying the Quotient Rule.
( )( )
( )( )
( )( ) ( )( )
( )
( )
d x x d x x
dx dx x xx x
d x
dx x
x x
x
x
2
2
2
2
4 3 1 3
2 1 32 5 3
1
2 1
2 1 1 1 2
2 1
3
2 1
Ex 4 cot 3d x
dx x
2 1
csc cot cot
csc csc cot
x x x xd x
dx x x
x x x x x
x
2 2
22
2 2 2
2
2
2
1 3 3 3 23
1 1
3 3 3 3 2 3
1
Ex 5 cos xd
xdx
5
cos sin cos ln
ln cos sin
x x x
x
dx x x
dx
x x
5 5 5 5
5 5
50
Remember that, in section 1.1, we said we would need the Product Rule to deal with the derivative of a function where the variable is in both the base and the Exponent. We can now address that situation.
Ex 6
cos xd
xdx
2
ln
ln cos
cos
cos
sin ln cos cos
cos ln cos tan
x x x
x x
x
d dx e
dx dx
e x x x xx
x x x x x
2 2
22
22
12
2
51
1.5 Homework Set A Find the derivative of the following functions.
1. 3 cosy t t 2. 85 2( ) 1 4 3g x x x x
3. 1 tan
sec
xy
x
4.
34 22 5 8 5y x x
5. 2xy xe 6.
2
sin xy
x
52
7. cosx xy e 8. 2 1
ry
r
9. 1
siny xx
10. 5 cos3xy e x
11. ( ) lnf x x x 12. ln x xy e xe
13. Find the equation of the line tangent to 2 xy x e at the point 11, e .
53
14. If ( )ln
xf x
x , find '( )f e . 15. ( )
xh t
x
172
2
1
1, find '( )h t
16. y x x 2 25 , find 'y 1
17. ( ) sin tanf x x x x
5
4 72 , find 'f x .
54
18. Find the equation of the lines tangent and normal to sin ln2
y x x
when
x e
19. Find the equation of the line tangent to sin 4
2x x
y e when 0x
20. Find the equation of the lines tangent and normal to 1
siny xx
when
4x
.
55
1.5 Homework Set B
1. Find the first derivative for the following function: 2 2 4sin 5tx t e t t
2. Find the first derivative for the following function: 5 4tan 3tx t e t
3. Find the first derivative for the following function: 2 2 15
3
x xy
x
4. Find the first derivative for the following function: 2 45tx t e t t
56
5. 2
3
7 5
sin
xd e x
dx x
6.
sinln cot
y yde e
dy
7. 2 2 1sin
ln
d xx x
dx x
8. If 1 5h and ' 1 3h , find ' 1f if 4
lnf x h x x h x
9. Find 'g z if 118
5
1 ln
zeg z
z
10. 2 2cosxd e
x xdx x
57
11. 2 2 3
4
d x x
dx x
12.
2
5
cos 3
x
xd
dx e
13. 5 ln 5 4ln
d xx x
dx x
14. Find 'g t if 15
2
2
4
1
tg t
t
15.
2
cosxde x
dx
16. 1 tan
ln 4
d x
dx x
58
17. sin tand
t tdt
18.
1 ln
csc
d x
dx x
19. 45 ln sinxd
e xdx
20. 2 22
5 sin ln 3 11
pd pp p e p
dp p
21. 4 3tan 5xde x x x
dx
22. 5 2
ln 3 7
d x
dx x
59
23. 2
4
9 20
d x
dx x x
24. 2sin 4 2
d dx
dx dx
25. 5 3
3
12 19
3
d c c c
dc c
26. Given the table of values below, find ' 3g if sing x f h x h x
x f x 'f x h x 'h x
2 2
1
2
-7
3 4
2
3
1
3
2 3
4
-1
60
27. If 32
g
, ' 34
g
, and sin3 3cos
3
g xf x x g x g x e
, find
' 3f
61
1.5 Homework Set A
1. 3 cosy t t 2. 85 2( ) 1 4 3g x x x x
2' 3cos siny t t t t 74 2 2'( ) 4 1 4 3 17 9 21g x x x x x x
3. 1 tan
sec
xy
x
4.
34 22 5 8 5y x x
tan 1
'sec
xy
x
43 2 2' 8 2 5 8 5 4 30 5y x x x x
5. 2xy xe 6.
2
sin xy
x
2 2' 1 2xy e x
3
cos 2sin'
x x xy
x
7. cosx xy e 8. 2 1
ry
r
cos' cos sinx xy e x x x
3
2 2
1'
1
y
r
9. 1
siny xx
10. 5 cos3xy e x
1 1 1
cos sindy
dx x x x 5 5cos3 3sin3xdy
e x xdx
11. ( ) lnf x x x 12. ln x xy e xe
1 2ln
'( )2 ln
xf x
x
1
dy x
dx x
13. Find the equation of the line tangent to 2 xy x e at the point 11, e .
1 1
1y xe e
62
14. If ( )ln
xf x
x , find '( )f e . 15. ( )
xh t
x
172
2
1
1, find '( )h t
'( ) 0f e
'( )
t th t
t
162
182
68 1
1
16. y x x 2 25 , find 'y 1
7
2'(1)y
17. ( ) sin tanf x x x x
5
4 72 , find 'f x .
'( ) sin tan sin cos tan secf x x x x x x x x x x
4
4 7 6 3 7 2 75 2 2 2 2 28
18. Find the equation of the lines tangent and normal to sin ln2
y x x
when
x e
Tangent: 1y e x e Normal: 1y e x e
19. Find the equation of the line tangent to sin 4
2x x
y e when 0x
3y
20. Find the equation of the lines tangent and normal to 1
siny xx
when
4x
.
Tan: 2 4 4
2 28
y x
Normal:
4 2 42 2
4y x
1.5 Homework Set B
1. Find the first derivative for the following function: 2 2 4sin 5tx t e t t
2 2 2 4 2 4' 2 5 cos 5 2sin 5tx t te t t t t t
63
2. Find the first derivative for the following function: 5 4tan 3tx t e t
25 3 4 4' 12 sec 3 5tan 3tx t e t t t
3. Find the first derivative for the following function: 2 2 15
3
x xy
x
1dy
dx
4. Find the first derivative for the following function: 2 45tx t e t t
3 2' 5 20 2tx t t t tte
5. 2
3
7 5
sin
xd e x
dx x
6.
sinln cot
y yde e
dy
3 2 3
2 3
14 sin 7 5 cos
sin
x xe x x e x x
x
sin2
sincsc
cos ln cotcot
yy y
y y
y
e e ey e e
e
7. 2 2 1sin
ln
d xx x
dx x
2
3 2 22
ln 12 cos 2 sin
ln
x x xx x x x
x x
8. If 1 5h and ' 1 3h , find ' 1f if 4
lnf x h x x h x
7503
' 1 ln55
f
9. Find 'g z if 118
5
1 ln
zeg z
z
10. 2 2cosxd e
x xdx x
590
119
118 5 5 ln 1'
1 ln
ze z z zg z
z z
3 2 22
12 sin 2 cos
xe xx x x x
x
64
11. 2 2 3
4
d x x
dx x
12.
2
5
cos 3
x
xd
dx e
2
2
8 5
4
x x
x
2 2
5
5cos 3 2 sin 3
x
x x x
e
13. 5 ln 5 4ln
d xx x
dx x
14. Find 'g t if 15
2
2
4
1
tg t
t
5
42
1 ln5 ln 5 4
5 4 ln
x xx x
x x
142
162
150 4'
1
t tg t
t
15.
2
cosxde x
dx
16. 1 tan
ln 4
d x
dx x
2
sin 2 cosxe x x x 2
2
1 sec ln 4 tan
ln 4
x x x x
x x
17. sin tand
t tdt
18.
1 ln
csc
d x
dx x
2sin sec cos tant t t t 1 cot cot ln
csc
x x x
x x
19. 45 ln sinxd
e xdx
45 3cot 20 ln sinxe x x x
20. 2 22
5 sin ln 3 11
pd pp p e p
dp p
22 2 2
2 22
6 15 1 2 cos 5sin
3 1 1
p p p p pp e p e p e
p p
65
21. 4 3tan 5xde x x x
dx
22. 5 2
ln 3 7
d x
dx x
3 2 2 4 3tan 4 15 1 sec 5x xe x x e x x x
2
5 3 7 ln 3 7 5 2
3 7 ln 3 7
x x x
x x
23. 2
4
9 20
d x
dx x x
24. 2sin 4 2
d dx
dx dx
2
5x
32cos 8 4x
25. 5 3
3
12 19
3
d c c c
dc c
3238
3c c
26. Given the table of values below, find ' 3g if sing x f h x h x
x f x 'f x h x 'h x
2
2
1
2
-7
3
4
2
3
1
3
2 3
4
-1
3 3
' 3 12
g
27. If 32
g
, ' 34
g
, and sin3 3cos
3
g xf x x g x g x e
, find
' 3f
81
' 34
f
66
1.6: Higher Order Derivatives What we have been calling the Derivative is actually the First Derivative. There can be successive uses of the derivative rules, and they have meanings other than the slope of the tangent line. In this section, we will explore the process of finding the higher order derivatives. Second Derivative--Defn: The derivative of the derivative. Just as with the First Derivative, there are several symbols for the 2nd Derivative:
Higher Order Derivative Symbols
Liebnitz: d y
dx
2
2 = d squared y, d x squared;d y
dx
3
3;…
n
nd ydx
Function: "f x = f double prime of x; '''f x ; IVf x ;… nf x
Combination: y'' = y double prime
OBJECTIVE Find higher order derivatives.
Ex 1 d
x x x xdx
2
4 3 22 7 3 2 5
d d dx x x x x x x x
dx dxdx
dx x x
dx
24 3 2 4 3 2
2
3 2
7 3 2 5 7 3 2 5
4 21 6 2
x x 212 42 6
67
Ex 2 Find d y
dx
3
3 if sin y x 3
sin
cos cos
sin sin
cos cos
y x
dyx x
dx
d yx x
dx
d yx x
dx
2
2
3
3
3
3 3 3 3
3 3 3 9 3
9 3 3 27 3
More complicated functions, in particular Composite Functions, have a complicated process. When the Chain Rule is applied, the answer often becomes a product or quotient. Therefore, the 2nd Derivative will require the Product or Quotient Rules as well as, possibly, the Chain Rule again.
Ex 3 xy e23 , find y''.
x x
x x
x x
x
dye x xe
dx
d yx e x e
dx
x e e
e x
2
2
2
2
2 2
2 2
2 2
2
3 3
3 3
3 3
3
6 6
6 6 6
36 6
6 6 1
Ex 4 siny x 3 , find y''
' sin cos
'' sin sin cos sin cos
sin cos sin
y x x
y x x x x x
x x x
2
2
2 2
3
3 6
3 2
68
Ex 5 lnf x x x 2 3 1 , find ’’f x .
’
’’
xf x x
x x x x
x x x xf x
x x
x x x x
x x
x x
x x
2 2
2
22
2 2
22
2
22
1 2 32 3
3 1 3 1
3 1 2 2 3 2 3
3 1
2 6 2 4 12 9
3 1
2 6 11
3 1
Ex 6 g x x 24 1 , find ’’g x .
’
’’
xg x x x
x
x x x x
g x
x
xx
x
x
x x
x
x
122
122
1 12 22 2
122
2122
122
2
2 2
322
322
1 44 1 8
2 4 1
14 1 4 4 4 1 8
2
24 1
164 1 4
4 1
4 1
4 1 4 16
4 1
4
4 1
69
1.6 Homework Set A In #1-5, find the first and second derivatives of the given function.
1. 5 2( ) 6 7f x x x x 2. 2( ) 1h x x
3. 2
33 1y x 4. ( ) tan3H t t
5. 3 5( ) tg t t e 6. If 23xy e , find ''y .
70
7. If 3siny x , find ''y . 8. If ( ) cosf t t t , find '''(0).f
1.6 Homework Set B
1. Find 'f x , ''f x , and '"f x if ln secf x x .
2. Find dy
d and
2
2
d y
d for 10cot 2 1y
3. For the function tan2y x , show that 2
22
8sec 2 tan 2d y
x xdx
.
71
4. Find "f x for 2ln 3 1f x x x
5. Find the first, second, and third derivative for 2 5 6 xf x x x e
6. Find the first, second, and third derivative for tan 3 2f
72
7. A fourth differentiable function is defined for all real numbers and satisfies each of the following:
2 5g , ' 2 2g , and " 2 3g
If the function f is given by 1
2k x
f x e g x
, where k is a constant.
a. Find 1f , ' 1f , '' 1f
b. Show that the fourth derivative of f is
14 16 '''' 2k x
k e g x
73
1.6 Homework Set A
1. 5 2( ) 6 7f x x x x 2. 2( ) 1h x x
4'( ) 5 12 7f x x x 2
'( )1
xh x
x
3"( ) 20 12f x x
2
322
1
"( )
1
xh x
x
3. 2
33 1y x 4. ( ) tan3H t t
2
133
2
1
dy x
dxx
2'( ) 3sec 3H t t
32
2 433
2 2
1
x xd y
dxx
2 "( ) 18sec 3 tan3 H t t t
5. 3 5( ) tg t t e 6. If 23xy e , find ''y .
2 5'( ) 5 3tg t t e t 23 2'' 6 6 1xy e x
5 2 "( ) 25 30 6tg t te t t
7. If 3siny x , find ''y . 8. If ( ) cosf t t t , find '''(0).f
2 2'' 3sin 2cos siny x x x '''(0) 3f
1.6 Homework Set B
1. Find 'f x , ''f x , and '"f x if ln secf x x .
' tanf x x
2'' secf x x
2''' 2sec tanf x x x
74
2. Find dy
d and
2
2
d y
d for 10cot 2 1y
220csc 2 1d
dy
2
2
280csc 2 1 cot 2 1
d
d y
3. For the function tan2y x , show that 2
22
8sec 2 tan 2d y
x xdx
.
2
2
2
22
2
2sec 2
4sec2 sec2 tan 2 2
8sec 2 tan 2
dy
dx
d y
dx
d y
dx
x
x x x
x x
4. Find "f x for 2ln 3 1f x x x
2
22
2 6 11''
3 1
x xf x
x x
5. Find the first, second, and third derivative for 2 5 6 xf x x x e
' 2 5 xf x x e
'' 2 xf x e
''' xf x e
6. Find the first, second, and third derivative for tan 3 2f
2' 3sec 3 2f
2'' 18sec 3 2 tan 3 2f
2 2 2''' 54sec 3 2 sec 3 2 2tan 3 2f
75
7. A fourth differentiable function is defined for all real numbers and satisfies each of the following:
2 5g , ' 2 2g , and " 2 3g
If the function f is given by 1
2k x
f x e g x
, where k is a constant.
a. Find 1f , ' 1f , '' 1f
b. Show that the fourth derivative of f is
14 16 '''' 2k x
k e g x
a. 1 6f , ' 1 4f k , 2'' 1 12f k
b. 1
' 2 ' 2k x
f x ke g x
2 1'' 4 '' 2
k xf x k e g x
3 1''' 8 ''' 2
k xf x k e g x
4 1'''' 16 '''' 2
k xf x k e g x
76
1.7: Derivatives of Inverse Functions We already know something about inverse functions. Exponential and Logarithmic functions are inverses of each other, as are Radicals and Powers. Inverse Functions—Defn: Two functions wherein the domain of one serves as the range of the other and vice versa. --Means: Two functions which cancel (or undo) each other. The definition gives us a way to find the inverse for any function, the symbol for
which is f 1. You just switch the x and y variables and isolate y.
Ex 1 Find f 1 if f x x 3
f x x y x 3 3
So, for f 1, x y 3
y x
f x
3
1 3
This is little more interesting with a Rational Function.
Ex 2 Find f 1 if x
f xx
1
2
x x
f x yx x
1 1
2 2
So, for f 1,
yx
y
y x y
x xy y
x y xy
x y x
xy
xx
f xx
1
1
2
2 1
2 1
2 1
2 1 1
2 1
12 1
1
77
Note that - f f x x1 . For Example, f 1
02
and - f
1 10
2.
General inverses are not all that interesting. We are more interested in particular inverse functions, like the natural logarithm (which is the inverse of the exponential function). Another particular kind of inverse function that bears more study is the Trig Inverse Function. Interestingly, as with the Logarithmic Functions, the derivatives of these Transcendental Functions become Algebraic Functions.
Inverse Trig Derivative Rules
-
-
-
sin
cos
tan
u
u
u
du D
dx u
du D
dx u
du D
dx u
1
2
1
2
12
1
1
1
1
1
1
-
-
-
csc
sec
cot
u
u
u
du D
dx u u
du D
dx u u
du D
dx u
1
2
1
2
12
1
1
1
1
1
1
OBJECTIVE Find the derivatives of inverse trig functions.
Proof that sin
dx
dx x
1
2
1
1.
sin sin y x y x 1
sin
cos
cos
xD y x
dyy
dx
dy
dx y
1
1
Note the use of implicit differentiation when we took the
derivative of sin y
78
But this derivative is not in terms of x, so we are not done. Consider the right triangle that would yield this SOHCAHTOA relationship:
1-x 2
1x
y
Note that for sin y to equal x, x must be the opposite leg and the hypotenuse is 1 (SOH). The Pythagorean Theorem gives us the adjacent leg. By CAH,
cos y x 21
Therefore,
sin
cos
dx
dx y
x
1
2
1
1
1
Note that in the proof above, when we wanted to take the derivative of sin y, we had to use The Chain Rule (since the y is a function other than x, and the derivative
of y is dy
dx.
Ex 4 - tan
dx
dx
1 43
-
tan d
x xdx
x
x
x
1 4 32
4
3
8
13 12
3 1
12
9 1
79
Ex 5 - sec
dx
dx
1 2
- sec
dx x
dxx x
x
x x
x x
1 2
22 2
22 2
4
12
1
2
1
2
1
80
1.7 Homework Set A Find the derivative of the function. Simplify where possible.
1. 1sin xy e 2. 1tany x
3. 1sin 2 1y x 4. 2 1( ) 1 tan ( )H x x x
5. 1 2cos 1y x x x 6. 2( ) arctanxf x e x x
7. 1sin 2y x 8. 21csc 1y x
81
9. 1 11cot tany x
x
10. 21cos 1y x x x
11. 1sec
0x
y for xx
12. 2 1ln 4 tan2
xy x x
1.7 Homework Set B
1. 1 3cos zy e 2. 1 2tan 1y x
82
3. 1 214sin 4
2y x x x
4. 1 1cos
1
xy
x
5. 1 1sec 4 csc 4y x x 6. 1 2 2( ) sint
f t c c tc
7. 2( ) arccosf x x x 8. 1( ) ln tan 5f x x
9. 1 1sin 5 cos 5g w w w 10. 1 2( ) sec 9f t t
83
11. 2 1ln 1 coty u u u 12. 2
1 2tan
1
x
x
ey
e
84
1.7 Homework Set A
1. 1sin xy e 2. 1tany x
21
x
x
dy e
dx e
312 2
1
2
dy
dxx x
3. 1sin 2 1y x 4. 2 1( ) 1 tan ( )H x x x
1
22
1dy
dxx x
1'( ) 1 2 tan ( )H x x x
5. 1 2cos 1y x x x 6. 2( ) arctanxf x e x x
1cosdy
dxx
2
2'( ) 2 arctan
1x x
f x e x xx
7. 1sin 2y x 8. 21csc 1y x
2
2
1 2
dy
dx x
2 2
2
1 2
dy x
dx x x x
9. 1 11cot tany x
x
10. 21cos 1y x x x
0dy
dx
2
2
3
1
dy x
dx x
11. 1sec
0x
y for xx
12. 2 1ln 4 tan2
xy x x
2
2 2
11
1
1 sec x
x
dy x
dx x
1tan
2
dy x
dx
1.7 Homework Set B
1. 1 3cos zy e 2. 1 2tan 1y x
3
6
3
1
z
z
dy e
dx e
2
1
1
dy
dx x x
85
3. 1 214sin 4
2y x x x
4. 1 1cos
1
xy
x
2
2
12 2
4
dy x
dx x
1
1
dy
dx x x
5. 1 1sec 4 csc 4y x x 6. 1 2 2( ) sint
f t c c tc
0dy
dx
2
2 2'( )
c tf t
c t
7. 2( ) arccosf x x x 8. 1( ) ln tan 5f x x
2
2'( ) 2 arccos
1
xf x x x
x
2 1
5'( )
1 25 tan 5f x
x x
9. 1 1sin 5 cos 5g w w w 10. 1 2( ) sec 9f t t
' 0g w 2 2
'( )9 8
tf t
t t
11. 2 1ln 1 coty u u u 12. 2
1 2tan
1
x
x
ey
e
213
cot1
dy uu
du u
2
2
1
x
x
dy e
dx e
86
1.8: Local Linearity and Approximations Before calculators, one of the most valuable uses of the derivative was to find approximate function values from a tangent line. Since the tangent line only shares one point on the function, y-values on the line are very close to y-values on the
function. This idea is called local linearity—near the point of tangency, the function curve appears to be a line. This can be easily demonstrated with the graphing calculator by zooming in on the point of tangency. Consider the graphs
of .y x 425 and its tangent line at x 1, .y x 75 .
x
y
x The closer you zoom in, the more the line and the curve become one. The y-values on the line are good approximations of the y-values on the curve. For a good animation of this concept, see
http://www.ima.umn.edu/~arnold/tangent/tangent.mpg
Since it is easier to find the y-value of a line arithmetically than for other
functions—especially transcendental functions—the tangent line
approximation is useful if you have no calculator.
OBJECTIVES Use the equation of a tangent line to approximate function values.
87
Ex 1 Find the tangent line equation to ( )f x x x x 4 3 22 1 at x = –1 and use it
to approximate value of ( . )f 0 9 .
The slope of the tangent line will be '( )f 1
'( )
'( )
f x x x x
f
3 24 3 4
1 3
[Note that we could have gotten this more easily with the nDeriv function on our calculator.]
( )f 1 1, so the tangent line will be
( )
or
y x
y x
1 3 1
3 2
While we can find the exact value of ( . )f 0 9 with a calculator, we can get a
quick approximation from the tangent line. If x = –0.9 on the tangent line, then
0.9 0.9 3 0.9 2 .7f y
This last example is somewhat trite in that we could have just plugged –0.9 into
( )f x x x x 4 3 22 1 and figured out the exact value even without a calculator. It
would have been a pain, but a person could actually do it by hand because of the operations involved (there is only basic arithmetic involved). Consider the next example, though.
88
Ex 2 Find the tangent line equation to xf x e 2 at x0 and use it to
approximate value of .e0 2 .
Without a calculator, we could not find the exact value of .e0 2 . In fact, even the calculator only gives an approximate value.
' xf x e 22 and ' xf e 20 2 2
f e 00 1
So the tangent line equation is y x 1 2 0 or y x 2 1
In order to find .e0 2 , we are looking for .f 0 1 . Therefore, we plug in
x = 0.1
. . .e 0 2 2 0 1 1 1 2
Note that the value that you get from a calculator for .e0 2 is 1.221403… Our approximation of 1.2 seems very reasonable.
89
1.8 Homework Set A
1. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x and use it
to get an approximate value of 1.9f .
2. Find the tangent line equation to 2
5( ) at 1F x x x
x and use it to get an
approximate value of 1.1F .
3. Find all points on the graph of 22sin siny x x where the tangent line is
horizontal.
90
4. Find the equation of the tangent line at x = 2 for 2ln 3g x x . Use this
to approximate 2.1g .
5. Find the equation of the line tangent to 25 2 tan 1g x x x when
1x . Use this tangent line to find an approximation for 1.1g .
6. Use a tangent line to find the approximate the value of .02f if
sin2 x xf x e x . Then use your calculator to find an actual value for .02f .
91
7. Find the equation of the tangent line for 2ln sin2
f x x x
for x = 1.
Plug in x = 1.1 into both the original function and the tangent line. Explain why the values are so similar.
8. Given the relation, 4 23 3x y y , which passes through the point (1,2),
find each of the following:
a. dy
dx
b. 1,2
dy
dx
c. The equation of the line tangent to the relation, passing through (1,2)
d. An approximate value for y when x = 1.01
92
1.8 Homework Set A
1. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x and use it
to get an approximate value of 1.9f .
21 75 2y x 1.9 13.5f
2. Find the tangent line equation to 2
5( ) at 1F x x x
x and use it to get an
approximate value of 1.1F .
21
4 12
y x 1.1 2.95F
3. Find all points on the graph of 22sin siny x x where the tangent line is
horizontal.
2 ,3 , 2 ,12 2
n n
4. Find the equation of the tangent line at x = 2 for 2ln 3g x x . Use this
to approximate 2.1g .
4 2y x 2.1 .4g
5. Find the equation of the line tangent to 25 2 tan 1g x x x when
1x . Use this tangent line to find an approximation for 1.1g .
7 4 1y x 1.1 7.4g
6. Use a tangent line to find the approximate the value of .02f if
sin2 x xf x e x . Then use your calculator to find an actual value for .02f .
.02 2.04; .02 2.021...f f
93
7. Find the equation of the tangent line for 2ln sin2
f x x x
for x = 1.
Plug in x = 1.1 into both the original function and the tangent line. Explain why the values are so similar.
1y x 1.1 .1; 1.1 .090f f
The results are so similar because of local linearity. The tangent line is extremely close to the curve near the point of tangency, so values of y for the tangent line serve as good approximations for values of f near the point of tangency.
8. Given the relation, 4 23 3x y y , which passes through the point (1,2),
find each of the following:
a. 34
3 2
x
y
dy
dx
b. 1,2
4dy
dx
c. The equation of the line tangent to the relation, passing through (1,2)
2 4 1y x
d. An approximate value for y when x = 1.01
2 4 1y x
94
Chapter 1 Test
1.
xdx x
dx xx
45 333 4
1 13 2 3
56
2. tan cot x
yx
1 1 3
3; find y' .
3. xD x x
3 5
4 77 5 2 1
95
4. Find the equation of the line tangent to 3 4
2 1f x x at x = 1 and use it to
estimate f (0.98).
5. ln z
g zz
2
32
2; Find g'(w) and g'(4).
6. Find the equation of the line tangent to lny x 5
24 3 at x 1.
96
7. Find "f
6 if sin 3f t t 27 .
8. cosxdx e
xdx
2
7 3 52
12
9. Find dy
dx and the equation for the tangent line for the relation
3 2 65 7x x y y passing through the point (3,1).
97
Chapter 1 Test
1.
xdx x
dx xx
45 333 4
1 13 2 3
56
lnxx x x x
714 44 33 4 315 3 3
2 9 5
2. tan cot x
yx
1 1 3
3; find y' .
dy
dx x 2
18
9
3. xD x x
3 5
4 77 5 2 1
x x x x x 2 4
3 4 7 7 314 7 5 2 1 77 25 6
4. Find the equation of the line tangent to 3 4
2 1f x x at x = 1 and use it to
estimate f (0.98).
1 24 1y x 0.98 0.52f
5. ln z
g zz
2
32
2; Find g'(w) and g'(4).
'w w
g ww w
2
2
2 4
3 2 2 'g
5
364
6. Find the equation of the line tangent to lny x 5
24 3 at x 1.
80 1y x
7. Find "f
6 if sin 3f t t 27 .
"f
126
6
98
8. cosxdx e
xdx
2
7 3 52
12
cos sinxx x e
x xx
9 4 5
4
1 142 25
9. Find dy
dx and the equation for the tangent line for the relation
3 2 65 7x x y y passing through the point (3,1).
25 3
2 6
dy x
dx y
Tangent Line: 1 1 3y x
