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WebWorkHomework 5: Problem 3Previous ProblemProblem ListNext Problempoint) Let f(x) = 6V1 - 2. If h # 0, then the difference quotient can be simplified ash) - f(x)V...

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WebWorkHomework 5: Problem 3Previous ProblemProblem ListNext Problempoint) Let f(x) = 6V1 - 2. If h # 0, then the difference quotient can be simplified ash) - f(x)VBx + Ch+ VxwhereB, and C are constants: (Note: It'$ possible for one more tnese constants t0 be 0.) Find the constants.and €Use your answer iom adove to find f' (x) = lim [(x+h)_ f(x)Finally; find eachtollowing:f'(I)f' (2)and f' (3) =Note: You can earn partial credit on this problomn; Preview My Answers Subm

WebWork Homework 5: Problem 3 Previous Problem Problem List Next Problem point) Let f(x) = 6V1 - 2. If h # 0, then the difference quotient can be simplified as h) - f(x) VBx + Ch+ Vx where B, and C are constants: (Note: It'$ possible for one more tnese constants t0 be 0.) Find the constants. and € Use your answer iom adove to find f' (x) = lim [(x+h)_ f(x) Finally; find each tollowing: f'(I) f' (2) and f' (3) = Note: You can earn partial credit on this problomn; Preview My Answers Submit Answers You have attempted this problem umes. You havo attempts remaining:



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Compute the following limit for each function in Problems $81-88 .$ $$\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$$ $$ f(x)=3 x+1 $$

For the given problem, we want to find the limit of the difference question. Also known as taking the derivative, where we know f of X is equal to three over X squared. And this is going to equal Messaging Me three over X plus eight square -3 over X. Right? In this case we want to get a common denominator. So we're going to multiply this by X plus H squared. And we're gonna multiply this by X squared. It's become three x squared and it becomes three expressed beach squared. Okay? So then we end up getting um that this is three times X squared Plus six XH plus three H squared. We can combine our terms giving us this right here. We'll factor out the negative eight year, giving us this right here and this When we send H20, though, we end up getting our final answer. Much more simplified.

Section 84 Problem number one We're given a multi bearable function. F is a function of X and y and we're ask defying partial derivatives so fine the derivative of X shooting the Druid of F with respect to X So the partial derivative with respect to x The way you do this is you treat X is your variable and everything else is a constant. So if x were the variable When I take the derivative, I get to X waas to why, okay, because the derivative this last term has no X, no external xn. It's in the driven to zero and then to find the derivative with respect toe why of X and y you treat Why is your variable? Everything else is a constant. So this derivative is going to be two acts plus three why squared

Section eight, not four. Problem number two Partial derivatives We have a function of X and Y two x squared plus three y squared over. Ask defying the partial derivatives with respect to X and with respect a y. So the derivative with respect to acts you basically look at the function and realize that you're only variable. Here is the X variable. You want to take the derivative there I get four X plus zero And then when I take the derivative with respect to why remember that why is my only variable then this becomes six y?

Okay, F of X is equal to expert minus four X. We want to use uh the limit of the difference quotient to find F prime of three. So f prime of three is going to be the limit, as a judge approaches zero of F. Evaluated at three plus H minus F evaluated at three divided by H. So for this function F of x F prime of three will equal the limit of this expression as H approaches zero. So now we just have to go ahead and uh find out what is that for? Three plus H what is that of three? Uh substitute those in and then try to uh do a little bit of algebra and then eventually take the limit F of three plus H f of X is X squared minus four X. So F f three plus H is going to be three plus H squared substituting three plus H N for X or three plus eight square minus four times X. We're plugging in three plus H for xr minus four times three plus each. Then we need to subtract F of threes are substituting three in for X. We're going to subtract ah well f of three is going to be three squared minus four times straight and that all has to get put over each. So F prime of three will be the limit of this expression as H approaches zero. So expanding uh this out three plus H squared is nine plus six. H plus each square, then minus four times three minus 12 minus four times H minus four H minus three square. That subtract nine and then minus of minus 12 is plus fall all over each. So our derivative f prime of three is going to equal to limit of this function as H approaches zero. Now we can do some canceling things are going to get a little bit easier now. Nine minus nine cancels minus 12 plus 12 cancels us. What that leaves us with is the limit as a church approaches zero. In the numerator uh we have six H minus four H. Which is two H plus H squared. And in the denominator we have H. And so we're going to factor out an H. Out of the numerator. So factoring an H. Out of the numerator, the numerator gets rewritten as H times two plus H. H. Times two is 28 pluses plus H. Times H. Is eight square. And that gets put over each and multiplying by H. And dividing by age. Where we can cancel out uh those ages as long as Hs and zero. We can cancel mountain the numerator and denominator. We're taking a limit as H approaches zero. H is not going equals zero. So we can cancel dhs. Now the limit of this expression as H approaches zero, his H approaches zero. Uh We're going to have two plus H approaching two, so to limit of two plus H as H approaches zero is going to be two plus zero, which is to, so this is equal to the derivative of F. A crime evaluated at three. This is F prime when X is through.


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