Problem

Suppose $ y = f(x) $ is a curve that always lies…

01:57

Question

Answered step-by-step

Problem 95 Hard Difficulty

(a) If $ n $ is a positive integer, prove that
$ \frac {d}{dx} (\sin^a x \cos nx) = n \sin^{a-1} x \cos (n + 1)x $
(b) Find a formula for the derivative of $ y = \cos^a x \cos nx $ that is similar to the one in part (a).

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Heather Zimmers

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Video Transcript

let's find the derivative of this function and show that it is what they gave us in the book as a formula to prove. So we're going to use the product rule. We have the first factor and the second factor. So the first is a sign of X to the n power times, a derivative of the second. So the derivative of co sign in X would be negative sign of annex times in. So that's the chain rule derivative of the outside times, the derivative of the inside. Now we have plus the second co sign of an X Times, the derivative of the first and the derivative of sign of X to the n power would be end times sign of X to the n minus one Power times the derivative of sign which is co sign so coastline of X. Okay, now we're going to factor out the common factors. So we have a term here and we have a term here and they both have a common factor of end. So we're going to factor that out and they also both have a common factor of sign of X to the n minus one. So we'll factor that out. So what remains then in the first factor is that extra sign of X as well as the negative sign of in X. And what remains in the second factor is the co sign of Annex Times, a co sign of X. Okay, now I'm going to re arrange that last quantity a little bit. I'm going to write it as co sign in X Times Co sign X minus Sign in x Times Sign X Now hopefully that looks somewhat familiar because hopefully when you learned trig identities in your pre calculus class, you learned some angle addition Identities so co sign of Alfa plus beta equals co sign Alfa coastline Beta minus sign Alfa Signed beta. That's what we have. If we let Alfa B r x, where are we? Right here and we let beta be Are we let Alfa BRN X and we let bait of your ex. So then what we have is and times a sign of X to the n minus one times The coastline of l bless beta is a co sign of an X plus X. And then what we could do with that is we could factor the X out of both of those terms there. So we've in times e sign of X to the n minus one power times, a co sign of and plus one times X. And that's what we set out to prove. Now we're asked to do something just like that for this one. So again we're going to use the product rule. So we have the first co sign of X to the n power times, the derivative of the second. And we need to use the chain rule. And we're going to get negative sign of annex times and the derivative of the inside, plus the second co sign of an ex times the derivative of the first. So again, using the chain rule, we're going to bring down the end raised co signed to the n minus one and then multiply by the derivative of co sign, which is negative sign. Now we have these two terms and let's see what we can factor out of both. So they both have an end civil factor that out and they both have coastline of X to varying powers. Weaken factor out co sign of X to the n minus one power and they both have a negative, so we can factor that out. Okay, so what do we have left then? In the first term, we still have our leftover co sign X and we have a sign of nxe. And in the second term, we have the coastline of NXE and we have the sign of X. So hopefully when we look at this expression, we remember again those trig identities. The sign of Alfa Plus Beta is sign Alfa Coastline Beta plus co sign Alfa signed beta. So if we let Alfa equal and X and Beta equal X, then we have the sign of an X plus X. So we'll rewrite our quantity inside the brackets as a sign of N x plus x. And finally we could factor x out of both of those terms and we get our formula here, which will be negative end times co sign of X to the n minus one Power times a sign of and plus one x

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

Suppose $ y = f(x) $ is a curve that always lies…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

James StewartCalculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

James Stewart

Calculus: Early Transcendentals