WEBVTT 00:00:02.400 --> 00:00:09.500 Now we have the tools, limits and continuity, to develop a major concept of calculus, the derivative. 00:00:09.502 --> 00:00:14.900 Let's take a look at a graph. We're going to use quadrant one here, just because everything is positive. 00:00:14.902 --> 00:00:22.250 It makes it a little easier to follow some of the notation. What we're going to do here is we're going to 00:00:22.252 --> 00:00:28.698 draw the graph of a parabola just because it's a convenient function, and we're going to label some 00:00:28.700 --> 00:00:45.301 points on the parabola. This point right here I'm going to label as x on the x-axis and on the y-axis, f of x. 00:00:45.303 --> 00:00:54.200 Now I'm going to take another point further to the right, let's say some quantity away, 00:00:54.202 --> 00:01:06.301 and we're going to call this quantity h. This point also on the graph would be labeled x plus h 00:01:06.303 --> 00:01:18.100 on the x-axis and f of x plus h on the y-axis. Now you have to follow the notation carefully 00:01:18.102 --> 00:01:25.300 to get the concept. What we want to do is develop something that you cannot do with algebra. 00:01:25.302 --> 00:01:34.201 We want to find the slope of the tangent line to this curve at this one point. We start out by using 00:01:34.203 --> 00:01:44.501 what we know from algebra. We start with what's called a secant line, and we draw a line 00:01:44.503 --> 00:01:53.100 across the graph in these two points. We can find the slope of this line using what we already know. 00:01:53.102 --> 00:02:01.401 We're going to call this m sub s, for the slope of the secant line. We know the definition is 00:02:01.403 --> 00:02:14.600 y-two minus y-one, which is the change in y, over x-two minus x-one. For our purposes we will change 00:02:14.602 --> 00:02:28.901 the notation to read slope of the secant line is, and y-two will be the y-value here which is f of x plus h. 00:02:28.901 --> 00:02:38.301 y-one is the y-value of this first point, f of x. So what do we have in the numerator? 00:02:38.303 --> 00:02:48.200 We have the change in y, y-two minus y-one, or the rise of the function, divided by the run, 00:02:48.202 --> 00:03:02.601 which is x-two minus x-one. x-two would be x plus h, and x-one is x. Ok, we see right away that 00:03:02.603 --> 00:03:09.300 in the denominator we have x plus h minus x, so when I use this again, I'm just going to use h 00:03:09.302 --> 00:03:16.901 in the denominator. Now here's what you can do with calculus that you can't do with algebra. 00:03:16.903 --> 00:03:28.601 We're going to let h get smaller and smaller and smaller until h approaches zero. What that does in effect 00:03:28.603 --> 00:03:44.800 is bring this point closer and closer to this point, and as h gets smaller and smaller, this point gives us a 00:03:44.802 --> 00:03:53.901 line that comes to approximate the tangent line at this one point. See I'm letting h get smaller and 00:03:53.903 --> 00:04:05.201 smaller and smaller, and when the limit as h approaches 0 is taken, this point gets so close to this point 00:04:05.203 --> 00:04:13.300 that what we actually have is the tangent line. So let's write that symbolically. 00:04:13.302 --> 00:04:27.701 The slope of the tangent line is the limit as h approaches 0 of the slope of the secant line which is 00:04:27.703 --> 00:04:40.400 f of x plus h minus f of x, and in the denominator I'm just going to simplify and use h. So what we have now 00:04:40.402 --> 00:04:48.900 because of the concept of limit is that we have a way to find the slope of the tangent line to a curve 00:04:48.902 --> 00:04:57.500 at a point. This rather important concept has a name. It's called the derivative. 00:04:57.502 --> 00:05:06.501 Let's look at the definition. For the function f, the derivative of f at x, and you can see how this is 00:05:06.503 --> 00:05:15.200 going to be written. f prime is how this is read. The notation is f prime. The derivative of f at x is, 00:05:15.202 --> 00:05:20.500 and look at what we have here. This part is called the difference quotient. It comes from the slope, 00:05:20.502 --> 00:05:27.900 and then we have the limit as h goes to 0. So the slope of the tangent line to the curve at this point 00:05:27.902 --> 00:05:37.200 is called the derivative, and it's noted as f prime of x. Notice of course that you have to add the little 00:05:37.202 --> 00:05:45.101 piece at the end that says "if the limit exists" because as we've seen with limits they don't always exist. 00:05:45.103 --> 00:05:55.900 Ok, a few other points. The derivative of f at x gives the instantaneous rate of change of f at x. Now why is that? 00:05:55.902 --> 00:06:05.500 Well that's what the slope is. The slope is the rate of change. It's the change in y over the change in x. 00:06:05.502 --> 00:06:15.701 If f prime of x, which is the derivative, exists for each x in the open interval from a to b, then f is said to 00:06:15.703 --> 00:06:25.400 be differentiable over the interval (a, b). Now let's take a look at some examples using the definition. 00:06:25.402 --> 00:06:46.703 Example 1. The directions will say something like this. Given f of x, find f prime of x. Now let's look 00:06:46.705 --> 00:06:59.200 for just a second. f prime of x is the limit as h approaches 0 of f of x plus h minus f of x divided by h. 00:06:59.202 --> 00:07:15.501 So given f of x, find f prime of x, and we're going to add using the definition. 00:07:15.503 --> 00:07:20.702 Right now that's all you have, but that gives you an inkling of the fact that in the future there may be 00:07:20.704 --> 00:07:38.099 some easier way to do this, but we have to work our way there. a. f of x equals 5x minus 7. 00:07:38.101 --> 00:07:44.002 Now let's take a look at this and see if we can figure out what the answer should be before we begin. 00:07:44.004 --> 00:07:51.700 This is a linear function. The slope is 5, so the slope of the tangent line to a linear function would be 00:07:51.702 --> 00:08:00.302 the line itself having the slope 5. So the answer here as we work through this definition should be 5. 00:08:00.304 --> 00:08:13.005 Let's see if we can get this to happen. f prime, which is the derivative of f at x, is defined to be the 00:08:13.007 --> 00:08:30.605 limit as h approaches 0 of f of x plus h minus f of x divided by h. Now let's apply that definition 00:08:30.607 --> 00:08:44.402 to our function, and remember it's if the limit exists. Ok, so f prime of x equals the limit as h approaches 0, 00:08:44.404 --> 00:08:52.003 and for our function f of x plus h would be f of, and remember what that means. You replace every 00:08:52.005 --> 00:09:05.002 value of x with x plus h, so we would have 5 times x plus h minus 7. Alright, now that takes care of this part, 00:09:05.004 --> 00:09:11.803 and so I'm going to put this in brackets. This is not necessary, but it does help to visualize what's happening. 00:09:11.805 --> 00:09:28.000 So this is f of x plus h for our function. Now it says subtract f of x. Well, f of x is just 5x minus 7. 00:09:28.002 --> 00:09:36.706 Ok, and then divide the whole thing by h. Alright now this is a limit, and what we'd like to do is be able 00:09:36.708 --> 00:09:43.698 to just substitute in h equals 0 and get a value, but as you can see when we try to do that we get a 00:09:43.700 --> 00:09:50.300 zero denominator. So what do we do? Well the first thing we try to do is algebraic simplification. 00:09:50.302 --> 00:10:16.603 So let's see. We could distribute the 5 and then the negative, and it's still all divided by h. Notice that 00:10:16.605 --> 00:10:25.203 I keep writing limit, limit, limit because we have not yet taken the limit. Now that equals the limit 00:10:25.205 --> 00:10:42.401 as h goes to 0 of, 5x minus 5x is 0, negative 7 plus 7 is 0, and that leaves us 5h divided by h. Now as 00:10:42.403 --> 00:11:01.504 long as h is not zero, we can cancel these terms, and so what we get now is the limit as h goes to 0 of 5 is 5 00:11:01.506 --> 00:11:07.600 which is exactly what we expected the answer to be. Now let's talk about what we have found here. 00:11:07.602 --> 00:11:20.602 We have found f prime of x which is the derivative of f at x, and it is the instantaneous rate of change of f at x. 00:11:20.604 --> 00:11:25.000 It is also the slope of the tangent line to this curve. 00:11:25.002 --> 00:11:36.101 Let's look at example 1b. Same directions. Given f of x find f prime using the definition. 00:11:36.103 --> 00:11:53.600 The function we're going to use here is f of x equals 2 x squared minus 4x plus 6, and we're going to 00:11:53.602 --> 00:12:06.499 find the derivative f prime using the definition. We start with f prime of x equals the limit as 00:12:06.501 --> 00:12:22.203 h approaches 0 of f of x plus h minus f of x divided by h, and now we apply that defintion to our example. 00:12:22.205 --> 00:12:37.199 So we have f prime of x equals the limit as h approaches 0 of f of x plus h, so f of x plus h would be 00:12:37.201 --> 00:12:55.600 2 times x plus h squared minus 4 times x plus h plus 6. So that takes care of f of x plus h. 00:12:55.602 --> 00:13:13.205 Minus f of x, which is 2 x squared minus 4x plus 6, and all of this is divided by h. Now we'd like to be 00:13:13.207 --> 00:13:20.304 able to sub in here h equals 0, but again we get a zero denominator, so we're going to have to algebraically 00:13:20.306 --> 00:13:34.802 simplify this. So we have f prime of x equals the limit as h goes to 0 of ... Now this term, this binomial 00:13:34.804 --> 00:13:44.602 squared, would be x squared plus 2xh plus h squared, and then the next order of operations would 00:13:44.6044 --> 00:13:57.602 be to distribute the 2, so I would have 2 x squared plus 4xh, and again I'm distributing a 2, this 2 00:13:57.604 --> 00:14:06.999 through here, so I would have 2 h squared. Alright, that takes care of this term. Then I distribute the 4. 00:14:07.001 --> 00:14:18.601 So I have minus 4x minus 4h, and then I have plus 6, and now I'm going to distribute this negative into here, 00:14:18.603 --> 00:14:30.401 and now I have negative 2 x squared plus 4x, distributing the negative, minus 6. It's a very common 00:14:30.403 --> 00:14:36.501 error because you're focusing on the rest of the problem to forget to distribute the negative into this 00:14:36.503 --> 00:14:49.201 function, so let's check it. Negative 2 x squared plus 4x minus 6. And again, this is all divided by h. 00:14:49.203 --> 00:15:03.801 We have 2 x squared minus 2 x squared, so that's 0. We have negative 4x, positive 4x. That's 0, and 00:15:03.803 --> 00:15:12.301 we have 6 minus 6 which is 0. So we had a lot of things cancel out. By the way, make an observation. 00:15:12.303 --> 00:15:19.898 The only terms we have left now, let me show you what they are. We have f prime of x equals, and we 00:15:19.900 --> 00:15:28.790 continue to write limit because we haven't taken the limit yet. So what's left? We have 4xh. 00:15:28.792 --> 00:15:40.600 We have 2 h squared, and we have the negative 4h. Everything else is cancelled out, and of course this is all 00:15:40.602 --> 00:15:48.901 divided by h. Let me give you a little hint here. If you get to this point, all of your terms in the numerator 00:15:48.903 --> 00:15:54.298 should have an h in them. If there is a term here that doesn't, that means you did something wrong. 00:15:54.300 --> 00:16:00.002 Just go back and check your algebra. Now what we want to do is do something to get rid of this h 00:16:00.004 --> 00:16:07.102 so we can do our substitution of h equals 0. So we're going to need to factor an h out of here, and that's 00:16:07.104 --> 00:16:15.800 actually going to work because all the terms contain h as a factor. So we have the limit, still haven't 00:16:15.802 --> 00:16:35.800 taken the limit yet, as h goes to 0 of h times 4x plus 2h minus 4 all over h, and we have achieved 00:16:35.802 --> 00:16:47.102 our goal because h over h is 1 over 1, and we now have a function left, 4x plus 2h minus 4, where we 00:16:47.104 --> 00:17:02.701 can substitute in here and actually take the limit, and when we do, this h goes to 0, and we have left... 00:17:02.703 --> 00:17:21.300 What do we have left? 4x minus 4, and that is the derivative of 2 x squared minus 4x plus 6. 00:17:21.302 --> 00:17:25.790 Alright, let's take a look at another example. 00:17:25.792 --> 00:17:40.600 c. f of x equals 3. Now this is a constant function. When we take the derivative, f prime of x, we write 00:17:40.602 --> 00:17:57.798 the limit as h approaches 0 of, and following the definition, f of x plus h minus f of x divided by h. 00:17:57.800 --> 00:18:06.100 When we attempt to apply this to our function, we don't have a lot of substituting here to do, so what we do is 00:18:06.102 --> 00:18:16.800 we write f prime of x is the limit as h goes to 0 of ... Now because this is a constant function. 00:18:16.802 --> 00:18:33.700 It is constantly 3, so f of x plus h is 3, and f of x is 3, divided by h. So what we have here is 00:18:33.702 --> 00:18:45.100 the limit as h approaches 0 of 0 over h. Now 0 over h is a perfectly good number. It's zero. 00:18:45.102 --> 00:18:57.602 It's the 0 in the denominator that is undefined, so this is the limit as h approaches 0 of 0, and so 00:18:57.604 --> 00:19:07.901 the limit of a constant function, a constant function is constant, and we get zero. 00:19:07.903 --> 00:19:16.000 If you think about this and think of it in terms of the equation of a tangent line, this is a horizontal line. 00:19:16.002 --> 00:19:22.100 What is the slope of a horizontal line? Zero, and so this all works out. 00:19:22.102 --> 00:19:28.500 Let's look at example 2 .This is a different type of example. We're not going to just find the derivative 00:19:28.502 --> 00:19:39.000 using the definition. Instead we're going to actually do something with it. Example 2. 00:19:39.002 --> 00:20:28.501 Find the equation of the tangent line to the graph of f of x equals 2 x squared minus 4x plus 6 at x equals 1. 00:20:28.503 --> 00:20:41.402 Now if you recall, back in example 1b we used this very function, and what we had here was, 00:20:41.404 --> 00:20:48.700 I'm going to write solution here so we can differentiate between what's given and what we're doing. 00:20:48.702 --> 00:20:58.201 We know that if this is the function back in 1b we found using the definition that the derivative of this 00:20:58.203 --> 00:21:07.600 function was 4x minus 4, using the limit definition and algebraically simplifying. So what we're going 00:21:07.602 --> 00:21:12.100 to do now here is we're going to recall that when we developed this definition earlier it is 00:21:12.102 --> 00:21:22.899 the slope of the tangent line to a curve at a point. So this is the slope function. That's what it's called. 00:21:22.901 --> 00:21:30.302 Now you may be used to thinking of the slope as a number, but this is a slope funtion until we actually 00:21:30.304 --> 00:21:38.500 put in the value of x equals 1. Now ultimately we want to find the equation of a line, so what we're going 00:21:38.502 --> 00:21:43.902 to do is remember the point-slope form of a line, and then everything is going to be based off of that. 00:21:43.904 --> 00:21:54.200 It's y minus y-one equals m, which is the slope, times x minus x-one. So our point is going to be 00:21:54.202 --> 00:22:04.501 (x-one, y-one). Now there are three pieces of information here. There is the x-value which is given, 00:22:04.503 --> 00:22:14.200 so that is going to be 1. There is the y-value which is not given but can easily be found, so we're 00:22:14.202 --> 00:22:23.300 going to find this by substituting 1 in the original function. So this value is going to be f of 1. 00:22:23.302 --> 00:22:32.100 This is the new part that you can now do since you know about the derivative. The slope of the 00:22:32.102 --> 00:22:42.500 tangent line to this curve is going to be this function evaluated at 1, so this is going to be f prime 00:22:42.502 --> 00:22:51.401 evaluated at 1. So let's find these two values. Alright, we have the derivative which we got earlier 00:22:51.403 --> 00:23:04.600 from the definition, so f prime of 1 is going to be 4 times 1, which is 4, minus 4 is 0. So we're going to 00:23:04.602 --> 00:23:18.900 put equals 0. Then f of 1 is going to be the value when I plug this into here. So f of 1 is going to be 00:23:18.902 --> 00:23:29.100 2 minus 4, so that would be negative 2, plus 6 is 4, so this value is 4. So what do I have? 00:23:29.102 --> 00:23:36.602 I have y minus 4 equals 0, and what does this mean? This means that we're going to end up with a 00:23:36.604 --> 00:23:44.401 horizontal line because when we put 0 in here the slope of a horizontal line is 0, and then we have 00:23:44.403 --> 00:23:54.200 x minus 1. Now this is technically the equation of a tangent line, but the form is not very nice, so what 00:23:54.202 --> 00:24:01.600 we're going to do here is rewrite it. When we distribute the 0, that term is wiped out, and we end up with 00:24:01.602 --> 00:24:13.200 y equals 4 which is a horizontal line, and that is the equation of the tangent line to this graph at the point x equals 1.