All right, so we need some more practice with the difference quotient, which in a previous video we talked about what that means, what that looks like, and it is a way to define slope. In this case, this would be slope as a function of x, not slope as a number, so not something we can compute. So it's very, very important that we have this formula in our memory banks because you will be expected to know it.
Okay, so the difference quotient, again, that's your f composed with the quantity x plus h minus the original f of x all over h. So very important that we know this formula. And we're just going to do some practice with finding difference quotients of certain functions. And I'm going to use three particular functions just as examples. And we'll do some manipulation with them because the manipulation I'm going to show you in the video is something that you will need to be able to find the derivative.
of the function. So for the first one, just a nice simple polynomial. And so our difference quotient again is F composed with X plus H minus F of X all over H.
We would really love for you to be able to get through this process without partitioning out your work and writing out side work or scratch work. We'd like to be able to follow the work without having to go somewhere else on the page. So we're going to do this. together. Excuse me, I have a little frog in my throat today.
So we're going to take our function f of x, and everywhere we see an x, we're going to replace it with x plus h. Okay, so for the first term, f of x plus h, I'm going to go ahead and put this in brackets, because this is what I'm doing right now. So everywhere I see an x, I'm going to replace it with the quantity x plus h. So I'll have x plus h quantity squared plus x plus h. And that will be my f of x plus h term minus f of x, which is just what's defined here.
And again, I want to make sure I put that in either parentheses or brackets. I like brackets because I like to change from parentheses to brackets and vice versa. All of that is over h.
Now, it's really important that we don't just square each term in this quantity. When we see a quantity squared, that means we have that quantity times itself. So we need to multiply that out.
These parentheses aren't particularly necessary. I only put them there to identify that we were replacing this term with x plus h. So I'll remove those now.
But it is very important that we distribute our negatives. So we have minus x squared minus x. And again, that is all over h.
Now, I'd like to go in and cancel things as I can. All right, if I can, as I can. So I have x and negative x. So those two terms will cancel.
And in the first term, I have to distribute. So I'll have x squared, x squared, plus xh plus x. h. So there's two of those and then h squared.
I've canceled this x. I still have this h. I still have this negative x squared term here and that's all divided by h.
And again I want to see if there's anything I can cancel as I work through this. And I see I have a positive x squared and a negative x squared. So those terms are going to cancel as well. All right, and so now I have 2xh plus h squared plus h all divided by h. And I notice that I have a common factor of h in the numerator, so I can go ahead and factor the h out, and that will leave me with 2x plus h plus 1 all divided by h.
And we've had this conversation a lot lately where we really want to cancel the H's here. And that would be okay, provided that we understand that H cannot be zero because I can't divide by zero. So that's our cleaned up version of the difference quotient for H not equal to zero. All right. So let's go ahead and look at another example.
This one's going to be a root function. Nothing we can't handle. Let's go ahead and rewrite our difference quotient. And you know what? I'm just thinking.
in the middle of my work here that it would be fun to actually give this function a different name. Let's call this one g, shall we? I need to change the video mid-video.
So the difference quotient for g will then be g of x plus h minus g of x all over h, because not all functions are going to be called that. Alright, so here's our difference quotient for g. So again, for g of x plus h, I want to take the original function and replace the x in here with x plus h.
Don't forget the plus 1 and the square root, obviously. So there's my f of x plus h. And this is an example where maybe these brackets and parentheses are not necessary, but I'm trying to use them as an identifying feature.
You know, they're showing you where you have g of x plus h and then g of x, which is the original function. All right. And then all of that over h. All right. And so now we have the square root of x plus h plus one.
I'm just going to go ahead and dump all these parentheses and brackets because they're not particularly necessary. And then I say to myself, is there something? thing I can do to rewrite this with these square roots, right?
And what you're going to be doing this with is, you know, evaluating a limit. And so we're going to have to use things that we've been using so far this semester in terms of limits with roots like this. And so we're going to go ahead and rationalize our numerator by multiplying by the conjugate of the numerator over itself. So this is not a new tool.
It's just a tool we've used in the past, right, in the recent past for a different purpose. Okay, so we'll multiply by the conjugate there, which means we're going to need to add the roots. And then don't forget when you multiply something by its conjugate, this is a nice review, it's the first term squared minus the second term squared.
Remember, it's difference of two squares. So there's no need to actually go through and... distribute, or if you want to use the acronym FOIL for this situation, you could.
There's no need to do that because two of those terms would eventually cancel out. We have h left in our denominator, and we are going to leave this as factored. So h times the quantity square root of x plus h plus 1 plus the square root of x plus 1. And I'm not going to distribute the h in, and I'll show you why in just a moment. All right, so now we have x plus h plus 1, because when we square a square root, that cancels. Minus, now be very careful here, you're subtracting the quantity x plus 1, because that is the result of squaring the square root, but you're subtracting all of it.
Okay, so those parentheses are super necessary. All right, and now our denominator, we're just going to go ahead and copy it as is. We like it the way it is. I know it looks messy, but we have a reason.
All right, now having done that, let's go ahead and distribute that negative through. Okay, so we've got x plus h plus 1 minus x minus 1. And again, that's all divided by h times, leaving it as is. I promise I have a reason.
I got a little messy down there. Sorry about that. And then I noticed that we have x minus x and 1 minus 1. So I can cancel those four terms and I'm left with h in the numerator. I'm just going to pull this down. So let's go ahead and rewrite what we have here.
So that's going to be h divided by h times the quantity square root of x plus h plus 1 plus the square root of x plus 1. And again, look at that. You can cancel out those h's as long as we make sure that we note that this is a true statement for all h not equal to 0. Hmm All right, again, this will be a useful tool for us to be able to rationalize the numerator this way. That basically means we're just going to shift the location of our roots from the numerator down to the denominator.
But again, when we're finding derivatives, this is going to be an extremely useful tool. All right, I'm going to go ahead and use up again for the next example. We're going to look at. a rational expression. All right, so for this one, again, we're looking for our difference quotient f of x plus h minus f of x all over h.
And so for the first term, we're looking at replacing this x with the quantity x plus h. Okay, so it's the same function, but we're replacing x with x plus h, and then minus f of x, so the function in its original form, 1 over x minus 2. And again, all of that is divided by h. So I'm just going to go ahead and dump all the parentheses, because in this situation, they're not extremely useful nor necessary.
So let's just go ahead and rewrite this as 1 over x plus h minus 2. Minus 1 over x minus 2 all over h. Oh, but now we've got that ugly complex fraction thing we don't love. And so this is the situation where, if you remember a couple weeks back, we would multiply both the numerator and the denominator by the least common denominator from the fractions within the fraction.
Okay, so in this case, we want to multiply by x plus h minus 2. I want to try to eliminate that one. But we also want to multiply by x minus 2. And again, whatever we multiply into the numerator, we must also multiply into the denominator. So that we are multiplying by 1 and not changing anything. So we're going to go ahead and distribute this in to the numerator. The nice thing is the denominator, there aren't multiple terms to distribute to.
So we'll just multiply. multiply that right off the bat. We're going to go ahead and skip a step because I think we've graduated to this point where if we're multiplying 1 over x plus h minus 2 times x plus h minus 2, that factor will cancel and we'll be left with x minus 2. So I'll go ahead and put that there. That's what's left in that term. And then here, if we're multiplying the 1 over x minus 2 times x minus 2, that factor will cancel and we'll be left with the x plus h minus 2 factor.
And again, we're going to make sure we put parentheses around these just in case we have some distribution to do, which I think most of you can see that right off the bat. And then I have in the denominator, just made an error here, the h and then our x plus h minus 2, I almost wrote plus 2, times x minus 2. So h times each of these factors. And then I see that I can distribute my negative through in the second quantity.
And so now we have x minus 2 minus x minus h plus 2, all divided by h times x plus h minus 2 times x minus 2. And then we look for things to cancel and we see x minus x. x and negative 2 plus 2. So those are going to cancel out. And now we have negative h over, oh, I ran out of space.
I need one more line. I'll go to the side here. Negative h is all we have left in the numerator.
And then we have h times x plus h minus 2 times x minus 2. Left in the denominator, again, we have the ability to cancel the h's as long as we say that h is not equal to 0 here. And so our simplified form is going to be negative 1 over the quantity x plus h minus 2 times x minus 2. So these are three of many different things that could happen when we're finding derivatives using the difference quotient, but these are the three that you'll definitely need for the first few days of this discussion.