4.1 Day Calculus - Ms. Carignan's Math Site · 2018. 10. 29. · AP Calculus (Ms. Carignan) Chapter 4 :More Derivatives (C30.4 & C30.7) Page 4 Ex #3: If y u u 10 52 where ux 132,

AP Calculus (Ms. Carignan) Chapter 4 :More Derivatives (C30.4 & C30.7) Page 1

I CAN TAKE DERIVATIVES USING THE CHAIN RULE.

VIDEO LINKS: a) https://goo.gl/8rSQrd b) https://bit.ly/2O2jQf1 c) https://goo.gl/cJnw2Z d) https://bit.ly/2OLghaf (Up to time 3:18)

A Calculus 4.1 Day 1—The Chain Rule P

THE CHAIN RULE:

When we have a function within a function, we need an additional rule called the CHAIN RULE. The chain rule must be used when we have a function whose derivative is not 1 within another function such as a power, a root, or a combination of those. Here are some examples of functions within functions (composite functions)

6

5 4y x 7

57 4y x 3

57 8 2y x x

We often rewrite the function by rewriting the function using a substitution of u for the “inner” function

6

let 5 4u x

y u

5

7

let 4

7

u x

y u

5

3

let 7 8 2u x x

y u

Once we have made the substitution, we take the derivative using the chain rule. Typically, our

functions contain x and y variables, so we usually find the derivative of y with respect to x, as indy

dx .

As we have changed the variable from x to u in the above questions, we need to adapt the formula to now take the derivative of y with respect to u while taking into account that u is based on a foundation of the x variable. The chain rule will produce the following:

dy dy du

dx du dx

This means: To find the derivative of the original function y, with respect to the original

function in terms of x, you need to 1) Find the derivative of the newly rewritten function y (containing the u) with

respect to the variable u. This is the dy

du part of

dy dy du

dx du dx

2) Multiply your answer in step 1 by the derivative of the equation u with

respect to the variable x. This is the du

dx part of

dy dy du

dx du dx

3) You now need to go back and replace any remaining u values with the equation that it represents in terms of x. We will often leave this answer fairly unsimplified and it may often contain negative or rational exponents.

If we think of the situation as actual composition of functions, we have the following rule: If ( ) ( ( ))F x f g x or ( ) ( )( )F x f g x then the derivative will be '( ) '( ( )) '( )F x f g x g x

https://goo.gl/8rSQrdhttps://bit.ly/2O2jQf1https://goo.gl/cJnw2Zhttps://bit.ly/2OLghaf


Ex #1: Find dy

dx if

3 10(2 4 3)y x x

STEP 1: Let u = Therefore our function now looks like

STEP 2: Find the derivative of each of the above boxes with respect to their proper variables:

du

dx

dy

du

STEP 3: To find the final answer, dy

dx, multiply the above two derivatives together. We usually multiply them in reverse

however and put dydu

first and du

dx second

dy

dx dy

du

du

dx =

STEP 4: Replace any value of u with the equation in . STEP 5: Simplify where appropriate. Combine like terms or expressions/ factor and reduce etc.

How can you write this rule differently?

THE CHAIN RULE: (Think back to the composition of functions: Chapter 10 of PC30)

If ( ) ( ( ))F x f g x , then '( ) '( ( )) '( )F x f g x g x .

In other words: (derivative of outside function)•(derivative of inside function)

THE POWER RULE COMBINED WITH THE CHAIN RULE:

If ( ) ( )n

F x f x , then 1

'( ) ( ) '( )n

F x n f x f x

Reminder: POWER RULE:

If ( ) nf x x ,

1'( ) nf x nx


Ex #2: Differentiate 5 23 (2 1)y x

b) 3 4

1

1f x

x


Ex #3: If 10 5 2y u u where 21 3u x , find

1x

dy

dx

by first using dy dy dudx du dx

.

Now, when 1,x u

So, 1x

dy

dx

=

Ex #4: A table of values for 𝑓, 𝑔, 𝑓′, 𝑔′ is shown below

x f(x) g(x) f’(x) g’(x)

3 5 6 8 4

4 1 3 5 2

5 3 9 7 10

a) If 𝐹(𝑥) = (𝑓°𝑔)(𝑥), 𝑓𝑖𝑛𝑑 𝐹′(4) b) If 𝐺(𝑥) = (𝑔°𝑓)(𝑥), 𝑓𝑖𝑛𝑑 𝐺′(5)


Ex #5: find the coordinates of the point (s) on the curve 𝑓(𝑥) = √𝑥2 − 4𝑥 + 33

at which the tangent line is

a) horizontal and b) vertical. https://www.desmos.com/calculator/ema32j8qcg

4.1 Day 1 Assignment: Textbook Page 158 # 13, 16, 19, 20, 33-38 PLUS Additional Assignment Below: #1, 5, 9, 13, 17, 21, 25, 26, 28

https://www.desmos.com/calculator/ema32j8qcg


SOLUTIONS TO ADDITIONAL ASSIGNMENT 4.1 DAY 1

I CAN TAKE DERIVATIVES OF TRIGONOMETRIC FUNCTIONS USING THE CHAIN RULE.

VIDEO LINKS: a) https://bit.ly/2vIGaQp b) https://bit.ly/2OLghaf (Start at time 3:18)

Trig functions: You must know all of these derivatives

by memory - they will not be given on the AP exam.

Ex #1: Find the derivative of the following:

a) 𝑓(𝑥) = sin(25𝑥) b) 𝑦 = 3 cos 7𝜃

c) 𝑓(𝑥) = 9 sin(2𝑥4) d) 𝑦 =1

√sin 10𝑥

A Calculus 4.1 Day 2—The Chain Rule and Trigonometric Functions P

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e) 4( ) sin 3f x x

Ex #2: Find 𝑑𝑦

𝑑𝑥 for the equation 𝑦 = 5 cot (

2

𝑥)

Ex #2:

a) Find the slope of the line tangent to the curve 𝑦 = sin5 𝑥 at the point where 𝑥 =𝜋

3

b) Show that the slope of every line tangent to the curve 𝑦 =1

(1−2𝑥)3 is positive.

https://www.desmos.com/calculator/ghoibua2fl

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SOLUTIONS TO ADDITIONAL ASSIGNMENT 4.1 DAY 1

4.1 Day 2 Assignment: Textbook Page 158 # 1, 3, 4, 9, 12, 16, 25, 31, 55, 63 PLUS 4.1 Day 2 Additional Assignment Below: #19, 26, 28, 33, 36, 37, 38, 39


REVIEW: Converting negative exponents to positive exponents

Ex #1:

5

2

2 1

1

x

x

Ex #2:

2

3

4 2

( 1)

3 ( 2)

x

x x

Ex #3:

3

3 221

( 1) 3 (2 3)2

x x x

Factoring (with rational/negative exponents):

Ex #4:

3 1 1

2 2 22 4 6x x x

Ex #5:

3 1

3 1 24 42 ( 2) ( 1) 4 ( 2)( 1)x x x x x x

I CAN TAKE DERIVATIVES USING THE CHAIN RULE WITH MULTIPLE CHAINS.

VIDEO LINKS:

a) https://bit.ly/2NCSpcA b) https://bit.ly/2PZGG4b c) https://bit.ly/2Q0ZDnr

Ex #1: Differentiate:

a) 3 42 1 2 3y x x

A Calculus 4.1 Day 3—The Chain Rule with Double or Triple Link Chains P

Find the common factor (if there is one) : _____.

Find the smallest exponent : _____

Remember that when you divide by the

common factor, you subtract your exponent.

https://bit.ly/2NCSpcAhttps://bit.ly/2Q0ZDnr


b) 6

2 1

2

ts t

t

c) 2( ) 1F x x x

d) 2( ) tan(tan(4 ))f x x


e) 2( ) (1 2sin(2 ))f x x

f) 2 3 2( ) cos ( )f x x x

Ex #2: Certain functions can be solved using different methods. To find the derivative of the following functions,

what rules could you use? Choose the easiest option and differentiate the functions.

a) 3

3

( 2)y

x

b)

2

4

( 3)( )

( 2)

x xf x

x

Ex #2: Find the value of ( ) 'f g at the given value of x

1

( ) 1f uu

, 1

( )1

u g xx

, x = -1


I CAN TAKE DERIVATIVES USING IMPLICIT DIFFERENTIATION.

VIDEO LINKS: a) https://bit.ly/2Moq348 b) https://bit.ly/2OMpkaO c) https://goo.gl/8rSQrd

Explicit Function: A function in which the independent variable can be written explicitly in terms of the

independent variable, for example 𝑦 = 𝑓(𝑥).

Given 𝑦 = 3𝑥2 − 2𝑥 + 1 𝐹𝑖𝑛𝑑 𝑦′

4.1 Day 3 Assignment: Textbook Page p. 153 # 10, 17, 27, 37 PLUS 4.1 Day 3 Additional Assignment Below: #2, 4, 5, 6, 7, 8, 10, 13, 14, 15, 18

SOLUTIONS

A Calculus 4.2 Day 1 – Implicit Differentiation P

https://bit.ly/2Moq348https://bit.ly/2OMpkaOhttps://goo.gl/8rSQrd


Implicit Function or Relation: A function or relation in which the dependent variable is not isolated on one

side of the equation. For example 𝑥2 + 𝑦2 = 1 represents an implicit relation, specifically a circle.

Given 𝑥2 + 𝑦2 = 1. 𝐹𝑖𝑛𝑑𝑑𝑦

𝑑𝑥

Ex #1: Find an expression for each derivative.

a) 𝑑

𝑑𝑥(2𝑥11) b)

𝑑

𝑑𝑥(6𝑦) c)

𝑑

𝑑𝑥(−2𝑦5)

Ex #2: Differentiate from left to right with respect to x :

a)

1

2 4(9 4 )d

x ydx

b) 3 4(2 )

dx y

dx

Review of Notations:

Recall 𝑦′ =𝑑

𝑑𝑥 which is the operator that tells you to take the derivative with respect to x.

If you see this operation 𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥∙ 𝑦, this is implicit differentiation

3( )

dx

dx means: find the derivative of 3x with respect to x.

3( )d

xdx

4( )

dy

dx means: find the derivative of

4y with respect to x. 4( )d

ydx

This is challenging as there are no x ’s

To take the derivative of a function that is defined implicitly, we take the derivative from left to right, and

wherever there is a value of y in the equation, we need to use the chain rule and multiply that term by dy

dx . We

normally use the chain rule when taking the derivative of x values but the chain rule of those terms ends up being

dx

dx which reduces to 1.


Ex #3: Apply the derivative operator to both sides of the equation, and solve for the derivative.

a) 𝑦 = 𝑥2 b) 𝑦2 = 𝑥4

Ex #4: If 2 2 169x y , find

dy

dx implicitly. Then, find the equation of the tangent line to this circle at (12,-5).

USING IMPLICIT DIFFERENTIATION when working with EQUATIONS containing a

mixture of x and y:

STEP 1: Differentiate both sides of the equation, from left to right, with respect to x.

STEP 2: Collect all the terms with dy

dx on one side of the equation

STEP 3: Factor out the dy

dx

STEP 4: Isolate dy

dx.



𝑑𝑡 given 𝑦4 − 𝑦 = 𝑡3 + 𝑡

Ex #6: Suppose 2 22 3x y y x . Find

dy

dx.


Ex #7: Find the slope and the equation of the tangent line (in standard form) to the curve 3 2 23 3x x y y at (-1,4)

I CAN TAKE DERIVATIVES USING IMPLICIT DIFFERENTIATION.

VIDEO LINKS: a) https://bit.ly/2Dn0kWw b) https://bit.ly/2ptGO0C c) https://bit.ly/2xHs0PT

Ex #1: Find (3,0)

dy

dx given 2𝑦 + 3 = 𝑥 + sin 𝑦

4.2 Day 1 Assignment: Textbook Page 167 #1-4, 9-12, 17-22

A Calculus 4.2 Day 2 – Implicit Differentiation Cont. P

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𝑑𝑥 given sin(𝑥𝑦) + 𝑥2 = 3𝑥 − 2𝑦2

Ex #3: u-Use implicit differentiation to solve for the second derivative.

2𝑥3 − 3𝑦2 = 8


Ex #4: Find 𝑑2𝑦

𝑑𝑥2 given 3𝑦2 − 𝑥 = 𝑥2 − 𝑦

I CAN TAKE DERIVATIVES OF INVERSE FUNCTION.

VIDEO LINKS: a) https://bit.ly/2PU6K0O b) https://bit.ly/2NCtNkd c) https://bit.ly/2puUSHd

RECALL:

A function is one-to-one if for every x, there is one and only one y.

Every function that is one-to-one has an INVERSE FUNCTION that is one-to-one.

We denote an inverse as 𝑓−1(𝑥)

o NOTE: if the slope of 𝑓(𝑥) = 2 then the slope of the inverse function is 𝑓−1(𝑥) =1

2.

to find an inverse, we switch x and y and then solve for y. If graphed the original and its inverse function are symmetrical about y=x

4.2 Day 2 Assignment: Textbook Page 167 #5-8, 23-36

A Calculus 4.3 Day 1 Derivatives of Inverse Functions (CALCULUS 30L OUTCOME) P

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Ex #1: Given 𝑦 = 𝑥3, find the inverse equation. Sketch each.

How do we take the derivative of an inverse???

If f is differentiable at every point of an interval I and 𝒅𝒚

𝒅𝒙 is never zero on I, then

f has an inverse and 𝒇−𝟏 is differentiable at every point on the interval 𝒇(𝑰).

This means that we can find the derivative of an inverse function as long as the

derivative of the original function does not equation zero on a given interval.


Ex #2(30L): If 𝑓(𝑥) = √𝑥 − 3, find (𝑓−1)′(𝑥) . Use both a developmental method and the above first

formula. Which do you prefer?

Ex #3(30L): If 𝑓(𝑥) = 𝑥3 + 2𝑥 − 10, 𝑓𝑖𝑛𝑑 (𝑓−1)′(𝑥)

We also know that an inverse will “undo” a function. 𝒇 (𝒇−𝟏(𝒙)) = 𝒙

If 𝒇′(𝒙) ≠ 𝟎 for any 𝒙 ∈ 𝑹, then 𝒇−𝟏(𝒙) is also differentiable, and we can say that

(𝒇−𝟏)′(𝒙) =𝟏

𝒇′(𝒇−𝟏(𝒙))

o Another way to write this is:

𝑰𝒇 𝒈 = 𝒇−𝟏𝒕𝒉𝒆𝒏 𝒈′(𝒙) =𝟏

𝒇′(𝒈(𝒙))

o A third formula is: if y = 𝒇−𝟏(𝒙), then 𝒅𝒚

𝒅𝒙=

𝟏𝒅𝒙

𝒅𝒚

o


AP Ex #4(30L):

Let f be a differentiable function such that 𝑓(3) = 15, 𝑓(6) = 3, 𝑓′(3) = −8 𝑎𝑛𝑑 𝑓′(6) = −2. The

function g is differentiable and 𝑔(𝑥) = 𝑓−1(𝑥) for all x. What is the value of 𝑔′(3)?

AP Ex #5(30L): Suppose 𝑓(2) = 3, 𝑓(3) = −1, 𝑓′(2) =1

2, 𝑓′(3) = 4. If possible find (𝑓−1)′(3)

AP Ex #6(30L): Given the following table find 𝑔′(3).

Let 𝑔(𝑥) = 𝑓−1(𝑥)

𝑥 1 2 3 4

𝑓(𝑥) -2 3 1 5

𝑓’(𝑥) 4 -3 4 2


AP Ex #7: Use the table to find the value of (𝑓−1(−2))′

𝑥 -3 -2 -1 0 1 2 3

𝑓(𝑥) 2 4 5 -2 -3 -4 -6

𝑓′(𝑥) 8 7 0 -1 -2 -3 -3.5

4.3 Day 1 Assignment (INTEGRAL CALCULUS 30L QUESTIONS)


I CAN TAKE DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS

VIDEO LINKS: a) https://bit.ly/2MXlCtn b) https://bit.ly/2NCtNkd c) https://bit.ly/2puUSHd

The graphs of each inverse function is shown below (recall from section 1.6 that we had to restrict the domains of the original function to make it one to one – this restricts the range of the new function). Please use a highlighter to better show the appropriate domain and range for each.

DEVELOPMENT OF THE FORMULA TO FIND THE

A Calculus 4.3 Derivatives of Inverse Trigonometric Functions (CALCULUS 30L OUTCOME) P

Remember: sin 𝜃 =𝑂

𝐻 cos 𝜃 =

𝐴

𝐻 tan 𝜃 =

𝑂

𝐴

AND: 𝜃 = sin−1 (𝑂

𝐻) 𝜃 = cos−1 (

𝐴

𝐻) 𝜃 = tan−1 (

𝑂

𝐴)

RECALL FROM SECTION 1.6:

If siny x then the inverse function is sinx y or 1siny x . This function is read as y equals the inverse

of sine, or y equals arcsin

If cosy x then the inverse function is cosx y or 1cosy x . This function is read as y equals the

inverse of cos, or y equals arccos

If tany x then the inverse function is tanx y or 1tany x . This function is read as y equals the inverse

of tan, or y equals arctan

Domain where siny x is one-to-one is 2 2

x

1the range for y=sin is ,domain is -1 12 2

x y x

Domain where cosy x is one-to-one is 0 x 1the range for y=cos is 0 , domain is -1 x 1x y

Domain where tany x is one-to-one is x

1the range for y=tan is , domain is x R 2 2

x y

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DEVELOPMENT OF THE FORMULA FOR THE DERIVATIVE OF ARCSIN sinx y OR 1siny x

DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS

12

1(arcsin ) sin

1

d dx x

dx dx x

12

1(arccos ) cos

1

d dx x

dx dx x

1 21

(arctan ) tan1

d dx x

dx dx x

12

1(arccsc ) csc

1

d dx x

dx dx x x

12

1(arcsec ) sec

1

d dx x

dx dx x x

1 21

(arccot ) cot1

d dx x

dx dx x

DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS INVOLVING THE CHAIN RULE

NOTE: THESE FORMULAS NEED TO BE MEMORIZED & WILL BE TESTED

Often the x in the above formulas is replaced by another function that we will call u. In that case we have to use the chain rule and add the derivative of u to the end of each of the above formulas.

(Examples could look like the following: 1 2sin ( 3 )y x x or 1 1 3cos (sin ( ))y x or 13

1tan

3 4

xy

x

)

In general we are given functions of the form 1siny u , 1cosy u , 1tany u etc

where u some other function

12 2

1 '(arcsin ) sin

1 1

d d du uu u

dx dx dxu u

12 2

1 '(arccos ) cos

1 1

d d du uu u

dx dx dxu u

1 2 21 '

(arctan ) tan1 1

d d du uu u

dx dx u dx u

12 2

1 '(arccsc ) csc

1 1

d d du uu u

dx dx dxu u u u

12 2

1 '(arcsec ) sec

1 1

d d du uu u

dx dx dxu u u u

1 2 21 '

(arccot ) cot1 1

d d du uu u

dx dx u dx u


Ex #1(30L): If 𝑦 = sin−1 8𝑥2, find 𝑑𝑦

𝑑𝑥. Show your answer using two methods: One using the formula and one using

implicit differentiation.

Ex #2(30L): Given 𝑦 = sec−1(3𝑥 − 4), find 𝑑𝑦

𝑑𝑥.


Ex #3(30L): If 𝑥 + 𝑦 = tan−1(𝑥2 + 3𝑦), find the derivative with respect to x.

Ex #4(30L): The table below give values for f(x) and its derivative. What is the value of 1( 3) 'f ? Use two

methods to find your answer (using basic principles of inverses and using the formula (𝒇−𝟏)′(𝒙) =𝟏

𝒇′(𝒇−𝟏(𝒙))

4.3 Day 2 Assignment (INTEGRAL CALCULUS 30L QUESTIONS) P 175 1-27 odd, 35-40

Memorize the Formulas from Today


I CAN TAKE DERIVATIVES OF LOGARITHMIC FUNCTIONS

Recall: Logarithmic functions are inverses of exponential functions.

Ex #1: Find the derivative of the following logarithmic functions

a) 5logy x b)

9

6logy x

c) 3712log 4y x x d) 3

8 2

2log

3

xy

x

e) 𝑦 = 3 log4(𝑥2 − 5)3

THE DERIVATIVE OF A LOG FUNCTION WITH BASE b: y = logbu

If logay u , then the derivative is 1 '

logln ln

a

d du uu

dx u a dx u a

NOTE: THIS FORMULA NEEDS TO BE MEMORIZED & WILL BE TESTED

A Calculus 4.4 Derivatives of Exponential & Logarithmic Functions (SOME CALCULUS 30L) )OUTCOME)

P


Ex #2: Find the derivative of each of the following functions

a) f(x) = ln(x9) b) y = 8ln(4x – 5) c) y = ln7

d) 25

( ) ln2 1

xf x

x

e) 3( ) ln 6 5f x x

g) f(x) = ln[(𝑥 − 3)(𝑥 + 3)]

THE DERIVATIVE OF A LOG FUNCTION WITH BASE e: y = lnu (or y = logeu )

If log lney u x , then the derivative is 1 '

logln ln

e

d du uu

dx u e dx u e

. Since we have established

that ln 1e , we are actually left with the following derivative formula :

Given lny u , 1 '

lnd du u

udx u dx u


Ex #3: Find the slope of the tangent line to the graph of the function y = 3lnx at the point x = 4

Ex #4: Find the derivative of the function 3 2( ) 2 ln 4f x x x . Write your answer in simplified factored form.

b) Use Logarithmic Differentiation to finddy

dxif

4

3

4 3

2 5

xy x

x


SOLUTIONS

4.4 Day 1 Assignment (& Memorize Formulas)


I CAN TAKE DERIVATIVES OF EXPONENTIAL FUNCTIONS

Recall: Exponential functions are of the form: 𝑦 = 𝑏𝑥, where b the base is any positive number

except for 1.

In calculus we use the number “e” which is a constant.

o e = 2.718281828…

When we differentiate exponential functions, we will be dealing with the following 2 types:

o Those involving base “a” – which is any number other than e

o Those involving base “e”

Ex #1: Find the derivative of the following functions:

a) 54 xy b) 2 52x xy c) 3 2 56 xy x

Ex #2: Find the derivative of the following:

a) 4xy e b)

23 57 x xy e c) 23( ) xf x x e

THE DERIVATIVE OF AN EXPONENTIAL FUNCTION y = au

If uy a where u is a function of x then the derivative is ln ln 'u u ud du

a a a a a udx dx

THE DERIVATIVE OF AN EXPONENTIAL FUNCTION y = eu

If uy e where u is a function of x then the derivative is 'u u ud du

e e e udx dx

NOTE: THESE FORMULAS NEED TO BE MEMORIZED & WILL BE TESTED

A Calculus 4.4 Derivatives of Exponential & Logarithmic Functions (SOME CALCULUS 30L) )OUTCOME)

P


d) 5

xy e e)

22

2

7 xey

x f)

4xy e g) 5y e

Ex #3: Find the following derivatives

a) 7 14log 4xy b)

2 4 3

5( ) 2 log 5x xf x x

c) 4ln xy e d) lnxy x e

e) 4

2log 32x

y

f) ln4xy e

PROPERTIES OF LOGARITHMS AND EXPONENTS

log ub b u ln or logu uee u e u logb ub u lnue u


g) 6ln xy e h) 4 6ln( ) ln x xf x e e

i) 𝑦 = log𝑎 𝑎𝑠𝑖𝑛𝑥 j) 𝑦 = ln 4𝑥2 ∙ 𝑒2𝑥

Ex #5(30L): If 𝑓(𝑥) = ln (𝑥 + 4 + 𝑒−3𝑥), then 𝑓′(0) is?


Ex #3(30L): Logarithmic differentiation:

a) Find 𝑑𝑦

𝑑𝑥 if 𝑦 = 𝑥𝑥

4.4 Day 1 Assignment & Memorize Formulas


SOLUTIONS TO 4.4 DAY 2 ASSIGNMENT

4.4 Day 3 Assignment Textbook P 192 #5, 7, 9, 11, 13, 15, 19, 21, 25, 33, 37, 45

& Memorize Formulas

Chapter 4 Review Assignment Textbook P 186 (Questions with a * are extra challenging)

#11-27 odd, 22(Use Logarithmic Differentiation), 30, 39, 41, 46, 53, 64*, 67ac, 71, 82*, 83* & Memorize Formulas

Documents

4.1 Day Calculus - Ms. Carignan's Math Site · 2018. 10. 29. · AP Calculus (Ms. Carignan) Chapter 4 :More Derivatives (C30.4 & C30.7) Page 4 Ex #3: If y u u 10 52 where ux 132,