WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about instantaneous slope and tangents, which are also called derivatives.
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While there are many other things that we could explore in limits, we now have enough of an understanding
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to move on to another major topic in calculus, the derivative.
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The **derivative** of a function gives us a way to talk about how fast the function is changing.
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It allows us to find the **instantaneous slope**, which is also called the **instantaneous rate of change**; they mean equivalent things.
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And this is a new idea, which we will go over in just a moment.
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At first glance, knowing a function's rate of change at any location may not seem that useful.
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But actually, it tells us a massive amount of information.
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It lets us easily find maximums and minimums; it lets us find increasing or decreasing intervals, and many other things.
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Knowing the derivative of a function is really, really useful.
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For example, if we have some function that gives the location of an object,
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the derivative of that function will tell us the object's velocity, because derivative tells us the rate of change.
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If we know something's location, and then we talk about the rate of change of that location,
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well, the rate of change of your location is what your speed is, effectively.
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That gives us velocity, since velocity is pretty much speed.
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This is really useful stuff; being able to talk about derivatives of a function is really, really useful, and it forms one of the cornerstones of calculus.
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Let's check it out: Long, long ago, when we first took algebra, we were introduced to the concept of slope.
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We can think of slope as the rate of change that a line has, how fast it is moving, in a way.
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That is, how far up does it go for going some amount horizontally?
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We define slope as the vertical change, divided by the horizontal, which is also the rise, divided by the run:
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rise over run, the amount that we have changed vertically, divided by the amount that we have changed horizontally.
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This tells us how fast our line is changing; it tells us the rate of change, the slope, how much it moves on a moment-by-moment basis for a line.
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For one step to the right, how much will we go up or down?
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The idea of slope makes a lot of sense for a line, because its rate of change is always constant.
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But if we wanted to apply the idea of slope to a non-line, something like, say, a parabola, for example?
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The first thing to notice is that, for most functions, slope is constantly changing--
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not necessarily for all functions (for a line, it isn't changing), but for anything that isn't a line, the slope won't be the same everywhere.
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The rate of change for a function varies depending on what location we consider.
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For example, on this one, how fast it is changing is totally different in this area; it is totally different from this area and totally different from this area.
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Each of those three areas is going in a very different rate of change.
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The way it is moving there is very different.
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In this area over here, it is mainly going down; in this area over here, it is mainly just going horizontally; and in this area over here, it is mainly going vertically.
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There is always some horizontal motion in this case, but it ends up changing how it is moving.
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So, it doesn't have a constant slope; its slope is changing for these things.
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We want to have some way of being able to talk about what the slope is at this place.
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What does it change to in this moment?
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So, we can't talk about a rate of change (slope) for the entire function, but we can look for a way to find the instantaneous slope.
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What is the slope at some specific point?
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Here, it is changing at this moment, at some current speed; it is going like this here.
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But here, it is going like this; or here, it is going like this; and here, it is going like this.
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We end up getting these different ways of being able to talk about how it is moving in this moment.
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Where is it going from one spot to the next spot--how is it changing at that place?
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Another way to work towards this idea of instantaneous slope at a point is through the notion of a tangent line.
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Now, let's have just a quick break from this: there is a slight relationship between the trigonometric function "tangent" and the "tangent" line of something.
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But it is really not worth getting into; it is not going to really help us understand things, and the connection is only really tenuous.
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For now, let's just pretend that they are two totally different ideas, and that they just happen to have the same name.
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Tangent, when we talk about taking the tangent of some angle, and when we talk about the tangent line on a curve--
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they are completely unrelated, at least as far as we are concerned right now.
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It is easier to think about it that way.
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Back to what we were talking about: for a circle, the tangent line to a point on the circle
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is a line that passes through the point, but intersects no other part of the circle.
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Consider this point right here on the circle: the tangent line will go through that point, but it will intersect no other part.
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We look at it, and we see that we get this right here.
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See how it barely touches--it just touches, feather-like, that one point on the circle; but it touches nothing else.
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I want you to notice how the tangent line is basically the instantaneous slope of the curve at that point.
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If we look in this region, right here, at our point, that is how much the curve is changing at that moment.
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So, it is as if the tangent line is going the same direction as the curve in that one location.
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We can take this idea of a tangent line and expand it to any curve.
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If we have some function f(x), when we draw its graph, we just have a curve on our paper.
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Now, we can consider some point on the graph and try to find a tangent line at that point.
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So, say we have some point, like right here, and we want to do the same thing
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of just barely feather-like touching that one location and going in the same direction as the graph is going.
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It will pass through that one place, but just barely going in the same direction as the curve is in that moment.
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We look at that, and we see how that has the instantaneous slope.
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Notice: the tangent line is the instantaneous slope of the curve at that point.
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This curve right here has what the slope is at this one place.
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But if we go to some other place, we would end up having a totally different tangent line for these different points.
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If a tangent line were to pass through these different points, it would have a totally different slope.
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The tangent line is going in the same direction as the curve is at that moment, at that single point.
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At a different point, it might end up having a totally different tangent line, having a totally different slope.
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So, how can we find this instantaneous slope--what can we do to work towards this slope at some point?
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Well, let's say we wanted to find the specific instantaneous slope for the function f(x) = x² + 1 at a horizontal location of x = 1.
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Here is the point that we are trying to find the instantaneous slope of.
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And notice that we are currently at the horizontal location x = 1.
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How could we do this--what could we do to approach it?
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Well, long ago, when we talked about the properties of functions,
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we noticed how we could talk about the average slope between two horizontal locations,
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x₁ and x₂, on a function, with this formula here.
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The average slope between these two horizontal locations, x₂ and x₁, is f(x₂) - f(x₁)/(x₂ - x₁.
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Well, why is that? Well, if we have some other point that is at some x₂, what height will that end up being at?
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Well, that will end up being at a height of...to find the height, we just evaluate the horizontal location.
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Your input is your horizontal location, and your output is your vertical location.
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So, that would come out to be f(x₂); what input did we have?
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Well, we put in x₁, so we would have an output of f(x₁).
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So, our top, the change, is going to be the difference: f(x₂) - f(x₁),
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because we went to f(x₂), and we came from f(x₁).
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The difference in our ending and our starting is f(x₂) - f(x₁).
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So, that tells us the top part of our average slope formula.
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The bottom part of it: well, if we go from x₁ to x₂, then that means that our horizontal motion is going to be x₂ - x₁.
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If we go to 10 from 2, we have traveled 8, 10 minus 2; and so, that is why we divide by x₂ - x₁.
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It is the rise (how much we have changed vertically), divided by how much we have changed horizontally.
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So, we have our function, f(x) = x² + 1; and we want to know what the slope at x = 1 is.
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What is the value? We have our slope average formula, m<font size="-6">avg</font> = f(x₂) - f(x₁) over (x₂ - x₁).
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Now, we want to know the slope at x = 1; so we can use this average slope formula to give us approximations.
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By using the average slope formula, we can get approximations for what the slope is near x = 1, for finding our instantaneous slope at x = 1.
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So, since we want to find it near x = 1 (we want to find it specifically at x = 1),
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let's set the first thing that we will use in our average slope formula as x₁ = 1; so we establish this as being x₁.
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For our first approximation, let's use a horizontal location that happens to be two units forward.
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So, we will have our second horizontal location be moved two forward; we go forward one, forward two; and that makes 3 x₂.
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Or, we could have x₂ = x₁ + 2; and since we are using x₁ = 1, we have 1 + 2, which comes out to be 3.
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So, we have x₁ and x₂; we are looking to find the average slope between these two points.
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All right, our function is f(x) = x² + 1; we are looking for our instantaneous slope at x = 1.
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We are working towards that by approximations right now.
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And our average slope formula is f(x₂) - f(x₁), over (x₂ - x₁).
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We know that x₁ = 1, because that is the point that we are interested in finding the slope for.
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So, we will just set that as sort of a starting place to work from.
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And we decided, for our first one, to go with x₂ = x₁ + 2, which was 3.
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Using our formula, we have m<font size="-6">avg</font> = f(x₂) - f(x₁) over (x₂ - x₁).
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x₂ is 3; so if we plug that into f(x), then we have f(x₂) = 3; so f(x₂) is f(3).
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f(3) would come out to be 3² + 1; 3² + 1 gets us 10.
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For f(x₁), f(x₁) would be f(1); f(1) would be 1² + 1, so that gets us 2.
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So, it is 10 - 2 on the top, and then x₂ (3) minus x₁ (1), so 3 - 1 on the bottom.
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We simplify the top; we get 10 - 2, which becomes 8; 3 - 1 becomes 2; 8 divided by 2 gets us 4.
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And if we draw it in, we end up getting this purple line right here: that is the slope if we set it equal to that average slope.
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It ends up passing through those two points; so we have an average slope of 4 between those two horizontal locations.
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OK, that is not a bad start; it is far from perfect--we can clearly see that this line right here is not, in fact, the tangent line.
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It is not the tangent line; it doesn't pass perfectly against that point; it doesn't just barely, feather-like, touch that one point.
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It isn't going in the same direction, but it does give us an approximation; it is a start.
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How can we make this approximation better?
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Well, we could probably think, "Well, the issue here is the fact that x₂ is too far away."
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We want x₂ to be closer; so we can improve it by bringing x₂ closer to x₁.
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This time, let's go only one away; we will do x₁ + 1, so that it is only one distance forward.
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So, we will now have x₁ + 1, or 2; since we are starting at 1, 1 + 1 gets us 2; now, we are only one horizontal distance away.
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It is still the same function, x² + 1; we are still looking for x = 1; it is still the same slope average formula.
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x₁ is still equal to 1, but now x₂ is going to be one forward from 1, so that is 2 for our x₂.
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We plug that into our average slope formula; f(x₂) is f(2) in this case, because that is our x₂ at this place.
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So, f(x₂) would be 2² + 1; 4 + 1 gets us 5;
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minus f(x₁): f(x₁) is still 1² + 1, so that still gets us - 2.
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Our bottom is now x₂ - x₁; our new x₂ is 2, in this case;
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2 - 1 simplifies to...5 - 2 on top becomes 3; 2 - 1 on the bottom becomes 1.
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And so, we get a slope of 3; and if we graph that, we end up getting this line right here.
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All right, nice--we are getting better; it is still not perfect, but the approximation is improving as we bring x₂ closer to x₁.
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So, as we bring our x₂ closer and closer and closer, we are going to end up getting better and better approximations.
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What we want to do is bring it really close.
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Before we can bring x₂ really close, though, we need to think about what we are doing in general,
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so that we can figure out an easy way to formulate talking about bringing x₂ really, really close.
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So, let's talk about what we have been doing, in general.
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We have our average slope formula; m<font size="-6">avg</font> is equal to f(x₂) - f(x₁), divided by (x₂ - x₁).
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That is the output of our second point, minus the output of our first point, divided by the horizontal location difference of our second and first points.
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So, what we did first was (since we are looking to use this m<font size="-6">avg</font> formula): we set x₁ at some value.
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In this case, we set it at the point that we are interested in; we wanted to find the slope of x = 1, so we set our first point,
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our first horizontal location, as x₁ = 1; the first horizontal location is 1, because we want to find out about that slope.
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Then, from there, we set x₂ some distance away from it.
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The very first time, we set it 2 distance away; so 1 + 2 became 3.
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The second time we did this, we had 1 + 1 (a horizontal distance of 1); 1 + 1 became 2.
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So, we want to bring x₂ closer and closer by putting less and less distance.
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What we really care about isn't so much the second point, but the distance to the second point.
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There are two different ways of looking at this.
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So, let's call this distance something; we will call it h.
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What we want to do is bring this h smaller and smaller and smaller.
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We want to bring x₂ closer and closer, so we want to make the distance between our point that we care about,
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and the point that we are referencing against for our average slope, to become closer and closer and closer.
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So, if we are calling this distance h, then we can say that x₂ is equal to x₁, our starting place, plus the distance away, h.
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So, x₂ = x₁ + h; with this in mind, we can now rewrite our average slope formula in terms of x₁.
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We started with m<font size="-6">avg</font> = f(x₂) - f(x₁), over (x₂ - x₁).
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But now, we have this new way of writing x₂: x₂ is equal to x₁ plus the distance forward, h.
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So, we can rewrite the formula in terms of x₁ and the horizontal distance, h, to be able to create a new formula--
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not a new formula, so much as a restatement of the old formula--but a new way of looking at it.
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So, we plug in x₂ here; it now becomes x₁ + h; so we have x₁ + h...
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f(x₁) is still the same; so f(x₁ + h) - f(x₁); x₂ now becomes x₁ + h, minus...still x₁.
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On the bottom, we have x₁ here and -x₁ here; so we can cancel them out.
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And so, we are left with just h here; on the top, though, we can't cancel anything out,
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because f(x₁ + h) and f(x₁)...we don't know how they compare until we apply some specific function to them.
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So, we are going to have to use the function before we can cancel anything out.
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We can't cancel out the x₁ part inside of the function,
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because we have to see how h interacts with x₁ before we can cancel anything out.
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OK, one last thing to notice: at this point, the only thing showing up is x₁.
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Only x₁ shows up in this formula: we have x₁ here, x₁ here, x₁ here, x₁ here, x₁ here, x₁ here.
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But no longer x₂: we don't have to care about it anymore, because instead we are thinking in terms of this distance h.
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That is how we swapped to x₁ + h.
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So, if we don't really care about x₁ versus x₂, then we can just rename x₁ as simply x,
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because at heart, we are all lazy; and it is easy to write out x, compared to x₁--just one less thing to write.
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So, we can now write our average slope as: the average slope is equal to f(x + h) - f(x), divided by h.
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All right, going back to our thing, we have that our function is x² + 1 (back to our specific example);
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and we want to find out what the slope is at a specific value of x = 1.
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So, for any h, for any distance away from our location x = 1, the average slope is going to be equal to f(x + h) - f(x), all divided by h.
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Let's see how that is the case: well, if we have x here, and we have x + h here,
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then the distance forward that we have gone is x + h - x, or simply x.
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So, that is why we are dividing by h, because we divide by the run; we divide by the horizontal change.
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And if we want to look at the two heights, well, the height that we will end at will be f(x + h).
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If we plug x + h in to get an output, it is going to be f acting on (x + h).
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What is the first one? Our first location is going to be plugging in x, so that would be f(x), f acting on our input of x (so f(x)).
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What is the distance that we end up having? Well, that will be f(x + h) - f(x).
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And so, that is why we end up having f(x + h) - f(x) on the top.
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So, for any distance h that we end up going out, that tells us what the average slope is
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between our horizontal location x and the h that we end up choosing--whatever distance we end up wanting to use.
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So, as we make our h smaller and smaller, we will be able to get better and better approximations.
00:18:03.900 --> 00:18:06.600
We want to see why we end up getting better and better approximations.
00:18:06.600 --> 00:18:18.300
Notice: let's look at a couple of different points; we choose different points, and we end up running lines through these different slopes.
00:18:18.300 --> 00:18:27.300
We end up seeing that we get closer and closer to the actual tangent we want.
00:18:27.300 --> 00:18:30.200
We are getting closer and closer to the tangent we want.
00:18:30.200 --> 00:18:38.900
So, as we bring our h closer and closer, as we make our h shorter and shorter and shorter, we get better and better approximations.
00:18:38.900 --> 00:18:44.400
So, as h grows smaller and smaller, our slope approximation becomes better and better.
00:18:44.400 --> 00:18:49.000
Now, what would be great is if we could somehow set h equal to 0.
00:18:49.000 --> 00:18:54.700
We want to have the smallest amount of space we can possibly have; the less distance we have, the better our approximation.
00:18:54.700 --> 00:18:58.100
But if we were to set h equal to 0, then we would be dividing by h.
00:18:58.100 --> 00:19:05.500
It is f(x + h) - f(x), divided by h; so if h equals 0, we would divide by 0.
00:19:05.500 --> 00:19:09.200
We can't divide by 0, because that doesn't make sense; it is not defined.
00:19:09.200 --> 00:19:15.500
So, what we want is...if only there was some way that we could somehow have the same effect as dividing by 0,
00:19:15.500 --> 00:19:17.900
but not have that issue where we are actually causing it to divide by 0.
00:19:17.900 --> 00:19:23.500
If only we could look at what it was going to become the instant before it ended up breaking...
00:19:23.500 --> 00:19:26.500
Limits! That is what that whole thing that we were studying with limits was about!
00:19:26.500 --> 00:19:35.700
Limits give us a way to talk about what it will become before it breaks--what it is going to become just before it ends up dissolving and not actually making sense.
00:19:35.700 --> 00:19:41.800
So, we want that infinitesimally small (as h gets really, really, really, really close to 0) thing that ends up happening.
00:19:41.800 --> 00:19:45.500
What we are looking for is the limit as h goes to 0.
00:19:45.500 --> 00:19:52.200
As this becomes really, really close to actually being on top of that point, what value do we end up getting out?
00:19:52.200 --> 00:19:55.700
That is going to be the best sense of what the instantaneous slope is.
00:19:55.700 --> 00:20:02.100
By getting it really, really, really close, we will be able to get our best idea of what the slope is at that exact place.
00:20:02.100 --> 00:20:08.800
All right, with this idea in mind, let's take the limit as h goes to 0 of our average slope formula for f(x).
00:20:08.800 --> 00:20:14.000
So, our average slope formula is m<font size="-6">avg</font> = [f(x + h) - f(x)]/h.
00:20:14.000 --> 00:20:19.300
So, what we do is take the limit as h goes to 0, because the limit as h gets smaller and smaller...
00:20:19.300 --> 00:20:21.700
our average slope will give us a better and better approximation.
00:20:21.700 --> 00:20:27.900
As h goes to 0, as h becomes infinitesimally close to it, we will end up getting the best possible approximation.
00:20:27.900 --> 00:20:34.800
The limit just before the instant it touches--that is the best possible approximation we can get for what the slope is at that place.
00:20:34.800 --> 00:20:41.000
So, at this point, we can now plug in f(x) = x² + 1 into our specific f(x) up here
00:20:41.000 --> 00:20:48.600
and start trying to work towards what the formula is for what the slope will be at that place, at our location x.
00:20:48.600 --> 00:20:53.700
So, if we have f(x + h), and we are plugging it into f(x) = x² + 1,
00:20:53.700 --> 00:21:00.800
we are plugging in x + h into something squared plus 1; so we get (x + h)² + 1,
00:21:00.800 --> 00:21:06.400
because that is what our function does; so it is (x + h)² + 1 for our first portion;
00:21:06.400 --> 00:21:13.200
and then minus...when we plug in just x, we end up just getting x² + 1, so - (x² + 1).
00:21:13.200 --> 00:21:16.600
And the bottom is just h, because we don't have anything to affect the bottom yet.
00:21:16.600 --> 00:21:24.400
Limit as h goes to 0...well, we can expand (x + h)²: (x + h)² becomes x² + 2xh + h².
00:21:24.400 --> 00:21:30.700
The +1 still remains; and -(x² + 1) is -x² - 1.
00:21:30.700 --> 00:21:36.100
At this point, we see that we have positive x² and negative x², so they can cancel each other.
00:21:36.100 --> 00:21:39.600
We have +1 and -1, so they can cancel each other.
00:21:39.600 --> 00:21:44.200
And we are left with 2xh + h² on top, all divided by h:
00:21:44.200 --> 00:21:48.800
the limit as h goes to 0 of 2xh + h², all divided by h.
00:21:48.800 --> 00:21:55.100
Great; if we had just plugged in 0 initially, it would end up breaking; we would have 0/0, so we don't end up getting anything out of it.
00:21:55.100 --> 00:21:59.400
But at this point, we can now cancel things; that is one of the things we talked about when we wanted to evaluate limits.
00:21:59.400 --> 00:22:02.300
Remember the lesson Finding Limits: how do we find these limits?
00:22:02.300 --> 00:22:05.900
We get them to a point where we can cancel stuff, so we can see what is going on.
00:22:05.900 --> 00:22:11.000
So, at this point, we can cancel; we have 2xh + h², over h, in our limit;
00:22:11.000 --> 00:22:17.000
so we see that we can cancel this h; that will cancel the h here; and over here, it will cancel the squared and turn it to just h to the 1.
00:22:17.000 --> 00:22:21.600
So, at this point, we have it simplified to the limit as h goes to 0 of 2x + h.
00:22:21.600 --> 00:22:30.100
And as h goes to 0, 2x won't be affected; but the h will end up canceling out as it just drops down to 0, and we will be left with 2x.
00:22:30.100 --> 00:22:33.700
Now, what point did we care about? We cared about the horizontal location.
00:22:33.700 --> 00:22:43.500
We wanted to show slope at x = 1; so we now have this nice formula to find out what the slope is at some horizontal location.
00:22:43.500 --> 00:22:50.300
So, we can plug in our x = 1, and we have the instantaneous slope when the horizontal location is x = 1.
00:22:50.300 --> 00:22:56.200
We know what the slope of f(x) is at the single moment, that single horizontal location, of x = 1.
00:22:56.200 --> 00:23:06.600
2 times 1...we plug in our value for our x; 2 times 1 comes out to be 2; we have found what the slope is.
00:23:06.600 --> 00:23:10.900
Let's check--let's see it graphically: if we check this against the graph, we see
00:23:10.900 --> 00:23:16.000
that a slope of 2 at the horizontal location x = 1 produces a perfect tangent line.
00:23:16.000 --> 00:23:24.500
This slope of 2 at the horizontal location x = 1, right here, produces a perfect tangent line to the curve at that point.
00:23:24.500 --> 00:23:33.000
If we draw a line that goes through that point, that has this slope of 2, we end up seeing that it does exactly what we are looking for.
00:23:33.000 --> 00:23:39.100
It has this bare feather-like touch, just barely on that curve; it just barely touches that one point,
00:23:39.100 --> 00:23:48.700
and it goes off in the direction that the curve has at that one instant, at that one horizontal location, at that one point; cool.
00:23:48.700 --> 00:23:55.100
So, this leads us to define the idea of a derivative; we can do what we just did here, for this one specific function, in general.
00:23:55.100 --> 00:24:02.900
We define the derivative: the **derivative** is a way to find the instantaneous slope of a function at any point.
00:24:02.900 --> 00:24:14.100
The derivative of the function f at some horizontal location x is f prime of x (we read this f with this little tick mark as "f prime of x"),
00:24:14.100 --> 00:24:21.700
and it is the limit as h goes to 0 of f(x + h) - f(x), all divided by h.
00:24:21.700 --> 00:24:29.500
So, our average slope gives us a better and better sense of what is happening at that instant at that horizontal location x.
00:24:29.500 --> 00:24:36.100
As our h shrinks down, we get a better and better sense of what is going on from this average slope thing right here.
00:24:36.100 --> 00:24:42.000
And since we have a better and better sense, the best sense that we will have is the instant before it actually ends up breaking on us at h = 0.
00:24:42.000 --> 00:24:44.100
So, we take the limit as h goes to 0.
00:24:44.100 --> 00:24:49.100
Now, of course, this limit has to exist; if the limit doesn't exist, then the derivative doesn't end up working out.
00:24:49.100 --> 00:24:54.100
But as long as the limit does exist, we manage being able to find out what that instantaneous slope is.
00:24:54.100 --> 00:25:01.400
We call the process of taking a derivative **differentiation**; this is called differentiation--we take the derivative through differentiation.
00:25:01.400 --> 00:25:05.000
You can differentiate a function to get its derivative.
00:25:05.000 --> 00:25:10.900
And when we write it out, it is just this little tick mark right here, f'(x).
00:25:10.900 --> 00:25:18.100
You just put this little tick mark right next to your letter that is the letter of the function; and that says the derivative of that function.
00:25:18.100 --> 00:25:24.300
Once we have the derivative, f'(x), for some function f(x), we can find the instantaneous slope
00:25:24.300 --> 00:25:31.700
at some specific horizontal location x = a by simply plugging it into our general derivative formula, f'(x).
00:25:31.700 --> 00:25:40.400
If we want to know the instantaneous slope at some horizontal location a, we just plug it into f'(x), and we have f'(a),
00:25:40.400 --> 00:25:46.300
just like we did before--we figured out that, in general, for x² + 1, f prime became 2x;
00:25:46.300 --> 00:25:49.900
and then we wanted to know what it was at the specific horizontal location of 1.
00:25:49.900 --> 00:25:57.900
So, we took f'(1); we plugged in 1 for our x, and we got simply 2 as the instantaneous slope at that location.
00:25:57.900 --> 00:26:05.500
The derivative is a very, very important idea in calculus; it makes one of the absolute cornerstones that calculus is built upon.
00:26:05.500 --> 00:26:08.900
And so, there end up being a number of different ways to denote it.
00:26:08.900 --> 00:26:16.700
Given some y = f(x), that is, some function f(x)--or we could talk about it as the vertical location y--
00:26:16.700 --> 00:26:19.200
we can denote the derivative with any of the following.
00:26:19.200 --> 00:26:22.200
We can talk about the derivative with any of the following symbols:
00:26:22.200 --> 00:26:33.400
f'(x), dx/dy, y', d/dx of f of x...and there are even some other ones.
00:26:33.400 --> 00:26:38.500
In this course, we will end up using f'(x) for the limited period of time that we actually talk about derivatives.
00:26:38.500 --> 00:26:42.000
But these other ones will end up being used occasionally, as well.
00:26:42.000 --> 00:26:45.200
And there are reasons why they end up being used; they actually make sense in calculus.
00:26:45.200 --> 00:26:48.400
We don't quite have time to talk about it right now, but as you work through calculus,
00:26:48.400 --> 00:26:53.200
as you study calculus, see if you can start to understand why we are talking about it as dx/dy.
00:26:53.200 --> 00:26:58.000
It has to do with these ideas of infinitesimals; but I will leave that for you, working in your calculus course.
00:26:58.000 --> 00:27:02.300
All right, the important thing to know is that, while there are all of these different ways to talk about it--
00:27:02.300 --> 00:27:07.600
we have just 4 right here, but there are even more, occasionally (but you will only end up experiencing really...
00:27:07.600 --> 00:27:11.600
these two right here are really likely the most common ones you will end up seeing,
00:27:11.600 --> 00:27:16.200
and this is the only one we will use in this course), they all do the same thing.
00:27:16.200 --> 00:27:23.100
They represent some function; they represent the derivative that tells us the instantaneous slope for a horizontal location.
00:27:23.100 --> 00:27:28.600
We plug in a horizontal location, and it tells us what the instantaneous slope is, what the slope is,
00:27:28.600 --> 00:27:33.700
what the rate of change is at that one specific horizontal location.
00:27:33.700 --> 00:27:39.900
The important idea of all of this that I really want you to take away (we will work to it) is that,
00:27:39.900 --> 00:27:44.500
as you progress in calculus, you are quickly going to learn a wide variety of rules.
00:27:44.500 --> 00:27:51.200
You are going to learn a lot of different rules, a lot of different techniques, that will make it really easier (much easier) to find derivatives.
00:27:51.200 --> 00:27:57.400
Things that at first seem complicated will actually end up becoming pretty easy, as you learn rules and get used to using rules in calculus.
00:27:57.400 --> 00:28:04.100
And in actuality, you will very seldom, if ever, use that formal definition that we just saw--
00:28:04.100 --> 00:28:08.300
that limit as h goes to 0 of f(x + h) minus f(x), over h.
00:28:08.300 --> 00:28:14.800
That thing won't actually end up getting used a lot; we will mainly end up using these rules and techniques that you will learn as you go through calculus.
00:28:14.800 --> 00:28:20.300
So, if that is the case--if we end up not really using it that much--why did we learn it? What was the point of learning it?
00:28:20.300 --> 00:28:23.200
It is because the important part, the reason why we are talking about all of this,
00:28:23.200 --> 00:28:27.000
is to give you a sense of what the derivative represents before you get to calculus.
00:28:27.000 --> 00:28:33.800
What we really want to take away from this is that the derivative is a way to talk about instantaneous slope.
00:28:33.800 --> 00:28:40.500
It is a way to talk about how something is changing, and equivalent to instantaneous slope is the instantaneous rate of change.
00:28:40.500 --> 00:28:45.800
How is the function changing in this one moment, at this one point, at this one horizontal location?
00:28:45.800 --> 00:28:49.400
How is it changing--what is the slope right there of the function?
00:28:49.400 --> 00:28:55.400
It is what it is right there in this function at some specific location; that is the idea of a derivative.
00:28:55.400 --> 00:29:00.300
So, as you learn these rules, no matter how many rules you learn for finding derivatives,
00:29:00.300 --> 00:29:11.000
never forget that, at heart, what a derivative is about is a way to talk about a function's moment-by-moment change.
00:29:11.000 --> 00:29:15.500
It is this idea of how the function is changing right here, right now.
00:29:15.500 --> 00:29:18.000
What is the slope at this specific place?
00:29:18.000 --> 00:29:20.900
You will end up learning a lot of rules; you will end up learning a lot of techniques.
00:29:20.900 --> 00:29:27.500
And it is easy to end up getting tunnel vision and focusing only on the rules and techniques and getting the right numbers out, getting the right symbols out.
00:29:27.500 --> 00:29:32.400
But even as you are ending up working on this, try to keep that broad idea of what you are thinking about.
00:29:32.400 --> 00:29:35.800
It is how the thing is changing--how your function is changing, on the whole.
00:29:35.800 --> 00:29:42.900
You will have to understand how to get those correct values, how to get those correct symbols, when you differentiate--when you take the derivative.
00:29:42.900 --> 00:29:48.800
But if you forget that the idea of all of this is to talk about the rate of change, you are missing the most important part.
00:29:48.800 --> 00:29:56.600
The most important part that makes all of this actually have meaning--to be useful--is this idea of how the function is changing here and now.
00:29:56.600 --> 00:30:00.100
That is why we care about the derivative--not just so that we can churn out numbers;
00:30:00.100 --> 00:30:06.100
not just so that we can churn out symbols; but so that we can talk about how this thing is changing right here and right now.
00:30:06.100 --> 00:30:10.500
And by understanding that the derivative represents how this thing is changing right here and right now,
00:30:10.500 --> 00:30:15.900
and thinking about it in those terms, you will be able to understand all of the larger ideas that we get out of a derivative,
00:30:15.900 --> 00:30:20.800
of what the derivative represents, and all of the interesting things that a derivative tells us about a function.
00:30:20.800 --> 00:30:24.100
If you just try to memorize the techniques and rules, and that is all you focus on,
00:30:24.100 --> 00:30:27.700
you won't have a good sense for what you are doing, and it will become very difficult in calculus.
00:30:27.700 --> 00:30:33.200
But if you keep this idea in mind of what it represents, it will be easy to understand how things fit together.
00:30:33.200 --> 00:30:36.300
It will make things a lot more comfortable and make things make a lot more sense.
00:30:36.300 --> 00:30:41.200
So, the important part of all of this, that I really want you to take away, is to keep the derivative in mind
00:30:41.200 --> 00:30:48.500
as a way of talking about how the function is changing at some specific location: what is its slope right there?
00:30:48.500 --> 00:30:50.000
All right, we are ready for some examples.
00:30:50.000 --> 00:30:54.700
Let f(x) = x³, and consider the location x = -2.
00:30:54.700 --> 00:30:58.900
We have some function x³, and we are considering the location x = -2.
00:30:58.900 --> 00:31:08.200
Approximate the slope (that is the instantaneous slope), using our slope average function f(x + h) - f(x) divided by h, and the following values for h.
00:31:08.200 --> 00:31:17.100
For our first one, h = 2: if our x is at -2, then our x is going to be at -2 for h = 2.
00:31:17.100 --> 00:31:29.600
x + h, this portion right here, will be equal to...well, if it is -2 for x, and h is 2, then 2 + -2 comes out to be...
00:31:29.600 --> 00:31:34.500
well, let's write it the other way around: x + h, so -2 + 2.
00:31:34.500 --> 00:31:39.900
-2 + 2 will come out to be 0; so now, let's plug it into our average slope formula.
00:31:39.900 --> 00:31:48.200
The average slope is equal to f(x + h) (that was 0), minus f(x) (that is -2, the point we are concerned with),
00:31:48.200 --> 00:31:51.500
divided by h, the distance we are going out (that is 2).
00:31:51.500 --> 00:32:03.700
We start working this out; the average slope, f(0)...well, if it is x³, that is 0³ - (-2)³, all divided by 2.
00:32:03.700 --> 00:32:16.700
That comes out to be 0 minus...-2 cubed is -8...over 2; 0 - -8 becomes +8, so we have 8/2, which equals 4.
00:32:16.700 --> 00:32:24.200
So, this first approximation at h = 2 is: we end up getting an average slope of 4.
00:32:24.200 --> 00:32:30.100
The next one: h = 1--so once again, our first place will be x = -2, and then we are working from there.
00:32:30.100 --> 00:32:39.800
x = -2: so x + h will be equal to -2 + 1 (we are going one distance out from -2), which simplifies to -1.
00:32:39.800 --> 00:32:56.500
So, our average slope between -2 and -1 is going to be f(x + h), so f(-1), minus f(-2), over positive 2.
00:32:56.500 --> 00:33:07.400
What is f(-1)? Well, that is going to end up being -1 cubed...we already worked out that - f(-2) becomes -(-2)³,
00:33:07.400 --> 00:33:14.700
which means - -8, which became just +8; so we can just write that as + 8 right now, skipping to that part...divided by 2.
00:33:14.700 --> 00:33:19.000
-1 cubed becomes -1 times -1 times -1, or just -1, plus 8...
00:33:19.000 --> 00:33:25.400
Oh, I'm sorry; it is not divided by 2; I'm sorry about that; our h is 1--that is what we are using as our h.
00:33:25.400 --> 00:33:31.200
I accidentally got stuck on h = 2; we are dividing by 1, because that is the distance out: 1.
00:33:31.200 --> 00:33:43.700
-1 cubed, plus 8, is -1 + 8; we have 7/1, which equals an average slope of 7 between x = -2 and going a distance of 1 out.
00:33:43.700 --> 00:33:58.400
Our final one: h = 0.1: our h is still equal to the same starting location, -2, but now we are going out to the very tiny x + h = 2 + 0.1, which gets us 1.9.
00:33:58.400 --> 00:34:00.500
So, it is just a tiny little bit forward now.
00:34:00.500 --> 00:34:09.900
Our average slope is going to be f(-2)...sorry, not -2; we do x + h first, and this should be -1.9, because it is -2 + 0.1.
00:34:09.900 --> 00:34:22.000
So, x + h is f(-1.9), minus f(-2), all divided by our h; that is 0.1.
00:34:22.000 --> 00:34:33.700
f(-1.9) is going to be -1.9 cubed; we already figured out that minus -2 cubed becomes +8; divide by 0.1.
00:34:33.700 --> 00:34:44.300
Negative -1.9, cubed, becomes negative 6.859, plus 8, divided by 0.1.
00:34:44.300 --> 00:34:51.000
So that simplifies, up top, to 1.141 divided by 0.1.
00:34:51.000 --> 00:35:00.500
We divide by 0.1, and that moves the decimal over, and we get 11.41; great.
00:35:00.500 --> 00:35:05.400
And so, that gives us our final average slope of h = the tiny distance of 0.1.
00:35:05.400 --> 00:35:09.700
So, notice: at the very large distance, we got 4; at h = 2, we got 4.
00:35:09.700 --> 00:35:17.500
At h = 1, we got 7; at h = 0.1, we got 11.41; we are slowly approaching some specific value for what it is.
00:35:17.500 --> 00:35:23.200
What we are trying to figure out, if we were to draw a graph here, is:
00:35:23.200 --> 00:35:33.500
here is x³; what is the slope at x = -2?
00:35:33.500 --> 00:35:37.800
What is the slope that goes through this one place?--that is what we are trying to approximate.
00:35:37.800 --> 00:35:42.800
What is the m of this tangent line, of that instantaneous slope there?
00:35:42.800 --> 00:35:48.100
The best that we have gotten so far, when we plugged in this fairly small value of 0.1, was 11.41.
00:35:48.100 --> 00:35:52.100
If we wanted to get more accurate values from this numerical way of figuring out
00:35:52.100 --> 00:35:58.400
what the slope starts to work towards, we can use just smaller and smaller h: 0.01, 0.000001...
00:35:58.400 --> 00:36:01.200
And we will be able to get better and better, more accurate, results.
00:36:01.200 --> 00:36:05.300
However, if we want to get it perfectly, we have to use the derivative--a problem about the derivative!--
00:36:05.300 --> 00:36:09.300
for f(x) = x³, then evaluating f'(-2).
00:36:09.300 --> 00:36:19.200
And so, once again, f'(-2) tells us the instantaneous slope of our graph at this value, -2.
00:36:19.200 --> 00:36:28.700
So, figuring out f(-2) will tell us the slope; the slope of this is going to be this value that we get from f'(-2).
00:36:28.700 --> 00:36:31.600
All right, so how do we figure out f prime?
00:36:31.600 --> 00:36:45.200
Well, remember: f'(x) is equal to the limit as h goes to 0 of f(x + h) minus f(x), all over h.
00:36:45.200 --> 00:36:58.900
In this case, our f(x) is equal to x³, so we can swap this out for the limit as h goes to 0 of f(x + h)...
00:36:58.900 --> 00:37:07.900
so f(x + h) becomes (x + h)³--we are plugging in (x + h) into how it works over here,
00:37:07.900 --> 00:37:15.600
so (x + h)³, minus...now we are just plugging in x, so simply x³, all over h.
00:37:15.600 --> 00:37:19.600
What is (x + h)³? Let's just work that out in a quick sidebar.
00:37:19.600 --> 00:37:33.400
(x + h)³: we can write that as (x + h)(x + h)²; that becomes x² + 2xh + h²...
00:37:33.400 --> 00:37:38.000
(x + h) times x² + 2xh + h²...we get x³;
00:37:38.000 --> 00:37:43.600
x times 2xh is 2x²h; h times x² is 1hx², or 1x²h.
00:37:43.600 --> 00:37:53.700
So, we get + 3x²h; then, x times h² is 1xh²; h times 2xh is 2xh²;
00:37:53.700 --> 00:37:59.800
so we get a total of 3xh², plus h³; h² times h gets us h³.
00:37:59.800 --> 00:38:04.800
Notice that we can also get that through the binomial expansion, if you remember.
00:38:04.800 --> 00:38:07.800
If you recently worked on the binomial expansion, you might recognize that.
00:38:07.800 --> 00:38:12.900
All right, so at that point, we can plug back into our limit, limit as h goes to 0...
00:38:12.900 --> 00:38:28.400
we swap out (x + h)³ for x³ + 3x²h + 3xh² + h³ - x³, all divided by h.
00:38:28.400 --> 00:38:33.200
At this point, we see that we have a positive x³ and a negative x³; they knock each other out.
00:38:33.200 --> 00:38:36.400
Now, we see that everything has an h; great.
00:38:36.400 --> 00:38:44.200
Currently, if we were to just plug in h goes to 0, we would get 0 + 0 + 0, divided by 0; 0/0...we can't do that.
00:38:44.200 --> 00:38:49.400
But we can knock out the dividing by 0, because everything up top now has a factor of h in it.
00:38:49.400 --> 00:38:54.600
So, we can knock out this h here; that knocks the h out here, turns this h² into an h¹,
00:38:54.600 --> 00:39:09.400
turns this h³ into an h²...and we now have the limit as h goes to 0 of 3x² + 3xh + h².
00:39:09.400 --> 00:39:15.500
The limit as h goes to 0...well, that is going to cause 3xh to just turn into 0.
00:39:15.500 --> 00:39:21.400
As h goes to 0, it will crush the x; as h² goes to 0, it will crush h² to 0.
00:39:21.400 --> 00:39:28.500
So, as h goes to 0, any h² will crush h² to 0; and that leaves us with just 3x²; great.
00:39:28.500 --> 00:39:34.000
So, f'(x) is equal to 3x².
00:39:34.000 --> 00:39:38.900
So now, we were asked to find what is the specific derivative at f'(-2).
00:39:38.900 --> 00:39:51.300
f'(-2) = what? Well, f'(x) = 3x², so f'(-2) will be 3(-2)².
00:39:51.300 --> 00:39:57.700
3 times -2 squared...that becomes 4, so we have 3 times 4; and we get 12.
00:39:57.700 --> 00:40:04.400
The instantaneous slope of the point, of the horizontal location -2, is a slope of 12,
00:40:04.400 --> 00:40:10.900
which, if you can remember from that last example that we were just working to... when we had .1, we managed to get to 11.41.
00:40:10.900 --> 00:40:14.600
As we were making our h smaller and smaller, we were slowly working our way
00:40:14.600 --> 00:40:21.600
to this perfectly instantaneous slope of 12, working our way to the derivative at -2; cool.
00:40:21.600 --> 00:40:25.200
The next one: Find the derivative of f(x) = 1/x.
00:40:25.200 --> 00:40:33.200
The same basic method works here: limit as h goes to 0 of...I'm sorry, f prime, the derivative...
00:40:33.200 --> 00:40:42.000
f'(x) is going to be equal to the limit as h goes to of f(x + h) - f(x), all over h.
00:40:42.000 --> 00:40:46.500
So, we start working this out: limit as h goes to 0 of f(x + h) - f(x)...
00:40:46.500 --> 00:40:53.300
Well, x + h...we will plug into our formula f(x) = 1/x, so f(thing) = 1/thing.
00:40:53.300 --> 00:41:03.200
So, if we plug in x + h for our thing, we have 1/(x + h), minus f(x) (is simply going to be 1/x), all divided by h.
00:41:03.200 --> 00:41:08.500
Oh, I forgot one important part: it equals the limit as h goes to 0 of that stuff.
00:41:08.500 --> 00:41:11.200
We have to keep up that limit, because it is very important.
00:41:11.200 --> 00:41:16.000
Limit as h goes to 0 of all of this stuff: now, from when we worked with finding limits,
00:41:16.000 --> 00:41:19.200
well, we have fractions on top of and inside of a fraction...
00:41:19.200 --> 00:41:24.300
Well, what we want is less fractions; we don't want fractions in fractions, so how can we get rid of the fractions up top?
00:41:24.300 --> 00:41:29.700
Well, notice: if we multiply by the top, we can cancel out those fractions with x times x + h.
00:41:29.700 --> 00:41:33.800
That will cancel out the right fraction and the left fraction on the top.
00:41:33.800 --> 00:41:41.200
We will be left with some stuff...but we have to multiply the top and the bottom of any fraction by the same thing; otherwise, we just have wishful thinking.
00:41:41.200 --> 00:41:50.800
x times (x + h) on the top and the bottom: we have...this is equal to the limit as h goes to 0 of 1/(x + h) times (x + h)...
00:41:50.800 --> 00:41:56.700
well, that (x + h) here will cancel out the (x + h) here, and we will be left with just x.
00:41:56.700 --> 00:42:07.800
We get x, minus quantity...here, the x cancels out the x here, and we are going to be left with x + h, minus (x + h), all over...
00:42:07.800 --> 00:42:11.200
well, we can't cancel out anything on the bottom; we just have h.
00:42:11.200 --> 00:42:15.400
And if you remember from our finding limits, it is actually going to behoove us to be able to keep the h there,
00:42:15.400 --> 00:42:19.400
because our goal, since we are going to have h go to 0...we can't divide by 0;
00:42:19.400 --> 00:42:23.300
so we need to somehow get that h at the bottom to cancel out and be canceled out.
00:42:23.300 --> 00:42:26.700
Otherwise, our limit will end up getting mixed up by that h still being there.
00:42:26.700 --> 00:42:30.300
So, h times x times x + h is on the bottom.
00:42:30.300 --> 00:42:37.000
At this point, we see that we have x - (x + h); well, put -h in here...
00:42:37.000 --> 00:42:46.900
Oh, I accidentally cut out the wrong thing there; I should not have crossed out the h; I should have crossed out the x.
00:42:46.900 --> 00:43:04.400
The -x here cancels out the positive x here; and we are left with the limit as h goes to 0 of -h, divided by h times x times (x + h); great.
00:43:04.400 --> 00:43:08.800
So, we have the h here on the bottom and the h here on the top.
00:43:08.800 --> 00:43:14.400
So, we can cancel out h here and h here; that leaves us with -1 on the top:
00:43:14.400 --> 00:43:22.600
the limit as h goes to 0 of -1, all divided by x times x + h.
00:43:22.600 --> 00:43:26.400
At this point, if h goes to 0, we won't have massive issues.
00:43:26.400 --> 00:43:39.800
Let's see one more: the limit as h goes to 0 of -1 over...we expand x and (x + h); we distribute that x over, and we get x² + xh.
00:43:39.800 --> 00:43:46.200
Now, as h goes to 0, that will cause the xh to cancel out; but it will have no effect on the x²,
00:43:46.200 --> 00:43:49.100
so things don't end up breaking, as long as x isn't equal to 0.
00:43:49.100 --> 00:43:56.600
But we are looking just in general; so -1/x² is what ends up coming out of this: -1/x².
00:43:56.600 --> 00:44:10.500
So, that means that we can write, in general, f'(x) is equal to what we ended up getting of all of this in the end, -1/x²; great.
00:44:10.500 --> 00:44:15.200
All right, finally, before we get to our final, fourth example, let's talk about the power rules.
00:44:15.200 --> 00:44:19.200
These are the sorts of rules that you will end up learning as you start working through calculus.
00:44:19.200 --> 00:44:25.400
One of the first and easiest rules to learn, that makes it so much easier to take a derivative, is the power rule.
00:44:25.400 --> 00:44:40.300
What it says is that, for any function f(x) = x^n, where n is just any constant number, the derivative of f(x) is f'(x) = n(x^n - 1).
00:44:40.300 --> 00:44:50.400
So, we take that power; we bring it down in front of it; we have x to the n, and then we bring it down in front;
00:44:50.400 --> 00:44:55.200
we have n times...and then we bring down our n to n - 1.
00:44:55.200 --> 00:44:58.900
So, you might not have noticed this yet, but this actually ended up holding true
00:44:58.900 --> 00:45:03.300
for all of the functions that we have worked through so far in this lesson, both in the early part of the lesson,
00:45:03.300 --> 00:45:06.700
where we were setting up these ideas, and in Example 2 and Example 3.
00:45:06.700 --> 00:45:11.900
When we took the derivative of f(x) = x² + 1, it came out to be simply 2x.
00:45:11.900 --> 00:45:15.800
Don't be too worried about the +1; we can think of it is being x⁰.
00:45:15.800 --> 00:45:19.300
So, when you bring down that 0 in front of it, it just cancels out and becomes nothing.
00:45:19.300 --> 00:45:21.300
So, we got 2x out of it.
00:45:21.300 --> 00:45:26.100
When we had f(x) = x³, when we took the derivative of that in Example 2,
00:45:26.100 --> 00:45:36.100
we ended up getting 3 times x²: the 3 went down in front, and we brought down 3 by 1 to 2; 3 - 1 becomes 2.
00:45:36.100 --> 00:45:46.400
And finally, Example 3, that we just worked on: f(x) = x^-1...we brought down the -1, and we got -x; and -1 - 1 becomes -2.
00:45:46.400 --> 00:45:48.800
So, we ended up seeing this inadvertently.
00:45:48.800 --> 00:45:54.800
And this ends up being true for any constant number n; it really works for anything at all.
00:45:54.800 --> 00:45:58.500
If you are curious about seeing this some more, try looking at what would happen if you tried to take the derivative
00:45:58.500 --> 00:46:05.300
through that formal h goes to 0 of x⁴, f(x⁴); try working through the formal definition of x⁴.
00:46:05.300 --> 00:46:12.900
And if you really want to try seeing how it works for any integer at all, or any positive integer, at least, try using the binomial theorem.
00:46:12.900 --> 00:46:15.800
And it even works for anything at all, as we have just seen for negative numbers.
00:46:15.800 --> 00:46:18.800
But you can end up seeing how it works for binomial theorem, as well.
00:46:18.800 --> 00:46:21.900
If you are curious about this, try checking it out; it is pretty easy to end up seeing,
00:46:21.900 --> 00:46:25.300
with the binomial theorem, that it ends up being true for any positive integer.
00:46:25.300 --> 00:46:37.000
All right, let's put this to use: using that power rule, find an equation for the tangent line to the function f(x) = x⁴ that passes through (3,81).
00:46:37.000 --> 00:46:39.900
If we are going to pass through (3,81), let's first get a sense of what is going on.
00:46:39.900 --> 00:46:42.700
Let's draw just a really quick sketch of what is going on here.
00:46:42.700 --> 00:46:49.700
x⁴ shoots up really, really quickly; we shoot up massively, very, very quickly, with x⁴.
00:46:49.700 --> 00:46:58.500
By the time we have made it to a horizontal x location of 3, we are putting out an output of 81; 3⁴ is 9 times 9, which is 81.
00:46:58.500 --> 00:47:01.900
So, we are at this very, very high point, very, very quickly.
00:47:01.900 --> 00:47:08.800
So, what we are going to want to find is: we want to find the tangent line to the function at this point, (3,81).
00:47:08.800 --> 00:47:15.600
We want to find something that ends up going like this; that is what we are looking for: what is the slope at this?
00:47:15.600 --> 00:47:21.800
Well, to find the slope at any given horizontal location, at any point, we end up taking the derivative.
00:47:21.800 --> 00:47:31.800
How can we take the derivative? If we have f(x) = x⁴, the power rule says that we can get the derivative,
00:47:31.800 --> 00:47:40.700
f'(x), by taking the exponent and bringing it down in the front; so we have 4 times x;
00:47:40.700 --> 00:47:47.000
and the exponent goes...subtract by 1; so 4 - 1 becomes 4x³.
00:47:47.000 --> 00:47:58.700
So, if we want to find out what the slope is at x = 3 (that is the horizontal location we are considering),
00:47:58.700 --> 00:48:05.600
if we want to find out what the slope is for our tangent line, the slope at x = 3 is going to be f'(3).
00:48:05.600 --> 00:48:13.000
f'(3) will be 4 times 3³; we plug it into our f prime, our derivative function.
00:48:13.000 --> 00:48:21.900
f'(3) = 4(3)³; 3 cubed is 27; 4 times 27 equals 108.
00:48:21.900 --> 00:48:26.800
So, we now know that the slope of our tangent line is 108.
00:48:26.800 --> 00:48:37.600
So, if we are going to work out the tangent line: our tangent line is going to end up using this slope, m = 108.
00:48:37.600 --> 00:48:45.500
So, any line at all can be described by slope-intercept form; it is a really good form to have memorized: y = mx + b.
00:48:45.500 --> 00:48:48.200
You always want to keep that one handy; it is always useful.
00:48:48.200 --> 00:48:57.700
y = mx + b: well, we just figured out that the slope...f it is going to be the tangent, it must have a slope of 108, because f prime,
00:48:57.700 --> 00:49:02.600
the instantaneous slope at 3, f'(3), comes out to be 108.
00:49:02.600 --> 00:49:08.300
So, we know that the instantaneous slope at the point we are interested in, (3,81), is 108.
00:49:08.300 --> 00:49:17.000
So, that means that we have y = 108x + some b that we haven't figured out yet.
00:49:17.000 --> 00:49:22.300
How can we figure out what b is to finish creating our equation for the tangent line?
00:49:22.300 --> 00:49:26.500
If you are going to figure out what any line is, you need to know
00:49:26.500 --> 00:49:32.500
what the slope is and what the y-intercept is--what m is and what b is, at least if you are using slope-intercept form.
00:49:32.500 --> 00:49:37.200
y = 108x + b; well, do we know any points on that line?
00:49:37.200 --> 00:49:43.700
Yes, we are looking at the tangent line; and we were told that the tangent line passes through the point (3,81).
00:49:43.700 --> 00:49:48.700
So, we can plug in the point (3,81), because we know that our tangent line
00:49:48.700 --> 00:49:53.600
has to pass through the one point that it barely, barely just touches on that curve.
00:49:53.600 --> 00:50:02.700
So, we plug in (3,81); that is 81 for our y-value; that equals 108 times 3 (for our horizontal value x) plus b.
00:50:02.700 --> 00:50:14.500
81 = 108 times 3 (that comes out to be 324), plus b; 81 - 324 is -243 for our b.
00:50:14.500 --> 00:50:27.400
So now, we know that -243 = b; we know what our slope is; so that means that we can describe the tangent line in general as y = 108x - 243.
00:50:27.400 --> 00:50:30.900
That is y = mx + b with our m and b filled in.
00:50:30.900 --> 00:50:38.700
And now, we have the tangent line that passes through the point (3,81) and is tangent to the curve created by x⁴--pretty cool.
00:50:38.700 --> 00:50:42.000
Calculus gives us a whole bunch of stuff that we can end up doing with the derivative.
00:50:42.000 --> 00:50:45.300
The derivative is this massively, massively useful, important thing.
00:50:45.300 --> 00:50:50.100
And we are just barely touching the surface of how incredibly useful and cool this thing is.
00:50:50.100 --> 00:50:54.600
As you work through calculus, you will end up learning a whole bunch of things that the derivative lets us do.
00:50:54.600 --> 00:51:00.200
It lets us learn about a function; it is really, really amazing how much information it gives us; calculus is really, really cool.
00:51:00.200 --> 00:51:03.800
I hope that, at some point, you get the chance to take calculus and get to see how many cool things there are.
00:51:03.800 --> 00:51:08.500
And remember: when you work with the derivative, what you are looking at is what the slope is of that location,
00:51:08.500 --> 00:51:11.600
of that point, of that horizontal location on your graph.
00:51:11.600 --> 00:51:13.000
All right, we will see you at Educator.com later--goodbye!