If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u

Similar documents
If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy. du du. If y = f (u) then y = f (u) u

Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

( 3x +1) 2 does not fit the requirement of the power rule that the base be x

Change of Variables: Indefinite Integrals

Example. Evaluate. 3x 2 4 x dx.

Chapter 5: Integrals

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 3.6 The chain rule 1 Lecture. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

MATH 162. Midterm Exam 1 - Solutions February 22, 2007

Practice Exam I. Summer Term I Kostadinov. MA124 Calculus II Boston University

July 21 Math 2254 sec 001 Summer 2015

WeBWorK, Problems 2 and 3

Math 180, Final Exam, Fall 2012 Problem 1 Solution

Math 226 Calculus Spring 2016 Practice Exam 1. (1) (10 Points) Let the differentiable function y = f(x) have inverse function x = f 1 (y).

function independent dependent domain range graph of the function The Vertical Line Test

Integration by Substitution

College of the Holy Cross MATH 133, Calculus With Fundamentals 1 Solutions for Final Examination Friday, December 15

Differentiation Review, Part 1 (Part 2 follows; there are answers at the end of each part.)

The Free Intuitive Calculus Course Integrals

UNIT 3: DERIVATIVES STUDY GUIDE

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION

CHAIN RULE: DAY 2 WITH TRIG FUNCTIONS. Section 2.4A Calculus AP/Dual, Revised /30/2018 1:44 AM 2.4A: Chain Rule Day 2 1

Math 152 Take Home Test 1

Core Mathematics 3 Differentiation

Integration by Parts

Spring 2015 Sample Final Exam

King Fahd University of Petroleum and Minerals Prep-Year Math Program

Solutions to Exam 1, Math Solution. Because f(x) is one-to-one, we know the inverse function exists. Recall that (f 1 ) (a) =

Solutions to Exam 2, Math 10560

Study 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, 81. Class Notes: Prof. G. Battaly, Westchester Community College, NY Homework.

Integration by Substitution

Math 226 Calculus Spring 2016 Exam 2V1

Math 222 Spring 2013 Exam 3 Review Problem Answers

Math 21B - Homework Set 8

Section 10.7 Taylor series

11.5. The Chain Rule. Introduction. Prerequisites. Learning Outcomes

Section 5.6. Integration By Parts. MATH 126 (Section 5.6) Integration By Parts The University of Kansas 1 / 10

February 21 Math 1190 sec. 63 Spring 2017

Calculus & Analytic Geometry I

b n x n + b n 1 x n b 1 x + b 0

dx dx [x2 + y 2 ] = y d [tan x] + tan x = 2x + 2y = y sec 2 x + tan x dy dy = tan x dy dy = [tan x 2y] dy dx = 2x y sec2 x [1 + sin y] = sin(xy)

3.4 The Chain Rule. F (x) = f (g(x))g (x) Alternate way of thinking about it: If y = f(u) and u = g(x) where both are differentiable functions, then

(e) 2 (f) 2. (c) + (d). Limits at Infinity. 2.5) 9-14,25-34,41-43,46-47,56-57, (c) (d) 2

Math 111 lecture for Friday, Week 10

Edexcel GCE Core Mathematics C3 Advanced

Mar 10, Calculus with Algebra and Trigonometry II Lecture 14Undoing the Marproduct 10, 2015 rule: integration 1 / 18

A special rule, the chain rule, exists for differentiating a function of another function. This unit illustrates this rule.

M152: Calculus II Midterm Exam Review

and lim lim 6. The Squeeze Theorem

JUST THE MATHS UNIT NUMBER DIFFERENTIATION 3 (Elementary techniques of differentiation) A.J.Hobson

Math 106: Review for Exam II - SOLUTIONS

A.P. Calculus BC Test Two Section One Multiple-Choice Calculators Allowed Time 40 minutes Number of Questions 15

Chapter 5: Integrals

Math 250 Skills Assessment Test

1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.

The Area bounded by Two Functions

Calculus I Exam 1 Review Fall 2016

Math 181, Exam 1, Study Guide 2 Problem 1 Solution. =[17ln 5 +3(5)] [17 ln 1 +3(1)] =17ln = 17ln5+12

Math 131 Final Exam Spring 2016

Chapter 12 Overview: Review of All Derivative Rules

Chapter 7: Techniques of Integration

1.4 Techniques of Integration

Math 230 Mock Final Exam Detailed Solution

2t t dt.. So the distance is (t2 +6) 3/2

Name: Instructor: 1. a b c d e. 15. a b c d e. 2. a b c d e a b c d e. 16. a b c d e a b c d e. 4. a b c d e... 5.

MATH 101: PRACTICE MIDTERM 2

The Chain Rule. Mathematics 11: Lecture 18. Dan Sloughter. Furman University. October 10, 2007

MATH 1A Midterm Practice September 29, 2014

2.4 THE CHAIN RULE THE CHAIN RULE

MATH 151 Engineering Mathematics I

Section 3.5: Implicit Differentiation

Chapter 13: Integrals

Solutions to Problem Set 2

Derivatives of Trig and Inverse Trig Functions

Solutions to Second Midterm(pineapple)

5.5. The Substitution Rule

Calculus I: Practice Midterm II

More Final Practice Problems

Campus Academic Resource Program Chain Rule

AB 1: Find lim. x a.

Math 229 Mock Final Exam Solution

Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations

Grade: The remainder of this page has been left blank for your workings. VERSION D. Midterm D: Page 1 of 12

A basic trigonometric equation asks what values of the trig function have a specific value.

MATH 31B: MIDTERM 2 REVIEW. sin 2 x = 1 cos(2x) dx = x 2 sin(2x) 4. + C = x 2. dx = x sin(2x) + C = x sin x cos x

SECTION A. f(x) = ln(x). Sketch the graph of y = f(x), indicating the coordinates of any points where the graph crosses the axes.

Math 181, Exam 2, Study Guide 2 Problem 1 Solution. 1 + dx. 1 + (cos x)2 dx. 1 + cos2 xdx. = π ( 1 + cos π 2

Math 147 Exam II Practice Problems

C3 Revision Questions. (using questions from January 2006, January 2007, January 2008 and January 2009)

2.5 The Chain Rule Brian E. Veitch

Math RE - Calculus I Trigonometry Limits & Derivatives Page 1 of 8. x = 1 cos x. cos x 1 = lim

JUST THE MATHS UNIT NUMBER INTEGRATION 1 (Elementary indefinite integrals) A.J.Hobson

Mathematics Edexcel Advanced Subsidiary GCE Core 3 (6665) January 2010

and verify that it satisfies the differential equation:

Chapter 3. Integration. 3.1 Indefinite Integration

Math 221 Exam III (50 minutes) Friday April 19, 2002 Answers

OCR (MEI) Mathematics Advanced Subsidiary GCE Core 3 (4753) January 2010

Homework Solutions: , plus Substitutions

Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued)

Transcription:

Section 3 4B The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then f (u) u The Chain Rule with the Power Rule Theorem: If a and n are a real numbers If y = a u n then a n u n-1 u or in English If y = a ( some function of x inside the brackets ) n then a n ( some function of x inside the brackets ) n-1 the derivative of the function inside the brackets Example 1 The Chain Rule with the Power Rule Find y if y = ( 5x 2) 3 if we let u = 5x 2 then y = ( u) 3 where u = 5x 2 and u = 5 n(u) n 1 u n 3 u n 1 64 748 ( 5x 2) 2 u 5 15( 5x 2) 2 Math 400 3 4B Chain Rule Page 1 of 11 2013 Eitel

Example 2 The Chain Rule with the Power Rule Find y if y = 2( 3x 2 + 5x) 4 Think of the outside function as 2(u) 4 take the derivative of the outside function 2 4(u) 3 3 8 3x 2 + 5x now multiply that by the derivative of the inside function if u = 3x 2 + 5x then u = 6x + 5 a n 67 8 2 4 u n 1 647 48 ( 3x 2 + 5x) 3 64 7 u 48 ( 6x + 5) 8( 3x 2 + 5x) 3 ( 6x + 5) Example 3 The Chain Rule with the Power Rule Find y if y = 4( tan x) 3 Think of the outside function as 4(u) 3 take the derivative of the outside function 4 3(u) 2 12( tan x) 2 now multiply that by the derivative of the inside function if u = tan x then u = sec 2 x a n 64 74 8 64 7 u 48 12 ( tan x) 2 sec 2 x u n 1 12tan 2 x sec 2 x Math 400 3 4B Chain Rule Page 2 of 11 2013 Eitel

Example 4 The Chain Rule with the Power Rule Find y if y = 3 ( 6x 2 1) = ( 6x 2 1) 1/3 Think of the outside function as (u) 1/3 take the derivative of the outside function 1 3 (u) 2/3 2/3 1 3 6x2 1 now multiply that by the derivative of the inside function if u = 6x 2 1 then u = 12x n 1 3 u n 1 6 47 4 8 ( 6x 2 1) 2/3 67 u 8 ( 12x) 1 3 12x ( 6x2 1) 2/3 4x ( 6x 2 1) 2/3 Example 5 The Chain Rule with the y = Sin (u) Find f if f (x) = sin( 3x 4 ) if y = f (u) then f (u) u Think of the outside function as sin(u) take the derivative of the outside function to get cos(u) cos 3x 4 now multiply that by the derivative of the inside function if u = 3x 4 then u = 12x 3 64 f 7 (u) 48 cos 3x 4 u 12x 3 f (x) = 12x 3 cos 3x 4 Math 400 3 4B Chain Rule Page 3 of 11 2013 Eitel

Example 6 The Chain Rule with y = e u Find f (x) if f (x) = e x 2 if y = f (u) than f (u) u Think of the outside function as e u take the derivative of the outside function to get e u e x2 now multiply that by the derivative of u if u = x 2 then u = 2x f (u) e x 2 u 2x 2x e x 2 Example 7 Find f (x) if f (x) = e x = e x1/2 if y = f (u) than f (u) u Think of the outside function as e u take the derivative of the outside function to get e u e x1/2 now multiply that by the derivative of u if u = x 1/2 then u = 1 2 x 1/2 f (u) e x1/2 67 u 8 1 2 x 1/2 e x 2x 1/2 e x 2 x Math 400 3 4B Chain Rule Page 4 of 11 2013 Eitel

Example 8 The Chain Rule with y = ln( u) Find y if y = ln( 3x 4 + 2x 3 ) if u = 3x 4 + 2x 3 then u =12x 3 + 6x 2 if y = ln(u) then u u 12x 3 + 6x 2 3x 4 + 2x 3 6x 2 2x +1 x 3 3x + 2 6 2x +1 x 3x + 2 Example 9 The Chain Rule with y = ln( u) Find y if y = ln y = ln 3x 2 5x 3x 2 5x 1/2 = 1 2 ln ( 3x2 5x) if u = 3x 2 5x then u = 6x 5 if y = ln(u) then u u 1 2 6x 5 3x 2 5x 6x 5 2 3x 2 5x 6x 5 2x 3x 5 Math 400 3 4B Chain Rule Page 5 of 11 2013 Eitel

Example 10 The Chain Rule with y = ln Find y if y = ln u = ln u lnv v x 2 1 sin(2x) y = ln ( x 2 1) ln sin(2x) if y = ln(u) then u u der ofx 2 2x x 2 1 der 64 of 7 sin(2 48 x) cos(2x) 2 sin2x 2x x 2 1 2 cos(2x) sin(2x) Example 11 The Chain Rule with y = ln Find y if y = ln y = ln x sin(2x) x sin(2x) e x2 ln e x 2 u = ln u lnv v y = ln(x) + ln(sin(2x)) x 2 ln ( e) y = ln(x) + ln(sin(2x)) x 2 if y = ln(u) then u u der of x 1 + x der 64 of 7 sin(2 48 x) cos(2x) 2 sin(2x) der of x 2 2x 1 x + 2 cos(2x) sin(2x) 2x Math 400 3 4B Chain Rule Page 6 of 11 2013 Eitel

Repeated use of the Chain Rule then f g If y = f ( g [ ( h { x ) ] ) ( [ ( h { x ) ] ) g [ ( h{ x ) ] h x ( { ) Repeated use of the Chain Rule Example 12 ([ ]) Find f (x) if f (x) = sin cos 4x d dx ([ ]) sin cos 4 x ([ ]) d dx cos cos 4 x cos ( 4x ) cos cos 4 x cos cos 4x ([ ]) sin(4 x) d dx 4x ( ) ( sin(4 x) )( 4 ) 4 cos cos 4x ( ) sin ( 4x ) an alternate notation for example 12 ( ) Find y if y = sin cos 4 x if u = cos 4x y = sin u then where u = cos 4x 6 the der 47 of 4 sin 48 u the 6 der 47 of 4 cos 8 4x cos cos 4x sin 4x ( ) 4 cos cos 4x ( ) sin ( 4 x ) the der of 4x 4 Math 400 3 4B Chain Rule Page 7 of 11 2013 Eitel

Example 13 Repeated use of the Chain Rule Find f (x) if f (x) = ( cos ( 4 x )) 3 d dx ( cos ( 4x ))3 ( ) 2 d dx 3 cos 4x cos ( 4x ) 3 cos 4x ( ) 2 sin(4 x) d dx 4 x 3 cos 4x ( ) 2 sin(4 x) 12 cos 4x 4 ( ) 2 sin 4x an alternate notation for example 13 Find y if y = ( cos ( 4 x )) 3 if u = cos ( 4x ) then y = u 3 where u = cos 4x 6 the 4 der 7 of 4 (u) 48 n ( ) 2 3 cos 4x 12 cos 4 x the 6 der 4 of 7 cos(4 48 x) sin 4x ( ) 2 sin 4 x the der of 4x 4 Math 400 3 4B Chain Rule Page 8 of 11 2013 Eitel

Example 14 The Chain Rule with the Product Rule Find f (x) if f (x) = 4 x 3x 2 2 = 4x ( 3x 2 2) 1/2 644 the first der 7 of 4 sec 448 1 4 x 2 3x2 2 1/2 6x 64 the 7 sec 48 1/2 + 3x 2 2 der of first 4 12x 2 ( 3x 2 2) 1/2 + 4( 3x 2 2) 1/2 factor out a ( 3x 2 2) 1/2 1/2 12x 2 + 4 ( 3x 2 2) 3x 2 2 [ ] [ ] 1/2 12x 2 +12x 2 8 3x 2 2 24 x2 8 3x 2 2 Math 400 3 4B Chain Rule Page 9 of 11 2013 Eitel

Example 15 The Chain Rule with the Quotient Rule Find f (x) if f (x) = 2x 3x 2 2 2x = ( 3x 2 2) 1/2 6 the 4 bottom 748 ( 3x 2 2) 1/2 der of top the top 64 der 4 of 7 bottom 4448 2 2x 1 2 ( 3x2 2) 1/2 6x ( 3x 2 2) 2 14 243 bottom squared 1/2 6x 2 ( 3x 2 2) 1/2 ( 3x 2 2) 2 2 3x 2 2 factor out a ( 3x 2 2) 1/2 [ 6x 2 ] ( 3x 2 2) 2 ( 3x 2 2) 1/2 2 3x 2 2 [ ] 5/2 6x 2 4 6x 2 3x 2 2 4 ( 3x 2 2) 5/2 Math 400 3 4B Chain Rule Page 10 of 11 2013 Eitel

Example 15 The Chain Rule with the Power Rule and the Quotient Rule Find f (x) if f (x) = x 2 2 2 2x + 3 6 the 4 derof 74 u4 n 8 the 64 bottom 74 8 2 x 2 1 2 ( 2x + 3) 2x + 3 der of top 64 the 7 top 48 2x x 2 2 ( 2x + ) 14 243 2 bottom squared der of bottom 2 2 x 2 2 2x + 3 4x 2 + 6x 2x 2 + 4 2x + 3 2 2 x 2 2 2x + 3 2x 2 + 6x + 4 ( 2x + 3) 2 4 x2 +12x + 8 2x + 3 3 Math 400 3 4B Chain Rule Page 11 of 11 2013 Eitel