Problem

Find the derivative of the function. $ y = \cot^…

01:15

Problem 36 Hard Difficulty

Find the derivative of the function.
$ y = x^2 e^{-1/x} $

Answer

$\frac{d y}{d x}=e^{-1 / x}(2 x+1)$

More Answers

00:24

Frank L.

Topics

Derivatives

Differentiation

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Discussion

You must be signed in to discuss.

Top Calculus 1 / AB Educators

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College

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

here we have a function that's a product. So we're going to end up using the product rule to differentiate it. And before I start that, I'm going to rewrite the negative one over X power. So we're gonna write this as X squared times e raised to the negative X to the negative one. That'll make it easier to differentiate. Okay, so for the derivative, we're going to use the product rule. So we start with the first X squared times, the derivative of the second. So now it's time to take the derivative of E to the negative x the negative one power. So we would start with E to the negative X to the negative one power and then multiply by that power. And so the derivative of negative X to the negative one would be positive one times X to the negative, too. Because you bring down the negative one, multiply it by the negative that's already there. And then you subtract one to get the new power. Okay, so so far, we have the first times the derivative of the second. Now it's going to be plus the second e to the negative X to the negative one times the derivative of the first and the derivative of the first would be X squared two X derivative of X Squared is two X. Now let's see what we can do with this. So notice that we have X squared times X to the negative. Second, if you add the powers on X there you get X to the zero and X to the zero is one. So that term is actually just e to the negative X to the negative one. And then the other term is two x times e to the negative x the negative one So we can factor e to the negative X to the native one out of both of those terms, and it's time to rewrite it back in its original form E to the negative one over X that's being factored out. So what we have left in the first term is just one on what we have in the other term is two x So we one plus two x times e to the negative one over x

Heather Z.

Oregon State University

Topics

Derivatives

Differentiation

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Top Calculus 1 / AB Educators

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College