3.1 The Derivative and the Tangent Line Problem Apply ... 03 Homework Complete Solutions V2 Part 1:...

Ch 03 Homework Complete Solutions V2 Part 1: S. Stirling Name _________________________________

Calculus: Early Transcendental Functions, 4e Larson

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3.1 The Derivative and the Tangent Line Problem

Apply & Practice 3.1 Set 1: P 123-124

#1-4 Estimating slope & concept.

#6, 8, 9 slope @ point use diff. quotient if necessary.

The slope as x 3

should get less

steep. Draw secant

lines if needed.

Slopes are

estimated by

using the grids to

approximate the

slopes.

Draw secant line

and tangent line if

needed.



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8. Find slope of the tangent line to 2( ) 5g x x

at 2,1

9. 2( ) 3f t t t at 0,0

You MUST use0

( ) ( )( ) lim

x

f x x f xf x

x

even if you know the slope!

So slope = 3/2 at any point.

Note: 2x and 22 5 2 1g

Numerator:

3 3 2 3 2 3x x x x

Gives 0/0, so you

would have to use

knowledge of the

graph here.

0

0

0

1 1 19 92 2 2lim

12lim

1 1lim

2 2

x

x

x

x x x

x

x

x

Apply & Practice 3.1 Set 2: P 124 & 126 YOU MUST USE

THE FORM THEY SPECIFY!

#12-20 (even) Use diff. quotient!

#71, 74, 76 Use alternate form!

#30 VIP equation of the tangent

line.



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Note: 22 2 1 3f

You MUST use( ) ( )

( ) limx a

f x f af a

x a

!

Note: 31 1 2 1 3f

Note: 1 1 3

3 3

x

x x

and multiplied by the reciprocal

Expanding Shortcuts are helpful here:

3 3 2 2 33 3a b a a b ab b and

2 2 22a b a ab b

Use Rational Root Thm. and

synthetic division to factor.

1 1 0 2 3

1 1 3

1 1 3 0



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30. Find the equation of the tangent line to the curve, ( ) 1f x x , at 5, 2 . (b) graph and (c) confirm

on calculator.

The equation of the tangent line:

1

5 24

y x

Again, AP loves this form! Don’t

change it to slope-intercept.

(c) nDeriv(√(X – 1), X, 5) returns .25 or 5X

dX -1

dx

And to get (b), do Y1 = √(X – 1) and Y2 = .25(X – 5)+2



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#37-40, 43-48 Given f find graph f ʹ

#49-52 writing

#53-56 sketch

#59 concept

#63, 68 make good sketches of what you see!

Constant slope horizontal line.

( )f x x ( ) 1f x

Negative slope when x < 0

Zero slope when x = 0

Positive slope when x > 0

2( )f x x ( ) 2f x x

At x = 0, slope

Slope decreases as x

Always a positive slope

( )f x x ( ) 1 2f x x

Slope constant –1 from left.

Slope constant 1 from right.

At x = 0, slope DNE.



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49. Compare to the difference quotient

5 3(1 ) 2 ( ) ( )x f x x f x

x x

( ) 5 3(1 )f x x x so 1x

guess ( ) 5 3f x x

check (1) 5 3(1) 2f or ( ) 2f x

So ( ) 5 3f x x and 1c

Also…

1

2

x

y or as long as the

base is between 0 and 1.

(Exponential decay.)

Or 1

yx

…

Or Any function that is

always decreasing.

Also…

2xy or as long as the

base is greater than 1.

(Exponential growth.)

Or lny x or any log

Or y x

Or Any function that is

always increasing.

Great AP problems!

50. Compare to the difference quotient

3

2 8 ( ) ( )x f x x f x

x x

3( ) ( 2 )f x x x so 2x

guess 3( )f x x

check 3( 2) ( 2)f or ( ) 8f x

So 3( )f x x and 2c

51. Compare to the alt. definition 2 36 ( ) ( )

6

x f x f c

x x c

so 6c and 2( )f x x

check 2(6) (6)f or

( ) 36f c

So 2( )f x x and 6c

52. Compare to the alt. definition

( ) ( ) 2 2 9

9

f x f c x

x c x

so 9c and ( ) 2f x x

check (9) 2 9f or

( ) 2 9f c

So ( ) 2f x x and 9c



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On , 2 , f increases quickly then levels off, slope large positive to

small positive.

At x = –2, the slope, f , is zero. Likewise at x = 2.

On 2,2 , f decreases slowly to quickly to slowly. Slope slightly negative to really negative to slightly negative.

On 2, , f increases slowly to quickly. Slope small positive to large positive.

Y1 = X^3/4 – 3X

Y2 = nDeriv(Y1, X, X)

Zeros of Y1:

X = –2, X = 2

On [4, 6] , g’ positive so g is increasing

You don’t have any values for g, so the function can be

any vertical translation (shift) of g.

Graph of the derivative!

Positive slopes g increasing

Negative slopes g decreasing

Almost the derivative!

Secant slope for a small x .



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Apply & Practice 3.1 Set 4: P 124

Alt. form of derivative

#75, 78, 80

#81-86 Differentiable? from graphs

#91, 92, 94, 96 Differentiable? analytically

#100-102 T or F

4x , ( 4)

( ) 1( 4)

xf x

x

4x , ( 4)

( ) 1( 4)

xf x

x

Also, use the graph of the absolute

value shifted right 4.

Think about the graph and get

numerator to be a constant

1x x x

x x x x x

,

0x , 1( )0

g x

0x , 1( )0

g x

x

y

x

y

13

3 23 3 33

3 1lim lim

33x x

x

xx

3x , 1( )0

f x

Think about the graph too!

Or infinite discontinuity, vertical asymptote.

Or infinite discontinuities, vertical asymptotes.

Definition:

, 0

, 0

u uu

u u



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Note: (1) 1f and 1 1

lim lim 1 (1)x x

x x f

So f continuous.

Always test for continuity first, because if it’s not continuous, it’s not differentiable.

Limit does not exist at x = 0, jump.

2

1 1

( ) (1) 1 0lim lim

1 1x x

f x f x

x x

2 2

21

1 1lim

1 1x

x x

x x

Makes the numerator a perfect square!

2

2 21 1

1 1 1lim lim

1 1 1 1 1x x

x x x

x x x x

21

1lim

1x

x

x

Now direct sub.

2

1 1 #

01 1

(A vertical tangent.)

The limit from the right does not exist since f is defined for x > 1.

Therefore, f is not differentiable at x = 1.

1x , ( 1)

( ) 1( 1)

xf x

x

1x , ( 1)

( ) 1( 1)

xf x

x

Also, use the graph of the absolute

value shifted right 1.

Definition:

, 0

, 0

u uu

u u

Also, use the graph. Semi-circle with a radius of 1.



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101. False. If the derivative from the left of a point

does not equal the derivative from the right of a

point, then the derivative does not exist at that

point. For example, if ( )f x x , then the

derivative from the left at x = 0 is –1 and the

derivative from the right at x = 0 is 1. At x = 0,

the derivative does not exist.

102. True. Differentiability implies continuity.

Check continuity first!

(2) 2(2) 2f

2lim 2 2(2) 2x

x

2

1lim ( ) 2 1 2

2xf x

So 2

(2) lim ( ) 2x

f f x

.



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3.2 Basic Differentiation Rules and Rates of Change

Apply & Practice 3.2 Set 1: P 136

#1-2 From graph, est. & find slope

#3-24 (mult. of 3), #7 derive

#27-30 derive

#33, 36, 38 find slope

#40-52 (even)



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Y1 = –.5+(7/5)X^3

nDeriv(Y1, X, 0) or 1

0X

dY

dX

Y1 = 2+3cos (X)

nDeriv(Y1, X, π) or 1

X

dY

dX

Y1 = – 4e^(X)

nDeriv(Y1, X, 1) or

1

1X

dY

dX

CAUTION! You do not have a product or quotient

rule (yet) so remember to rewrite functions in a form

that you can work with!



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#53, 56 equation of tang. line

#57, 58, 59, 61 horiz. tang.?

#63, 66 tang. lines

#75 show no horiz. tangents

#67-72 conceptual

#83-88 conceptual T or F

Tangent Line:

5( 1) 2y x AP

or 5 3y x

For AP → 1 1

12 2

y e t e

Hint: Horizontal tangents happen when the derivative

(the slope) of the tangent line is zero. So, find the

derivative, set it equal to zero, and solve!

For x = 0, 4 2(0) 8(0) 2 2y

For x = 2, 4 2(2) 8(2) 2 14y

For x = –2, 4 2( 2) 8( 2) 2 14y

Remember 2

1yx

would have a

horizontal asymptote at y = 0 and the

slope of this function never reaches zero.

For x = , sin( )y

To find the matching

y-value when x = ln 4, ln 44(ln 4)

4 ln 4 4

y e

Y1 = X^4 – X

Y2 = -5X – 3

nDeriv(Y1, X, -1)

or 1

1X

dY

dX

You should get –5!

You need point notation!

You may use your calculator

to find function values.

The numerator ≠ 0,

so the derivative ≠ 0.



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Hints: If the line is tangent to the function:

They share a point in common → equate the functions.

The slope of the function at that point = the slope of the tangent

line → equate the derivatives (slopes =).

Gives two equations with two unknowns → solve the system.

Use the substitution or elimination methods.

Analytically:

OR Since 1 cos 1x

3 1 3 cos 3 1x

2 ( ) 4f x and therefore ( ) 0f x .

2 3 4 2k 2 3 4 10k

2 4 4k

The easy way: Solve for k, then substitute into

the other equation.

2 4k x now substitute into 2 4 9x kx x

2

2 2

2

2 4 4 9

2 4 4 9

9

3

x x x x

x x x x

x

x

So when 3x

2 3 4 2k

So when 3x

2 3 4 10k

If you don’t think ahead!! The hard way!! UGH: Solve for x,

then substitute into the other equation. THE HARD WAY is

22

kx now substitute into 2 4 9x kx x

2

2 2

2

2

2 2 4 2 92 2 2

1 12 4 2 2 8 9

4 2

12 5 0

4

8 20 0

10 2 0

k k kk

k k k k k

k k

k k

k k

So 10k or 2k

x

y

Show that f x cannot = 0!



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Note how the write that the derivative is positive,

0f or ( ) 0f x . Practice this!

Think: 0f → “The function is increasing”

“rate of change (slope) is decreasing.” →

f goes from steep to less steep.

g(x) is f(x) translated up 6 units. So the slopes

of the tangent lines at each x-value would be

the same.

g(x) is f(x) stretched up by a factor of 5 and

reflected over the x-axis. So the slopes of the

tangent lines would be opposites and 5 times

steeper.

Greatest?

positive >

negative.

(c) There is a tangent line

parallel to the secant line.



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Note how they generalize this. It’s

actually easy to generate.

Other answers for the criteria are

possible for #71 & 72.

Really great problems!



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Apply & Practice 3.2 Set 3: P 138-139 #81 linear approximation #90, 91 avg. velocity vs. instant. rate #93, 96 falling objects

#97-100 graphs of position and velocity (watch units! Velocity in mph.)

#102, 103, 104, 107

#111, 113

You could now finish the activity sheet we started in Chapter 2!

Important point on linear approximations: The

more curved the function is, the less accurate the

linear approximation will be.



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10 cos cos 0 13 3 2

03 3 3

f f

= 31 0.47752

Compare: 0.886 0.478 0

Conclusion: The average rate of change is

between the instantaneous rates of change at the

endpoints. Check out the graphs to confirm.

Note: “average rate of change” ≠ “the average of

the instantaneous rates of change”.

1 4.7182.718 2.859

2

0 sin 0 0f

3sin 0.866

3 3 2f

1 2 4.718g e

Compare: 1 2.718 4.718

“dropped” so 0 0v

“hit the water” when ( ) 0s t



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97.

When sketching, you must label your axes (and

not x and y) and establish the scale on each.

2 mi 60 min mi30 hr4 min 1 hr

t is in minutes, rates must

be converted to mph!


You don’t really

need to find the

functions to graph

these! Endpoints should be

open dots!

There are sharp points

on the (time, distance)

graph, so the

derivative (velocity)

would not exist!

98.


on [0, 6)


on (8, 10]

FYI, to make conversions: 40 mi 1 hr 2 mi

min1 hr 60 min 3

on [2, 6)

Plot (6, 4), draw seg.

Change in miles is 0.

Change in miles is 2.

So from (8, 4), draw

segment to (10, 4+2).

Note: Constant rates of change, velocity, so

distance function are all linear pieces!



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On [2, 4) change in time is 4. 30 mi 1 hr 1 mi

1 hr 60 min 2 min

velocity

1 mi 4 min2 mi

2 min 1 change in distance.

Plot (0+4, 0+2) or (4, 2). Make segment.

On (4, 6) change in time is 2.

velocity is 0. No change in distance.

Plot (4+2, 2+0) or (6, 2).

On (6, 10] change in time is 4. 60 mi 1 hr mi

1 1 hr 60 min min

velocity

mi 4 min1 4 mi

min 1 change in distance.

Plot (6+4, 2+4) or (10, 6).

You don’t need to find the

functions to graph these!

Or think units: 15000 mi 1 gal $1.55

( )1 yr mi 1 gal

23.250 $

yr

C xx

x

C is the annual cost.

migal

$yr

The rate of change in the cost with

respect to the mpg (fuel efficiency).

The rate of change in the cost at 15 mpg is decreasing

at a faster rate than at 35 mpg. (See above.) Both

decreasing, but the absolute value is greater.

The driver who gets 15 mpg would

benefit more, because the rate of change

at x = 15 is larger in absolute value than

at x = 35.



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Let 3( ) 9y f x x x

So 2( ) 3 9f x x

Tangent line(s) through 1, 9 would satisfy the equation:

1 9y m x would need the slope = the derivative…

2

3 2

3 2

3 9 1 9

3 3 9 9 9

3 3 9

y x x

y x x x

y x x x

just simplifying

Now you’re stuck, but the tangent line and the function have to

share a point…so the y-values would be the same for some x:

Now 3 3 2

3 2

9 3 3 9

0 2 3

x x x x x

x x

Solve for x.

20 2 3x x So 0x or 32

x so you get two points.

(2 points on the curve where tangent lines could occur. So you will

have two lines.)

For x = 0: 3(0) (0) 9(0) 0f the point

2(0) 3(0) 9 9f the slope

Tangent line through 0,0 is 9 0 0

9

y x

y x

For x = 3/2: 3

3 3 3 27 27 8192 2 2 8 2 8

f

2

3 3 27 36 93 92 2 4 4 4

f

Tangent line through 3 81,2 8 :

9 3 81

4 2 8y x

xy

The given point is not on

the curve!

(2)f



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Intentionally left blank



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Note: The book does their original set up a tiny bit differently

than we do (but they are equivalent)! So check your final

answers, and if you don’t get it correct, look at the process.

*4. Not using the product rule: 51

2 2( ) 4g s s s just distribute!

312 2

4 5( )

2 2g s s s

Power rule

1

221

( ) 4 52

g s s s

Factor out fractions and negative exponents

2

12

4 5( )

2

sg s

s

Equivalent and easier!

Remember this from the summer assignments?


#4, 6, 10, 12, 13, 16, 20, 22, 24, 26

#27-57 (mult. of 3)

#65, 66

Don’t expand! Product rule is quicker!

On All of these, use function notation to

label your work!!

10. (my way)

12

( )1 1

s sh s

s s

1 12 2

21

2

11 12

( )

1

s s s

h s

s

1 1 12 2 2

2 21 1

2 2

1 11 12 2

1 1

s s s

s s

12

2 21

2

1 222

2 11

ss

ss



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*30. Alternate: multiply then quotient rule

(easier!!) Get one rational 1st! 5 4

4 1( )

1 1

x x xf x x

x x

5 4 5 4

2

1 1( )

1

x x x x x xf x

x

4 3 5 4

2

5 4 1 1

1

x x x x x

x

5 4 3

2

4 2 4

1

x x x

x

3 2

2

2 2 2

1

x x x

x

22 1

1 1 1 1

x

x x x x

2

2, 1

1x

x

They’re gonna use Product rule…BIG mistake!

They skipped

lots of algebraic

steps here!

33. Probably easier not using Product Rule.

Just expand. Easiest is the Chain Rule, but you

haven’t learned that yet.

Distributing will get

you down to 2 terms,

so easier. Product

Rule would be a

disaster!

Just use the

Difference Rule now,

keeps the expressions

easy to simplify.



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Could do this with fractional exponents!

Note 1 2

2

1ds s

dx s

Used 2 21 cot csc . You may not need to simplify it this far.

*54. Alternate: Group factors 1st!

( ) 2sin cosf x x x

2

( ) 2cos cos 2sin cos

2cos 2sin cos

f x x x x x

x x x

Either of these expressions are good.

They used 2 2cos sin cos 2 .

To see of this is equivalent to 2cos2x , put both into

Y= and check that you are getting the same values in

the Table. Or graph, and it should be the same.

*57. Alternate: Fractional exponents & Product Rule

12

12

1( )

44

xxe

f x e xx

312 2

32

32

1 1 1( )

4 4 2

12 1

8

2 1

8

x x

x

x

f x e x e x

e x x

e x

x

Remember this from

the summer

assignments?

They skipped

a lot of steps!

At π, on the unit circle, 1,0

2 2

1 0 1 1( )h

At π/2, on the unit circle, 0,1

( / 4) 1 0 1f



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Apply & Practice 3.3 Set 2: #68, 71, 80, 82, 91, 94, 97, 100, 101, 103, 106, 107

#109 – 117, 119, 121, 122, 137, 138, 139

The main idea? Find the derivative and

set it equal to zero.

cos sin 0xe x x

since 0xe just look at cos sin 0x x

cos sinx x in quadrant II, at 3

4x

2 1

29 3

y x

Rate = derivative.

Tan negative in

QII and QIV

Always write your answer as a point! x- and y- value!! 3 3

4 43 3 2sin

4 4 2f e e



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Translation:

point (2, 0)

f decreasing on (–, 2)

f increasing on (2, )

Translation:

f always positive

f always

decreasing



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(b) The particle speeds up (accelerates) when

a > 0 and slows down when a < 0.

Answers will vary.

s = position

v = velocity

a = acceleration

v +

a – slow dwn

v –

a – spd up

v –

a + slow

dwn

v +

a + spd

up

2

2

m3 36 3 27 sec

m3 2 3 6 sec

v

a

When velocity & acceleration have opposite signs,

the object is slowing down, speed decreasing

(opposite forces). Think gravity & positive velocity.



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Challenge (optional):

1) Set Up, use what you know:

x-intercept at (1, 0): 2(1) (1) (1) 0

0

f a b c

a b c

(2, 7) on graph: 2(2) (2) (2) 7

4 2 7

f a b c

a b c

Slope 10 at (2, 7): (2) 2 (2) 10

4 10

f a b

a b

2) Solve Simultaneously:

0

4 2 7

3 7

a b c

a b c

a b

Now

3 7

4 10

3

a b

a b

a

And

3(3) 7

2

b

b

And

3 2 0

1

c

c

3) The equation is

2( ) 3 2 1f x x x

The main idea? Find the derivative.

Slope through ,1

xx

x

and 1,5 must =

the derivative of f. (the slope of the tangent

line)

Slope through 2 points = tangent slope:

2

511

1 1

x

x

x x

cross mult.

2

2 2

2

2

1 5 1 1 1

5 10 5 1 1

4 9 5 1

4 10 4 0

x x x x

x x x x x

x x x

x x

2 1 2 0

1 , 22

x x

x

Find points and slopes, then lines:

1

21 12 1 1

2

f

2

11 42 1 1

2

f

1

4 12

y x

and

2

2 22 1

f

2

12 1

2 1f

1 2 2y x



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18. Alternative: 2( ) 2 1g x x x Chain Rule

1

2 21( ) 2 1 2 22

g x x x x

1

2 2

1( )

( 1)

xg x

x

= ± 1 b/c 2x x or x


#9-36 (mult. of 3), 17 derive

#39, 42 just find derive. by nderiv

#101, 150

Writing & Conceptual #139-145; #172-173

Whole class (if time): #152, 165

Rewriting with negative exponent

& using Chain Rule is easier &

quicker than the Quotient Rule.

Can use: 2u u

Then knowledge of

the graph.

Again Chain Rule is easier

than the Quotient Rule.

27. 42( ) 2f x x x

Main property, Product rule:

f a b

f a b a b

4 32( ) 2 2 4 2 1f x x x x x

4 32

3

3

( ) 2 2 4 2

2 2 2 2

2 2 3 2

f x x x x x

x x x x

x x x

For b , you should

apply the Chain Rule

4

3

2

4 2 1

b x

b x

30. 2 2116

2y x x

1

2 2 2116

2y x x

Main property, Product rule:

f a b

f a b a b

1 12 2 22 2

1 12 3 22 2

12 2 22

2 2 2

1 12 22 2

1 116 16 2

2 2

116 16

2

116 2 16

2

32 2 32 3

2 16 2 16

y x x x x x

x x x x

x x x x

x x x x x

x x

For b , you should apply

the Chain Rule

12 2

12 2

16

116 2

2

b x

b x x

#27 & 30, give your answers in simplified factored form.



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33.

2

2

5( )

2

xg x

x

Main property, Chain Rule:

2g x u

2g x u u

2

22 2

2

32

5 1 10 22

2 2

2 5 2 10

2

x x xg x

x x

x x x

x

For u , you should apply the

Quotient rule

au

b

2

a b a bu

b

2

5

2

xu

x

2

22

2 2

22

2

22

1 2 5 2

2

2 2 10

2

1 10 2

2

x x xu

x

x x x

x

x x

x

Similar to #33, but this time I did most of the work within the

problem and some in my head. Do what you need to do!

Straight forward Chain Rule.



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x

y

FYI only

FYI only

101. Alternative:

22

22

( ) 6 1 2

12 1

f x x x

x x

22 2

22 2 2

2 2 2

2 2

( ) 12 1 12 2 1 2

12 1 48 1

12 1 1 4

12 1 5 1

f x x x x x

x x x

x x x

x x

Gets it into factored form easier than the other

method. This will be helpful when trying to find

horizontal tangents.

Using the Product

Rule w/ Chain

Rule of course.

Using the

distributive

property.

(a&b) Use Y1 = (X^2 – 9)√(X+2)

Y2 = nDeriv(Y1, X, X)

Y2(2) 6.75 6.75 2 10y t

FYI only



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The language here is very important! “The derivative is negative, so f is decreasing”…

AP Loves these!

Note: at the endpoint of f, there

would be a vertical tangent line.

So on f , there is a vertical

asymptote there.

Note: Horizontal compression

would make the graph steeper and

the tangent lines steeper. 3 times

steeper.



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AP Loves these too!



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OR Use Y1 = (cos(πX)+1)/X

nDeriv(Y1, X, X) graph

Use transformations to confirm your results.


#45 Use the calculator.

#48, 50, 51 Sketch graphs too.

#57-99 (mult. of 3) Skip #93!

#103-105 Second derivatives.

#114, 118, 119 Tangent lines & evaluate



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#104 is on the next page

USE LOG PROPERTIES

to your advantage!!!

They factored and used the Product Rule only

once. If you didn’t, you would have to use the

Product Rule on each of two terms. That’s the

hard way.

Note: 3e is a

number, a scalar

multiple. No

Product Rule here.



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12 24

y x

3 2 2

2 4 2y x

2

( ) sec( )f x x



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You will need the identity: 2cos(2 ) 1 2sinx x

Apply & Practice 3.4 Set 3: #153, 158 Tangent lines & evaluate

#160 Application

#183-185 T or F

#123-129, 131-133, 135 Deriv. other bases.

#174 prove deriv. of |x|

#175 – 177 apply deriv. of |x|

#179, 181 a & d Do the linear approx. ONLY



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USE LOG PROPERTIES to your advantage!!!



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Exercises 179-181 Find linear approximation at the given a value.

3.1 The Derivative and the Tangent Line Problem Apply ... 03 Homework Complete Solutions V2 Part 1:...

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