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Lecture Comments (4)

1 answer

Last reply by: Angela Patrick
Sun Dec 15, 2013 7:26 AM

Post by sergey smith on October 3, 2013

on the second to last example isn't tan(3t) divided by the a term or 3? so wouldn't it equal tan(3t)/9 + C?

1 answer

Last reply by: Vicente Orellana
Wed Jul 17, 2013 8:08 PM

Post by Huseyin Kayahan on March 3, 2013

for example 1, isnt it supposed to be -cos2x/2x + C ?

Answer EngineGet answers to any question!Ask any question related to Calculus AB

Working on the solution...

Chain Rule

Tip: choose u so that dx cancels out rest of x terms
Don’t forget to substitute original expression back into u!

Chain Rule

Solve ∫sin(x − π) dx using u substitution

u = x − π
du = dx
∫sin(x − π) dx = ∫sinu du
∫sin(x − π) dx = −cosu + c
∫sin(x − π) dx = −cos(x − π) + c
−cos(x − π) = cosx
In reality, we could have simplified the original integral, but this works too

∫sin(x − π) dx = cosx + c

∫x sin(x²) dx

u = x²
du = 2x dx
∫x sin(x²) dx = ∫x sinu [du/2x]
∫x sin(x²) dx = [1/2] ∫sinu du
∫x sin(x²) dx = [1/2] (−cosu) + c

∫x sin(x²) dx = −[1/2] cosx² + c

∫(3x − 5)⁴ dx

u = 3x − 5
du = 3 dx
∫(3x − 5)⁴ dx = ∫u⁴ [du/3]
∫(3x − 5)⁴ dx = [1/3] ∫u⁴ du
∫(3x − 5)⁴ dx = [1/3] [(u⁵)/5] + c

∫(3x − 5)⁴ dx = [1/15] (3x − 5)⁵ + c

∫cos√{x² + 2x + 1} dx with x ≥ 0

There is a constraint that means the square root will never be negative
∫cos√{x² + 2x + 1} dx = ∫cos√{(x + 1)²} dx
∫cos√{x² + 2x + 1} dx = ∫cos(x + 1) dx
u = x + 1
du = dx
∫cos√{x² + 2x + 1} dx = ∫cosu du
∫cos√{x² + 2x + 1} dx = sinu + c

∫cos√{x² + 2x + 1} dx = sin(x + 1) + c given x ≥ 0

∫(6x + 3)(x² + x)⁷ dx

u = x² + x
du = (2x + 1) dx
∫(6x + 3)(x² + x)⁷ dx = ∫(6x + 3) u⁷ [du/(2x + 1)]
∫(6x + 3)(x² + x)⁷ dx = ∫[(6x + 3)/(2x + 1)] u⁷ du
∫(6x + 3)(x² + x)⁷ dx = 3 ∫u⁷ du
∫(6x + 3)(x² + x)⁷ dx = 3 [(u⁸)/8] + c

∫(6x + 3)(x² + x)⁷ dx = [3/8] (x² + x)⁸ + c

∫√{7x + 9} dx

u = 7x + 9
du = 7 dx
∫√{7x + 9} dx = ∫√u [du/7]
∫√{7x + 9} dx = [1/7] ∫√u du
∫√{7x + 9} dx = [1/7] ∫u^[1/2] du
∫√{7x + 9} dx = [1/7] [(u^[3/2])/([3/2])] + c
∫√{7x + 9} dx = [2/21] u^[3/2] + c

∫√{7x + 9} dx = [2/21] (7x + 9)^[3/2] + c

∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx

u = x³ + 3x − 11
du = (3x² + 3) dx
∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx = ∫[(x² + 1)/(u⁴)] [du/(3x² + 3)]
∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx = ∫[(x² + 1)/(3x² + 3)] u⁻⁴ du
∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx = [1/3] ∫u⁻⁴ du
∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx = [1/3] [(u⁻³)/(−3)] + c

∫[(x² + 1)/((x³ + 3x − 11)⁴)] dx = −[1/9] [1/((x³ + 3x − 11)³)] + c

∫cosx (sinx)⁶ dx

u = sinx
du = cosx dx
∫cosx (sinx)⁶ dx = ∫(sinx)⁶ cosx dx
∫cosx (sinx)⁶ dx = ∫u⁶ du
∫cosx (sinx)⁶ dx = [(u⁷)/7] + c

∫cosx (sinx)⁶ dx = [1/7] (sinx)⁷ + c

∫[(sec² √x)/(√x)] dx

u = √x
du = [1/2] [1/(√x)] dx
∫[(sec² √x)/(√x)] dx = 2 ∫sec² √x [1/(2√x)] dx
∫[(sec² √x)/(√x)] dx = 2 ∫sec² u du
∫[(sec² √x)/(√x)] dx = 2 tanu + c

∫[(sec² √x)/(√x)] dx = 2 tan√x + c

∫x² (x³ + 4)⁵ dx

u = x³ + 4
du = 3x² dx
∫x² (x³ + 4)⁵ dx = [1/3] ∫(x³ + 4)⁵ 3x² dx
∫x² (x³ + 4)⁵ dx = [1/3] ∫u⁵ du
∫x² (x³ + 4)⁵ dx = [1/3] [(u⁶)/6] + c

∫x² (x³ + 4)⁵ dx = [1/18] (x³ + 4)⁶ + c

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Answer

Chain Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Chain Rule

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Intro 0:00
Chain Rule 0:07

Example 1

Example 2 3:17
Example 3 5:09
Example 4 7:53
Example 5 9:40
Example 6 11:39

Mathematics: AP Calculus AB

Section 1: Functions
	Definitions & Properties of Functions	11:26
	Graphing	13:58
	Inverse Functions	6:47
	Polynomial Functions	5:04
	Trigonometric Functions	6:45
	Inverse Trigonometric Functions	5:58
	Trigonometric Identities	17:42
	Exponential Functions	5:53
	Logarithmic Functions	7:08
	Rational Functions	15:36
	Conic Sections	14:58
Section 2: Limits and Continuity
	Limit Definition & Properties	7:15
	Solving Limits with Algebra	8:01
	Rational Limit Rules	3:16
	One Sided Limits	9:57
	Special Trigonometric Limits	8:28
	Limits & Continuity	10:14
	Limits: Multiple Choice Practice	6:16
Section 3: Derivatives
	Derivative Definition & Properties	4:11
	Basic Rules of Differentiation	7:07
	Power Rule	7:14
	Trigonometric Rules	7:53
	Product Rule	11:11
	Quotient Rule	16:50
	Chain Rule	19:48
	Higher Order Derivatives	15:00
	Derivatives of Exponential Functions	13:14
	Derivatives of Logarithmic Functions	11:30
	Derivatives of Inverse Trigonometric Functions	16:54
	Implicit Differentiation	16:53
	Multiple Choice Practice: Derivatives	11:07
Section 4: Applications of Derivatives
	Tangent & Normal Lines	22:36
	Position Velocity & Acceleration	18:42
	Related Rates	26:22
	Minimum & Maximum	12:22
	Concavity	11:43
	Rolles Theorem	8:28
	Mean Value Theorem	9:39
	Differentials	12:25
	Applications of Derivatives: Multiple Choice Practice	13:21
	Applications of Derivatives: Free Response Practice	10:22
Section 5: Integrals
	Definition of Integrals	1:08
	Integrals of Power Rule	8:50
	Integrals Basic Rules of Integration	9:43
	Trigonometric Rules of Integrals	8:58
	Chain Rule	13:59
	Integrals of Exponential Functions	12:52
	Integrals of Natural Logarithmic Functions	13:00
	Integrals of Inverse Trigonometric Functions	8:29
	Integrals: Multiple Choice Practice	15:37
Section 6: Applications of Integrals
	Fundamental Theorem of Calculus	15:55
	Area Under A Curve	18:34
	Reimann Sums	10:35
	Trapezoid Rule	12:46
	Mean Value Theorem	11:22
	Second Fundamental Theorem of Calculus	4:44
	Area Between Curves	16:39
	Revolving Solids Washer Disk Methods	21:09
	Revolving Solids Cylindrical Shells Method	26:46
	Revolving Solids Known Cross Sections	27:41
	Differential Equations Eulers Method	17:54
	Differential Equations Slope Fields	16:30
	Application of Integrals: Multiple Choice Practice	14:19
	Application of Integrals: Free Response Practice	9:14
Section 7: Sample AP Test
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 1	17:50
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 2	17:32
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 3	22:14
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 4	19:35
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 5	25:43
	AP Calculus AB Practice Test: Section 2: Free Response Part 1	16:50
	AP Calculus AB Practice Test: Section 2: Free Response Part 2	21:36
	AP Calculus AB Practice Test: Section 2: Free Response Part 3	13:39

Related Books

	Cracking the AP Calculus AB Exam, 2015 Edition Author: Princeton Review ISBN: 804124809 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

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Chain Rule

Slide Duration:

Table of Contents

Section 1: Functions

Definitions & Properties of Functions

11m 26s

Intro