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Problem

Find the derivative of the function using the def…

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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.

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


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Daniel Jaimes

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 8

The Derivative as a Function

Related Topics

Limits

Derivatives

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Top Calculus 1 / AB Educators
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Lectures

Video Thumbnail

04:40

Limits - Intro

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

Video Thumbnail

04:40

Derivatives - Intro

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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Watch More Solved Questions in Chapter 2

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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.

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Calculus: Early Transcendentals

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Related Topics

Limits

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Top Calculus 1 / AB Educators
Grace He

Numerade Educator

Heather Zimmers

Oregon State University

Caleb Elmore

Baylor University

Joseph Lentino

Boston College

Calculus 1 / AB Courses

Lectures

Video Thumbnail

04:40

Limits - Intro

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

Video Thumbnail

04:40

Derivatives - Intro

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

Join Course
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