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College Calculus: Level I The Quotient Rule

Section 3: Derivatives, part 1: Lecture 4 | 19:17 min

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Lecture Comments (11)

	1 answer Last reply by: Dr. Jennifer Switkes Mon Nov 12, 2018 6:56 PM Post by Samatar Farah on November 2, 2018 The audio file for examples 4 and 5 don't exist.
	0 answers Post by Mohamed Al Mohannadi on September 11, 2016 We can further simply "Lecture Example 1." We can write -16 / (6x-5)^2 as - [ 4 / 6x - 5 ] ^ 2 Am I right?
	0 answers Post by Ashley Simon on May 5, 2015 at 3:38, why is 6 negative??
	0 answers Post by Chateau Siqueira on September 3, 2013 Where Can I find a Lecture about the Difference quotient? Thanks!
	1 answer Last reply by: Martina Alvarez Tue Dec 6, 2011 1:20 PM Post by naman ahmad on October 24, 2011 on the last additional problem, last step we had:(e^x+7)-(Xe^x)/(e^x+7)^2. we could have cancelled out the e^x+7 and remained with -Xe^x/e^x+7.
	1 answer Last reply by: StefÃ¡n Berg Jansson Fri Nov 25, 2011 7:10 PM Post by Xiaosong Gao on January 17, 2011 example 3 can be further simplified to: 3^(1/2)x/1+x^3
	1 answer Last reply by: Larry Davis Wed Sep 15, 2010 10:25 AM Post by Larry Davis on September 15, 2010 on ex 2 you used the wrongterms in the numerator. it should have been ex d/dx[X8]- X8 d/dx[ex]/[x8]2 gives you =ex(8-x)/X9

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The Quotient Rule

Memorize the Quotient Rule – you will be using it a lot!
A constant factor can be factored out in front of a derivative.
A function of the form can be re-written as if you prefer the Chain Rule to the Quotient Rule.
Avoid common errors by remembering to use the Quotient Rule whenever you are differentiating a quotient!

The Quotient Rule

Given f(x) = 3x and g(x) = 4x, find [d/dx] [f(x)/g(x)]

[d/dx] [f(x)/g(x)] = [(g(x)f′(x) − f(x)g′(x))/(g(x)²)]
= [(4x [d/dx] 3x − 3x [d/dx] 4x)/((4x)²)]
= [(12x − 12x)/(16x²)]

Find f′(x) if f(x) = [3x/(x² + 1)]

f′(x) = [3x/(x² + 1)]
= [((x² + 1) [d/dx] 3x − 3x [d/dx] (x² + 1))/((x² + 1)²)]
= [((x² + 1)3 − 3x(2x))/((x² + 1)²)]
= [(3x² −6x² + 3)/((x² + 1)²)]

[(−3 (x² − 1))/((x² + 1)²)]

Find the derivative of tan(x) using the quotient rule

[d/dx] tan(x) = [d/dx] [sin(x)/cos(x)]
= [(cos(x) [d/dx] sin(x) − sin(x) [d/dx] cos(x))/(cos²(x))]
= [(cos(x)cos(x) − sin(x) (−sin(x)))/(cos²(x))]
= [(cos²(x) + sin²(x))/(cos²(x))]
= [1/(cos²(x))]

sec²(x)

Find the derivative of cot(x) using the quotient rule

[d/dx] cot(x) = [d/dx] [cos(x)/sin(x)]
= [(sin(x) [d/dx] cos(x) − cos(x) [d/dx] sin(x))/(sin²(x))]
= [(sin(x) (−sin(x)) − cos(x) cos(x))/(sin²(x))]
= −[(sin²(x) + cos²(x))/(sin²(x))]
= −[1/(sin²(x))]

−csc²(x)

Find the derivative of sec(x) using the quotient rule

[d/dx] sec(x) = [d/dx] [1/cos(x)]
= [(cos(x) [d/dx] 1 − 1 [d/dx] cos(x))/(cos²(x))]
= [(0 − (−sin(x)))/(cos²(x))]
= [sin(x)/(cos²(x))]
= sec(x) [sin(x)/cos(x)]

sec(x) tan(x)

Find f′(x) if f(x) = [(x³)/((5x + 4)tan(x))]

f′(x) = [((5x + 4) tan(x) [d/dx] x³ − x³ [d/dx] ((5x + 4) tan(x)t))/(t((5x + 4)tan(x))²)]
= [((5x + 4) tan(x) (3x²) − x³ [d/dx] ((5x + 4) tan(x)))/((5x + 4)² tan²(x))]
= [((5x + 4) tan(x) (3x²) − x³ ( (5x + 4) [d/dx] tan(x) + tan(x) [d/dx] (5x + 4) ))/((5x + 4)² tan²(x))]
= [((5x + 4) tan(x) (3x²) − x³ ( (5x + 4) sec²(x) + tan(x) 5 ))/((5x + 4)² tan²(x))]

[((5x + 4) tan(x) (3x²) − x³ ( (5x + 4) sec²(x) + 5tan(x)))/((5x + 4)² tan²(x))]

Find f′(x) if f(x) = [(√x + 3)/(x⁴ −16)]

f′(x) = [(√x + 3)/(x⁴ − 16)]
= [((x⁴ − 16) [d/dx] (√x + 3) − (√x + 3) [d/dx] (x⁴ − 16))/((x⁴ − 16)²)]

[((x⁴ − 16) [1/2] x^−[1/2] − (√x + 3) (4x³))/((x⁴ − 16)²)]

Find f′(x) if f(x) = [(.2x⁶)/(.1x + cos(x))]

f′(x) = [((.1x + cos(x)) [d/dx] .2x⁶ − (.2x⁶) [d/dx] (.1x + cos(x)))/((.1x + cos(x))²)]

[((.1x + cos(x)) 1.2x⁵ − .2x⁶ (.1 − sin(x)))/((.1x + cos(x))²)]

Find f′(t) if f(t) = [(t²)/sin(t)]

The letter or symbol used for the variable is not important. What's important is consistency.
f′(t) = [(sin(t) [d/dt] t² − t² [d/dt] sin(t))/(sin²(t))]
= [(sin(t) (2t) − t² cos(t))/(sin²(t))]

[(t(2sin(t) − t cos(t)))/(sin²(t))]

Find f′(1) if f(z) = [(z − 3)/(2 − z)]

f′(t) = [((2 − z) [d/dz] (z − 3) − (z − 3) [d/dz] (2 − z))/((2 − z)²)]
= [((2 − z) 1 − (z − 3) (−1))/((2 − z)²)]
= [(2 − z + z − 3)/((2 − z)²)] =

−[1/((2 − z)²)]

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Answer

The Quotient 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

Statement of the Quotient Rule

Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

Intro 0:00
Statement of the Quotient Rule 0:07

Carrying out the Differentiation
Quotient Rule in Words

Lecture Example 1 1:19
Lecture Example 2 4:23
Lecture Example 3 8:00
Additional Example 4
Additional Example 5

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

Related Books

	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

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Professor Switkes

The Quotient Rule

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro