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QuestionFind the derivative of f(x) =3x 2*+4 by using the limit definition of derivative....

Question

QuestionFind the derivative of f(x) =3x 2*+4 by using the limit definition of derivative.

Question Find the derivative of f(x) =3x 2*+4 by using the limit definition of derivative.



Answers

Use the limit definition to find the derivative of the function. $$ f(x)=-2 $$

For this problem we're looking to calculate the derivative of X to the one third. Using the definition of the limit. So the definition of the limit is F. X. Plus coach minus F. Of X divided by H. and so plugging in extra 1/3 into that that is going to give us uh X plus H. Two. The one third minus X to the one third. Yeah and so that is a nasty set of foiling to get that done. But the big thing you can do is to multiply both sides by X plus H. Two the two thirds. Um in order to get rid of that. So I'm gonna do here is I'm gonna multiply both sides by explosive age. Two thirds plus X plus H. It's the 1/3 Multiplied by x. 0. 1 3rd Plus X. The 2/3. And this is just um what I'm doing here is I'm I'm just doing what the hints in the pram recommended which is multiplying the numerator and the denominator together So that gives us that. And simplifying is going to give us a church over H. multiplied by that same bit. So again that's exposed aged 2/3. Mhm. Plus expose age to the 1/3 multiplied by X. To the one third Plus Extra 2/3. And now we have isolated the H. Is enough where we can pull them out. So there's an H. On both terms we can get rid of that. And that is going to give us one Mhm. Over extra the two thirds plus extra two thirds Uh plus X. The 2/3. Mhm. And that is going to be equal to 1/3 X 22 3rd. Have you did the power rule for extra one third. You would get the same answer. It's just another way to show you that. It does work.

So here we have the function F of X is equal to x squared minus four. And we're gonna use the limit definition in order to find the derivative of this function. So our limit definition here is the limit as H approaches zero of F. Of X plus H minus or function F of X. All over age. I do recommend to have this memorized. It's very important to know. So what I speaking. So here we had the limit. Oops. Yeah. Yeah. The limit As a chip which is zero of our function of our modified function F of X plus H. So here we have export H squared minus four minus our original function. So we have minus X squared plus four. Since this negative is being distributed to the whole function over age. Next up we're going to want to distribute out in fact our explosive squared. So We'll have the limit as h approaches zero of x squared plus two X H plus H squared minus four minus x squared plus four. Again, all over age. Now we can go out and combine like terms here. So we have a positive X squared and a negative X square. So these can be crossed out. A -4 and a positive for these can also be crossed out. So that will leave us with the limit As a approaches zero of two, X. H plus H squared a great. Now we can go ahead and take an H out of all of our components on the in the numerator and the denominator. So that will leave us with the limit As a church purchase zero of two X plus each. Then we can go ahead and evaluate our limit. So plugging in a zero for every robbery everywhere we see an H. Will be left with two X, which is a derivative of our function.

So what we want to do is we want to find the derivative of the function F X. Using the definition of the derivative which I provided for right here. So let us uh see if we can do this. So looking at this, we want to write the limit As Delta X approaches zero of F of X plus delta X. Well, what that means is for our function F of X anywhere I see an X. I instead want to put an X plus delta X. And so that means that becomes seven times X plus delta X minus three minus F of X. Well, what's F of X? Well, that is seven X -3. And all of that will be divided by delta X. So continuing on, we want to distribute some things. And so I'm going to distribute this seven uh and to this parentheses which gives us the limit As Delta X Approaches zero. You don't want to forget that 27 X plus seven delta X minus three minus seven X minus three. Divided by sorry, I'll be a plus three divide by delta X. Because we call this negative distributes as well. So something's cancel out what cancels out? Well we have a seven X. And a negative seven X. So those are gone and then we have a negative three and a plus three. So those are gone. And so what that ends up equaling is the limit. As Delta X approaches zero of seven, Delta X. Put that in parentheses, they're divided by delta X. Well, what we see is that these delta X is just end up cancelling out leaving the limit as delta X approaches zero of seven. Well, there is no uh delta X to plug in zero. So that just ends up equaling seven. And that is our answer.

So what we want to do here is we want to find the derivative of f of X, which is equal to one over x squared using the definition of the derivative that we have here. And that is uh the limit as delta X approaches zero of F of X plus delta X minus F of X divided by delta X. And so in order to do this, we are going to have to compute this limit. So uh we're going to start off by saying, okay, what is F of X plus delta X? Well, what that means is that anywhere I see an X, I'm gonna put X plus delta X instead. So we have the limit as delta X. Purchase zero of one over X plus delta X squared minus F of X. Well, F of X is one over X squared. And don't forget, all of that is still being divided by delta X. So we're obviously going to have to simplify this. So what does that simplify too? Well, in order to subtract these fractions, we're going to have to find a common denominator. Well, a common denominator for them would be X squared times X plus delta X squared. And so what we would have is we would have the limit As Delta X Approaches zero. Just rewriting all of this, it would be X squared over X plus delta X square times X squared minus X plus delta X over X squared times X plus delta X squared. And all of that still being divided by delta X. So still more stuff to simplify well, because they're both being divided by delta X. What can actually happen is that we can uh put that delta X in the denominator of both fractions, giving us X squared over X plus delta X squared times X squared times delta X minus X plus delta X squared. This is all going to be squared. My apologies for that, X squared X plus delta X square times delta X. And so we can obviously combine those two fractions since their denominators are the same, giving us the limit as delta X approaches zero, uh, X squared minus X plus delta X squared over X plus delta X squared times X squared times delta acts. So we still need to compute this limit and we still need to simplify this fraction. So we're going to simplify the fraction even more. And what we see is that uh this uh X plus delta X is being raised to the second power. So we're gonna foil that. I'm gonna skip a few steps of boiling, trusting uh that you'll get the same answer as me with your algebra just for the sake of clarity and the length of this problem. And when we do everything we're supposed to do, we will get that. It should be too negative too. X times delta X minus delta X squared, divided by the denominator, which is X plus delta X squared times X squared times delta X. So what happens now? Well, what I would do is I would factor out a delta X from the numerator of my fraction. And in doing that I can cancel out the delta X down below and the delta except top. So factoring out the delta X up top gives us delta X times negative two X minus delta acts divided by it was going to rearrange the order down here. Delta X times X plus data axe squared times X squared. And we see that those delta X is then cancel, which is great for us because then what we end up having is the limit as delta X purchase zero of negative two X minus delta X over X plus delta X squared times X squared. And so what ends up happening now is we can finally compute that limit. We can finally plug in zero for delta X. To get that, it is equal to negative two X over X squared times X squared. And that is equal to negative two X Over X to the 4th, and that is equal to negative two over X to the third. And in fact, that is our derivative.


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