Problem

Find the derivative of the function. $ g(x) = (2…

02:21

Question

Answered step-by-step

Problem 42 Hard Difficulty

Find the derivative of the function.
$ y = \sqrt {x + \sqrt {x + \sqrt {x}}} $

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

all right here we have quite the extensive function that we're going to differentiate using the chain rule. And every time I have a square root function, I like to change it to a 1/2 power function before I differentiate. That allows me to use the product or the power rule. Excuse me. So that means I'm going to rewrite this as the quantity X plus the quantity X plus X to the 1/2 to the 1/2 to the 1/2 which really doesn't make anything look better. Um, and it's going to look pretty bad once I take the derivative, but then we'll kind of clean it up after that. So using the chain rule, we have white prime equals. We're going to work on the derivative of the outside function. So the outer 1/2 power function bring down the 1/2 and then take the entire inside and raise that to the negative 1/2. Now we move on to multiply by the derivative of the inside, and this is the insight. So the inside is a some and to find the derivative of a some you find the derivative of each term in the sun. So we start with the derivative of X, and that's one. And now we move on to adding the derivative of the second part of the some, and we're going to need the chain rule for that. So again, a 1/2 power. So bring down the 1/2 and raised the inside to the negative 1/2 and then we need to multiply by its inside derivative. So inside his X plus extra, the 1/2 which is also a some so it's derivative will be the the some of the derivatives. So the derivative of X is one plus the derivative of X to the 1/2 is 1/2 X to the negative 1/2. Now we made it to the very inside of everything. So the next thing we want to do is simplify our answer and change things back into square root signs. So we have this too, which is going to go in the denominator of the entire thing. And because this is to the negative power, it's also going to go in the denominator of the whole thing, and we're going to change it back to a square root sign. So all the 1/2 powers will be changed back to square root signs. So the denominator of the whole big answer is going to have a two in it. And then it's going to have the square root of X plus the quantity X whoops. How about if I go back and change that to a square root as well? I'm gonna make that notation look a little bit better while I'm erasing. Okay, so we're rewriting this thing right here. We're rewriting that. So it's a square root of X plus, the square root of X plus the square root of X. Okay, now we're gonna work on simplifying this thing, and it's our numerator. So it's actually, um Well, we've got the plus one. We can start with that, and then we're gonna work on this part in a very similar way. We just worked on the previous part, so we have the two on the bottom. We have a negative 1/2 power, so that part's gonna go on the bottom as well. So we're going to have the two times the square root of X plus square root X. So that takes care of all of this and then this part will be in the numerator. So in the numerator we have one plus. But now deja vu. This part is going to have a two on the bottom. And because of the negative power, the square roots going to be on the bottom. So one over two square root X So how is that for a tower of incredible numbers and algebra?

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

Find the derivative of the function. $ g(x) = (2…

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

Catherine Ross

Anna Marie Vagnozzi

Kayleah Tsai

Samuel Hannah

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kayleah Tsai

Samuel Hannah

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

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kayleah Tsai

Samuel Hannah

James StewartCalculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kayleah Tsai

Samuel Hannah

James Stewart

Calculus: Early Transcendentals