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Find the derivative of f (2) = Vzz += 2; using the limit definition (i.e. first principles)_...

Question

Find the derivative of f (2) = Vzz += 2; using the limit definition (i.e. first principles)_

Find the derivative of f (2) = Vzz + = 2; using the limit definition (i.e. first principles)_



Answers

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

We have problem number 18 equation is X equal toe to minus X square. We need to find its derivative using, uh, limit process so f dash X were formula for to find derivative, which is a brush x equal toe f express Dale Tex minus effects Divide by Dale Tax So and one thing limit. Delta extends to zero. So according to this, we can write our question as to minus explicit text. Hold Squire minus tu minus X squared, divided by the attacks. So to minus just right. Using a plus behold square, this will be X squared. Plus they'll tax whole square plus two X dale tax minus tau minus X squared, divided model texts and off course limit. Delta extends to zero We'll be having limit. Delta extends to zero. Open the back it toe minus X squared minus. Still takes hold square minus two Works deal. Tex minus two plus taxes Square by Dale Text. So let us cancel the common terms out two and minus two Xs Square in minus X square got canceled out, so we have limit. Delta extends to zero minus. They'll tax whole square minus two works. They'll tax by they'll tax now. This is limit. Delta extends to zero at a state deal. Texas Common. We'll be having minus geltex minus two works divided by the tax, so these two will get canceled out. Now we have to plug in zero in place of real tax. So we will be getting the Terra Votive a flashback sequel toe minus two works. Thank you so much.

We want to find the derivative of the function. Why? Or F. Of X equals negative to this question is challenging our newfound knowledge of differentiation. In particular. It's testing our ability to deploy simple differentiation techniques. The problem at hand we've learned for differentiation techniques in the section. They are one through four here one, the constant rule D. D. X equals zero to the power rule, D X. X to the end. Because an X to the m minus one as well as three. The constant multiple rule. And for the some difference rule and all differentiation problems of this nature. We're going to make sure that we've rewritten our problem in its simplest form that makes it easiest to proceed in the next step, differentiation. Once we differentiate, we simplify our result and express the answer. So we ready negative two is already in the most simplest form that you can proceed to differentiate. DDX negative two is zero by the constant multiple rule. Rather the constant rule one db x equals zero. Thus our solution is zero.

Problem number 37 is a problem that requires us to use the limit definition of the derivative to find the derivative of one over X plus two. So the form of the limit definition is given here. And the first thing we want to do is be sure to include accurate impatient. So we've got F prime of X equals the limit as H approaches zero of and so now we'll plug in our X plus H in place of X and then do minus F. Of X as its original functions all over H. So now we thought a complex fraction going on here. And what would help us would be if we had a common denominator in the numerator. So what we can do is multiply the first reaction by X plus two over X plus two and the second ranking by X plus eight plus two over X plus age plus two to create a common denominator. So you've got The limit as H approaches zero of expose to over X plus two times X eight plus two minus X plus H plus two over that same denominator. And then of course this is all still over H. So now we can see that this would create um X plus two minus x minus h minus two, distributing that negative the second fraction. So now my exes and my two's cancel out and I'm left with just negative H over explicit tube, explicit age Plus two over H. So then I can Flip and multiply here multiplied by one over H to cancel out. Which also then cancels out my ages. So then I'm left with Limit as H approaches 0 -1 over my I'm a denominator and as I take this limit and plug in zero in place of H, then I've just got negative one over X plus two square because this would also be exposed to eliminating that age. So exposed to times itself would give me X plus change square.

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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