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College Calculus: Level I Implicit Differentiation

Section 4: Derivatives, part 2: Lecture 1 | 30:05 min

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Lecture Comments (5)

	0 answers Post by Joyce Ferreira on October 27, 2017 Dear Prof. Switkes, It has been a while that I took a calculus class and I need to review partial differentiation, so I can be able to follow the lectures on my physical chemistry class. Please, I would liek to ask you what calculus lecture should I watch? Thank you. Best regards.
	0 answers Post by Zhe Yang on November 3, 2014 you did a wonderful job explaining it, thanks!
	0 answers Post by kathy park on December 16, 2012 Would you please check the second problem ? If you use product rule,d/dx[f(x)g(x)] = f'(x)gx) + f(x)g'(x), the solution is d/dx[4x^3y^8] = 12x^2y^8 + 4x^3*8y^7dy/dx = 4x^2y^7(3y + 8xdy/dx)
	0 answers Post by David Jarrett on June 26, 2012 Wow, thanks so much! This totally cleared everything up for me!
	0 answers Post by Ron Weldy on March 2, 2011 The problems we have at school are like cos(y)=x; (0,pie/2) or sin(y)=5x^4-5; (1,pie) I don't know I am stuck!

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Implicit Differentiation

This method allows you to determine the derivative as a function of both and in situations in which it is not convenient to solve explicitly for as a function of .
The key to these problems is to recognize that and to use the Chain Rule whenever appears.
You will be differentiating both sides of an equation with respect to .
You will often find yourself using the Chain Rule within a Product Rule computation.

Implicit Differentiation

Find [dy/dx] given 2y = x³

[d/dx]2y = [d/dx]x³
2 [dy/dx] = 3 x²

[dy/dx] = [3/2] x²

Find [dy/dx] given siny = x

[d/dx]siny = [d/dx]x
cosy [dy/dx] = 1
[dy/dx] = [1/cosy]

[dy/dx] = secy

Find [dy/dx] given e^y = x

[d/dx]e^y = [d/dx]x
e^y [dy/dx] = 1

[dy/dx] = e^−y

Find [dy/dx] given lny = x² + cosx

[d/dx]lny = [d/dx]x² + [d/dx]cosx
[1/y] [dy/dx] = 2x − sinx

[dy/dx] = y(2x − sinx)

Find [dy/dx] given √{y² + y} + [1/x] = 1

[d/dx]√{y² + y} + [d/dx][1/x] = [d/dx]1
[1/2] [1/(√{y² + y})] [d/dx](y² + y) + (− [1/(x²)]) = 0
[1/(2 √{y² + y})] (2y [dy/dx]+ [dy/dx]) − [1/(x²)] = 0
[1/(2 √{y² + y})] (2y [dy/dx]+ [dy/dx]) = [1/(x²)]
[dy/dx]([(2y + 1)/(2 √{y² + y})]) = [1/(x²)]

[dy/dx] = [(2 √{y² + y})/(x²(2y + 1))]

Find [dy/dx] given [(x²)/4] + [(y²)/9] = 4

[d/dx][(x²)/4] + [d/dx][(y²)/9] = [d/dx]4
[2x/4] + [2y/9] [dy/dx] = 0
[2y/9] [dy/dx] = − [x/2]

[dy/dx] = − [9x/4y]

Find [dy/dx] given lny + e^y = cosx

[d/dx]lny + [d/dx]e^y = [d/dx]cosx
[1/y] [dy/dx]+ e^y [dy/dx] = − sinx
[dy/dx]([1/y] + e^y) = − sinx

[dy/dx] = − [sinx/([1/y] + e^y)]

Find [dy/dx] given y² + 3y = cos⁻¹ x³

[d/dx]y² + [d/dx]3y = [d/dx]cos⁻¹ x³
2y [dy/dx]+ 3 [dy/dx] = − [1/(1 − (x³)²)] [d/dx]x³
[dy/dx](2y + 3) = − [(3x²)/(1 − x⁶)]

[dy/dx] = [(−3x²)/((2y + 3)(1 − x⁶))]

Find [dy/dx] given x² + (y + 3)² = 1

[d/dx]x² + [d/dx](y + 3)² = [d/dx]1
2x + 2(y + 3) [d/dx](y + 3) = 0
2x + 2(y + 3) [dy/dx] = 0
2(y + 3) [dy/dx] = −2x
[dy/dx] = [(−2x)/(2(y + 3))]

[dy/dx] = [(−x)/(y + 3)]

Find [dy/dx] given √{2y + siny} = x²

[d/dx]√{2y + siny} = [d/dx]x²
[1/2] [1/(√{2y + siny})] [d/dx](2y + siny) = 2x
[1/2] [1/(√{2y + siny})] (2 [dy/dx]+ cosy [dy/dx]) = 2x
[dy/dx][(2 + cosy)/(2√{2y + siny})] = 2x

[dy/dx] = [(4x √{2y + siny})/(2 + cosy)]

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Answer

Implicit Differentiation

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

Purpose

Implicit Function

Lecture Example 1

Lecture Example 2

Lecture Example 3

Lecture Example 4

Additional Example 5

Additional Example 6

Intro 0:00
Purpose 0:09

Implicit Function

Lecture Example 1 0:32
Lecture Example 2 7:14
Lecture Example 3 11:22
Lecture Example 4 16:43
Additional Example 5
Additional Example 6

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

Implicit Differentiation

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro