Problem

Find the derivative of the function using the def…

05:36

Question

Answered step-by-step

Problem 29 Hard Difficulty

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

$G(t)=\frac{1-2 t}{3+t}$

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 2

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

Video Transcript

Alright, so here we have a function G F T which is one minus two T over three plus T. And we need to find its domain as well as di prima T and the domain of di prima T. And we're going to use the formal definition of derivative domain by the way. Normally for any function is all real, unless you have some quote unquote bad excess, for example, we don't want to divide by or in this case T. S we don't want to divide by zero. So we're gonna do is we're gonna set three plus two equal to zero. Or we can say we cannot let it be zero. So therefore T cannot be -3. Therefore domain would be between minus infinity and -3 and -3 to infinity, basically all reels except keeping minus three. Alright, let's go ahead now and solve for our derivative. Using definition of derivative. Okay, so G private E is equal to the limit As H goes to zero of G. Of T plus H minus G ft. All over age. Alright, so that's our formal standard definition of derivative. Let's go ahead and apply it then to our function. Okay, so that means for G every time I see a. T. It's now T plus H for the first term. So I'll get one minus twice T plus H over three plus T plus H and then minus GFT So minus um just GT as is. Okay, so that's just all on top and on the bottom we still divide by eight. Okay, so this is super messy. one thing we can do is get a common denominator and try to clean it up that way. So let's do that. So I'm going to basically um I need more room I think. So I'm going to kind of move over here so I have plenty of room to write this out. Okay so we are going to multiply the left fraction by three plus T. Over three plus T. This one here in the right fraction by this denominator which is really um three plus T. Plus H. So um okay so we're gonna get a big mess but let's do it. Okay so I'm going to get on the so I'm going to first look at this term and I'm going to multiply by three plus T over three plus T. That will give me one the same time. I'm going to distribute that too just to kind of clear that up. Okay so on the top we multiply by three plus T. Ah we're going to have a common denominator so this whole big denominator will be three plus T plus H. Has three plus t. Okay but I need to know work with this one. I'm going to multiply top and bottom by three plus two plus eight so I will get minus 1 -2 T. Times three plus T plus H, wow. Right what a mess. So now I actually need to distribute unless I can recognize any terms that are going to cancel out. That will save me some work. Let's see if there's any obvious terms. It might be hard to do because I still need to kind of foil it all out. Um Let's see. Uh Well I I think I'm just going to foil it out. There's just so many terms. Okay so limit h well I'm gonna do it below because they're really going to need a lot of room here and that was a close you're going to see okay so here we go limit H. Goes to zero. We're going to do a huge foil. So I'm going to take three times everything there and then I'm going to take so basically all the parts and then I'm going to do the same thing with T. So let's keep track so I can have three minus 60 -6. H. That's doing the in red now I'm going to do the same thing but T gets multiplied by all three parts. So I'll get plus t -2 T. Squared -2 H. T. Alright so that took care of the left super foil. Uh This is crazy. Look how big this is. Okay now on the right we're going to do the same trick. I'm gonna put this I'm gonna do one times each of the terms and then I will do the minus two T. Times E. To the terms we have a minus in front. So we have to distribute that as well. So that gives me -3 minus t minus H. Now do the minus two T. With double negative six plus two t. Times so plus 60 plus two t squared Plus two th that is amazing. Look at all those terms. Okay let's look for cancelations. I get a plus two T squared and a minus two C squared. Goodbye. So we had a plus two th m minus two th another. Goodbye. Okay we have a plus 60 and a minus 60 and we have a plus 10 and minus T. This is exciting so much is going away and alright so then I think we're left with they live it as H goes to zero. Let's see. We have oh the three's cancel ye so we're down to um um minus seven H. So we have -7 h. All over. I forgot to divide but oh this is terrible I forgot to divide by age. Okay so I forgot that we're still dividing by age here so I'm just going to if we divide a fraction by a single number that that never ends up on the bottom so I'm going to put the h down here we have an age down there so I will put it there and squish in the rest of the denominator three plus t. Alright now this is where it's great because we need this final cancelation to have um hs canceled because we don't want to be dividing by zero. Now I can finally plug in zero. That gives me -7/3. Three plus T. The H is zero and then times three plus T. So three plus t squared, wow. So that is G prime of T g prime of T. Okay, domain is the same as the original function. We just can't have the denominator go to zero. So t cannot be minus three or you can think of it as minus infinity minus three and -3 to infinity as our domain. Okay. Woo lots of terms. All right. Have a great day. I hope that helped and see you soon.

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