Problem

Differentiate. $ V(t) = \frac {4 + 1}{te^t} $

02:18

Question

Answered step-by-step

Problem 21 Easy Difficulty

Differentiate.

$ f(t) = \frac {sqrt[3]{t}}{1 - 3} $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Mary Wakumoto

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

Video Transcript

All right here we want to take the derivative of this function. F of t cube at t over t minus 3. I'M going to go ahead and rewrite the top as t to the 1 third power and then we're going to go ahead and apply quotient rule. So just a reminder of how cotient roll works, potient rule and basically it helps us take the derivatives of fractions. So let's say i want to in general, have a function that looks like? U f, t over v. F t, then, if i want to find the derivative, then what i'm going to do is take the derivative of the top times the bottom minus the derivative of the bottom times, the top all over the bottom square. All right! So that's potient rule. Let'S go ahead and apply it to our function, so we want to go down here now and go ahead and solve it for our expressions. So primatu is going to be the derivative of the top so that we're gonna use power rule so we'll get 1. Third t to the minus 2 third times the bottom, which is t minus 3, that's a denominator and then i'm going to subtract and take the derivative at the bottom. While the derivative of t minus 3 is just 1 and then we multiply by the top, and then we go all over the bottom square, so that is your derivative. We can take or do our hand at planting it up a little see if it cleans up nice. Sometimes it cleans up nice. Sometimes it doesn't. Let'S tend to take a look and see what we can do to clean this thing up. Okay, so first, let's go ahead and distribute the left term. The 113 of the minus 2 thirds distribute over t minus 3. Let'S you see if that helps us out. If we do that, we'll get 1 third and then t to the 1 third- and let's get my pen to work here there we go okay, so that's the first part of the distribution, then we're going to distribute the minus 3, so the minus 3 times a Third is just 1 and then we'll get t to the minus 2 thirds and then we still have a minus t to the 1. Third, so it looks like we can combine those t to the 1 third, on top all right. Let'S try that, let's see what happens next, we're going to go ahead and just paint up a little bit more again, where we have already solved it, but let's just see half a little fun and clean it up a little bit. I have 1 third to the 1 third minus t to the 1. Third, that will be minus 2. Third, minus 2 thirds t to the 1 third minus t to the minus 2 thirds all over t minus 3 square. So again we have already solved it, but the rest will just be fun kind of tricks to clean it up a bit so 1 way to clean up kind of a messy. A situation with powers, especially negative powers, is to multiply everything top and bottom by um. Basically, the positive version that that negative 1 we're going to multiply everything by t to the 2 thirds and you'll see what happens. If we do this, so it's pretty full and what we'll do is that will help us get rid of any negative um. You know any negative exponents so pretty false: let's go ahead and do that so we're going to distribute this t of the 2 thirds over both parts of the top. So when we do that, we'll get minus 2 thirds and this is cool because i add exponents, so i get t to the 1 third plus t to the 2 thirds and that becomes just t. So that's really pull then, when i multiply to the minus 2 thirds times t to the 2 thirds- and i add the exponents i get to the 0, so i just get minus 1. So that's really cleaned up nicely, but we do have a t of the 2 thirds on the bottom. So i'll go ahead and add it to the bottom and there we have it and the last thing we could do if you want to just plant it up slightly as multiplied top and bottom by 3, and what that does. Is it just gets rid of that fraction in the fraction, and then we have? I guess it looks prettier this way it's no different than the original, but it does have kind of a nice nice compact field to the answer. So this is as cleaned up as i think it's going to get with me and anyway, there was a front problem: how soit helped have a wonderful day?

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

Differentiate. $ V(t) = \frac {4 + 1}{te^t} $

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

Catherine Ross

Caleb Elmore

Michael Jacobsen

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Michael Jacobsen

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

Calculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Caleb Elmore

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals