WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to start talking about the derivative formally.
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Let us jump right on in.
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We finally arrived at the how.
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When we started this course, we talked about this thing called the derivative.
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We talked about the what it is and we said that the derivative is a slope of the curve, is a rate of change of the curve.
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We also said that the how, we gave you an expression for the how.
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It involves the limit, we have gone through this process of talking in a detailed way about the limit and
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now we are going to come down to actually finding derivatives analytically.
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At least, the first step of finding derivatives analytically.
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Let us go ahead and work in blue here.
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We have arrived at the how.
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We have arrived at the how for derivatives.
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Given f(x), given some function f(x), the derivative which we symbolize with the prime symbol f’(x) is =
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the limit as h approaches 0 of f(x) + h - f(x)/ h.
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Basically what this says at is, what this says is, when are given some function f(x), you are going to form this quotient.
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You are going to take x + h, you are going to form that, whatever that is.
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You are going to subtract from it f(x), you are going to divide by h.
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You are going to simplify this expression and you are going to take the limit as h approaches 0.
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What you are going to get is a function, that function is your derivative.
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We symbolize it with f’.
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It is derived from the original function.
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Let us do an example.
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Example 1, find the derivative of f(x) = x³.
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f’(x) = the limit as h approaches 0 of f(x) + h – f(x).
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I think it is a good idea when you are doing these problems to actually start each problem by writing down the definition.
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That way it will stick in your mind, /h.
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That = the limit as h approaches 0 of, f(x) + h, f is x³.
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This is going to be x + h³, that is the first one, - f(x) which is x ^ / h.
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Clearly, we cannot just plug in h as 0 because it would give us in the denominator.
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We have to simplify this expression
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In this case, simplification means just expanding, adding, subtracting, multiplying, dividing.
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Doing whatever you have to do until we find a simplified expression, that we cannot do anything else to.
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And then, we will take the limit again and see what happens.
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This is equal to the limit as h approaches 0 of x³ + 3x² h + 3x h² + h³ - x³/ h which = the limit as h approaches 0.
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x³ go away, you are left with 3x².
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3x² h + 3x h² + h³/ h.
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We factor an h from the top, it equal the limit as h approaches 0 of h × 3x² + 3x h + h²/ h.
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H goes away, you are left with the limit as h approaches 0 of the function 3x² + 3x h + h².
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Now we plug in h, h was 0, you plug in 0.
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That goes to 0, that goes to 0, you are left with 3x².
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There you go.
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f’(x), your derivative of x³ is equal to 3x².
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Now recall, we have a derivative.
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A derivative is two things, a derivative is many things actually.
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But it is definitely these two things.
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A derivative is the slope of the tangent line to the curve, at a given x.
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Notice this is a function, because when you are following a curve, the tangent line, the slope of the line actually changes.
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The derivative is also an instantaneous rate of change.
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Remember, when we talked about rates of change,
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we said that there was an average rate of change that is the secant line between two points.
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We have some function.
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If I take two points and draw a line, we call that the secant line.
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The average rate of change, that is the slope of that line.
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But if I have a tangent line, from that point, the slope of that is the instantaneous rate of change.
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When I'm at that point, if I move a little bit to the left, to the right, how much is y going to change at that instant?
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It is an instantaneous rate of change.
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What we are hoping will become a conditioned response when you see derivative,
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is that you are thinking two things, depending on what the problem is.
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But you will think that this is the slope of the tangent line, to the curve at a given point.
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And it is the instantaneous rate of change of the function at that point.
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We just want it to be a conditioned response.
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Derivative, instantaneous rate of change, and slope of tangent line.
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Instantaneous rate of change of the function, at a given x.
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Let us talk about some notations for the derivative.
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We have seen f’(x), a lot of times we will leave off the x and we will just write f’.
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You will see y’(x), if you express it as y = x³.
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Then, the derivative is going to be y’ = 3x².
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We leave the x off, we do it as y’.
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Then, there is this one, dy dx.
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This notation, this one is derived from Δ y/ Δ x.
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Δ y/ Δ x is a slope.
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Dy dx is the slope of the tangent of line.
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It is the instantaneous slope.
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That is a symbol that we use to describe the derivative, as opposed to the average.
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Average, instantaneous.
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Secant line, slope of the tangent line.
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I really have to slow myself down.
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Whenever, we form the quotient of any change in y/ any change in x,
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no matter what those variables are, they could be t, s, q, p, whatever.
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Whenever we form the quotient Δ y/ Δ x and it pass to the limit,
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which is what we did right, it is the limit of this f(x) + h - f(x)/ h.
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This is a change in y/ a change in x.
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When we pass the limit, that just means when we take the limit of this expression as h goes to 0.
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When we pass the limit, the Δ's turn into d.
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This notation reminds us that we are talking about a slope and instantaneous rate of change.
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If x changes by a really tiny amount, y changes by that tiny amount, that is what that means.
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If you were to see dy/ dx = 3.
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It is the same as dy dx = 3/1.
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If I change x by 1, I change y by 3.
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The rate of change is the slope.
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Recall that the slope and rate of change are synonymous.
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You are going to hammer that point a lot, my apologies.
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Recall that a slope and a rate of change are synonymous.
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How do you spell synonymous, all of a sudden I forgot, are the same.
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Recall that a rate of change is, recall what a rate of change is.
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When we have a Δ y/ a Δ x, this is equivalent to saying,
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when we change the x value by a unit amount, how much does y change?
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When I change x by a unit amount, this number up on top gives me the amount by which y changes.
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When we say that the function is x³ and when we differentiate it to get 3x², we have f’(x) = 3x².
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We can write it as y’(x) = 3x².
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We can write it as dy dx = 3x².
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Let us choose a specific value.
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Let us choose a specific value for x.
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Let us say at x = 2.
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Therefore, dy dx at x = 2, this is the symbolism that we use.
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Now we can write it as f’ at 2, y’ at 2.
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Dy/ dx, we put a line there and 2, like that, if you want.
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I suppose it is not going to be the end of the world, if you did something like that,
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whatever notation make sense to you.
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As long as you understand it and the people that you are writing it for understand it.
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dy dx at x = 2, dy dx is 3x².
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It is going to be 3 × 2², it is going to be 12.
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This is the same as 12/1.
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This means, when we are at the point 2, we said that x = 2.
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f(2) is equal to 8 because f is equal to x³.
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When we are at the point 2,8, from that point, if we move one unit to the left or to the right,
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then f changes by 12 units, up or down.
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In this case, it is going to be down or up.
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That is what this means, dy dx = 12, means dy dx is 12/1.
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If I change x by 1 unit, I change y by 12 units.
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Here is what it looks like.
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Here I have my function y = x³, that is my red line.
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We said that dy dx which is the derivative is equal to 3x², that is a function dy dx.
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The reason it is a function is because the slope of the line, as you see, the slope along the curve changes.
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Therefore, depends on what x is.
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dy dx at x = 2, we said it is equal to 12, which is the same as 12/1.
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This is the slope of the tangent line.
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A tangent line is the one that is in the broken up black.
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Here is our 2, here is our point 2,8.
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At that point, the tangent in line is that line right there.
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It has a slope of 12, that is what the derivative means.
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The slope of the tangent line is 12 at x = 2.
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The instantaneous rate of change of f at x = 2 is 12.
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From this point, if I move that way or this way by one unit, my function is going to change by 12 units.
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That is what that means.
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The derivative gives me a function.
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When I put a specific x value in, it gives me the slope of the tangent line at that point.
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It also gives me the rate at which the function changes from that point.
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Notice that the derivative is a function of x itself.
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f(x), you take the derivative, you get another function, which we symbolize f’(x).
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Our example x³, we took the derivative and we got 3x².
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We often will graph both together.
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Both the function and the derivative.
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In fact, in your problems, sometimes given the graph of f(x), you are going to be asked to graph f’(x).
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Also sometimes given f’(x), sometimes you will be given the graph of the derivative of the function.
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You have to recover, sometimes given f’(x), given the graph of f’(x) not just f’(x).
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Given the graph of f’(x), you must recover the graph of the original f(x).
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Let us look at x³ and x².
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I’m sorry, let us look at x³ and its derivative 3x², together on the same graph.
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The red is the x³ and the broken up black is the 3x².
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Here we have f(x) is equal to x³.
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We have f’(x) which is equal to 3x².
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This is the slope of f(x) at any, I should say, at the various values of x.
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This is going to be very important.
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You want to take your time when dealing with these graphs,
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so it can take you a little bit just to sort of wrap your mind around the fact
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that you are talking about two things that are connected, but are somehow separate.
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Now f itself is a function of x.
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f’(x) is also function of x.
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Each one of those has a graph.
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This is x³, this is 3x².
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But the 3x² is the derivative of the x³.
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The derivative tells me what the slope of the graph is, at that point.
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As I move x negative to positive, again, we are always moving from negative values, working our way that way.
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Left to right, negative to positive.
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Notice the slope of the curve here, the slope is positive which is y, and 3x²,
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the derivative is the slope of the curve which is why it starts above the x axis.
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But notice as x moves that way, the slope of the function is decreasing towards 0.
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That is what this demonstrates.
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This shows that it starts up here, it is declining towards 0.
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Here the slope of the function itself is 0, which is why the derivative function is 0.
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Now the slope starts to become positive again.
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Positive, positive, positive, positive, positive.
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But it is not just becoming positive, it is becoming more and more positive.
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Therefore, the black line, the derivative is telling me that.
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It is telling me that it is positive, the line itself, the graph of 3x², the derivative is above the x axis and it is increasing.
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The function is the red, the derivative is that.
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The derivative is the slope.
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The slope is positive but it decreases towards 0.
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It starts to increase again, that is what this line is telling me.
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Let us do another example.
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Let us go ahead and work in blue.
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Let us do example number 2.
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Find the derivative of x⁴ - x², I wish I had not chosen such a complicated function.
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Find the derivative of x⁴ - x².
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We know that f’(x) is equal to,
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I always do that, after 35 years, I still forget to write down my limit.
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It equal the limit as h approaches 0 of f(x) + h - f(x)/ h.
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I’m not going to keep writing it over and over again.
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I’m just going to work with the function.
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I will just simplify it and then we will take the limit at the end.
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f(x) + h, this is going to be x + h⁴ – x + h².
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That is the f(x) + h – f(x).
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f(x) is x⁴ - x²/ h.
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Again, you plug in 0, you are going to have 0 in the denominator.
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You have to simplify it out.
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Let us multiply all of this out.
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We are going to expand that.
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This is going to equal x⁴ + 4x³ h + 6x² h²
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+ 4x h³ + h⁴ - x² + 2x h + h².
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And then, we are going to do this - this – that.
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Remember we have to distribute.
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It is going to be –x⁴ + x²/ h.
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We are going to get, x⁴ cancels x⁴.
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We are going to get 4x³ h + 6x² h² + 4x h³ + h⁴ - x² - 2x h - h² + x²/ h.
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-x² and +x² cancel, I’m going to factor out an h.
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It is going to be h × 4x³ + 6x² h + 4x h² + h³ - 2x - h/ h.
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The h is canceled, I'm left with my final 4x³ + 6x² h + 4x h² + h³ - 2x – h.
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Now I take the limit.
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Now we take the limit as h approaches 0 of this thing.
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It is going to be 4x³ + 6x h + 4x h² + h³ - 2x – h.
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h goes to 0, that goes to 0, that goes to 0, that goes to 0.
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You are left with that.
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f’(x)is equal to 4x³ - 2x.
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We have y’(x) = 4x³ - 2x.
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You will see dy dx = 4x³ - 2x.
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Let us take a look at f(x) and f’(x) in the same graph.
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The red is your f(x), this was your x⁴ – that.
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That is your derivative which was 4x³ - 2x.
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The derivative graph tells you what the slope is of the graph.
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The slope is negative but it is increasing.
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Notice here the slope is 0.
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Here the slope is 0, here the slope of the function is 0.
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Therefore, the derivative, this derivative graph, the black, 0, 0, 0.
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Here it is negative, so it is below the axis.
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Here the slope of the graph is positive.
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It is above the x axis.
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Here the slope of the graph is negative.
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This is negative, it is below the x axis, it hits 0.
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After this point, the slope starts to become positive again.
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It arises and it is above the x axis.
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The derivative is the description of how the slope is changing, as you move along the graph.
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This is a description of how the slope of the original f(x) is changing.
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It is a function of x in its own right.
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That is all it is saying.
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For example, at x = that value, the slope is this.
00:30:39.800 --> 00:30:45.500
What is the slope, the slope is that number.
00:30:45.500 --> 00:30:50.900
That is what is happening here.
00:30:50.900 --> 00:31:11.100
Let us do example 3, find the derivative of f(x) = √x.
00:31:11.100 --> 00:31:16.800
We actually did this in one of the example problems from a previous lesson, but let us do it again formally.
00:31:16.800 --> 00:31:25.600
We have f’(x) = f(x) + h – f(x).
00:31:25.600 --> 00:31:28.000
Again, I forgot my limit.
00:31:28.000 --> 00:31:32.800
I think it is unbelievable, I never remember to write down my limit.
00:31:32.800 --> 00:31:36.400
I remember to take the limit, but I never remember the write it down.
00:31:36.400 --> 00:31:47.900
The limit as h goes to 0 of f(x) + h - f(x)/ h.
00:31:47.900 --> 00:31:50.300
Again, I’m not going to write the limit over and over again.
00:31:50.300 --> 00:31:56.600
I’m just going to go ahead and work with the function.
00:31:56.600 --> 00:32:07.400
√x, f(x) + h is √x + h – √x, that is the f(x) divided by h.
00:32:07.400 --> 00:32:08.600
I have to manipulate it.
00:32:08.600 --> 00:32:12.800
I’m going to go ahead and multiply by the conjugate of the numerator.
00:32:12.800 --> 00:32:24.200
It is going to be √x + h + √x/ √x + h + √x.
00:32:24.200 --> 00:32:35.900
I end up with x + h - x/ h × √x + h + √x.
00:32:35.900 --> 00:32:38.600
That and that go away, leaving me just h on top.
00:32:38.600 --> 00:32:48.800
h and h cancel, I’m left with 1/ √x + h + √x.
00:32:48.800 --> 00:32:51.200
Now I take the limit.
00:32:51.200 --> 00:32:59.700
The limit as h goes to 0 of 1/ √x + h + √x.
00:32:59.700 --> 00:33:01.800
h goes to 0, this turns to 0.
00:33:01.800 --> 00:33:04.900
I'm left with 1/ √x + √x.
00:33:04.900 --> 00:33:11.500
I get 1/ √x, that is my f’.
00:33:11.500 --> 00:33:29.800
f’(x) = 1/ 2√x.
00:33:29.800 --> 00:33:37.600
I have got f’(x) = 1/ 2√x.
00:33:37.600 --> 00:33:42.700
y’(x), same thing, different notation, 1/ 2√x.
00:33:42.700 --> 00:33:50.200
And the last notation, dy/ dx = 1/ 2√x.
00:33:50.200 --> 00:34:01.600
Now I’m going to ask the question, what is dy dx at x = 3?
00:34:01.600 --> 00:34:20.000
Very simple, dy dx at x = 3 is equal to, just plug in 3, 1/ 2√3.
00:34:20.000 --> 00:34:29.300
What is the f(3)?
00:34:29.300 --> 00:34:36.500
f, we said that f(x), the original function is just √x.
00:34:36.500 --> 00:34:45.500
f(3) is equal to √3.
00:34:45.500 --> 00:35:26.900
The tangent line to the graph for x = 3, touches the graph at the point x f(x), which is equal to 3, that is our x value.
00:35:26.900 --> 00:35:38.000
f of that which is √3.
00:35:38.000 --> 00:35:42.900
It has a slope of 1/ 2√3.
00:35:42.900 --> 00:35:46.200
Notice how I did that.
00:35:46.200 --> 00:35:50.400
The curve itself, there is an x value.
00:35:50.400 --> 00:35:53.400
That x value is going to give me y value.
00:35:53.400 --> 00:35:57.400
The point is going to be your xy or your x f(x).
00:35:57.400 --> 00:36:01.600
The derivative at that point is the slope of the line.
00:36:01.600 --> 00:36:04.600
Now you have a slope and you have a point that it passes through.
00:36:04.600 --> 00:36:09.700
You can find the equation of that line.
00:36:09.700 --> 00:36:24.100
The equation of the tangent line is, we know that it is y - y1 = m × x - x1.
00:36:24.100 --> 00:36:43.600
Here it is going to be y - y1 is √3 = the slope which is 1/ 2√3 × x – 3.
00:36:43.600 --> 00:36:46.600
There we go.
00:36:46.600 --> 00:36:55.000
The red is the graph, the black is the tangent line to the graph at x = 3.
00:36:55.000 --> 00:36:59.300
At the value x = 3, I go up here.
00:36:59.300 --> 00:37:08.900
This point right here, this is my 3√3.
00:37:08.900 --> 00:37:15.500
This is y = √x.
00:37:15.500 --> 00:37:20.300
This line, the equation of this line, we just found it.
00:37:20.300 --> 00:37:30.000
This is y - √3 = 1/ 2√3 × x – 3.
00:37:30.000 --> 00:37:35.400
y - y1 = m × x – x1.
00:37:35.400 --> 00:37:36.600
This point is easy to find.
00:37:36.600 --> 00:37:38.400
Just plug in the x value and you get the y value.
00:37:38.400 --> 00:37:41.100
That is going to be the 3 and √3.
00:37:41.100 --> 00:37:47.500
The slope, that is from the derivative.
00:37:47.500 --> 00:38:03.100
f(x) = √x, f’(x) 1/ 2√x, it is that simple.
00:38:03.100 --> 00:38:07.600
I hope that made sense.
00:38:07.600 --> 00:38:09.900
Let me write out everything here.
00:38:09.900 --> 00:38:23.100
f(3) = 3, the tangent line touches 3√3.
00:38:23.100 --> 00:38:30.900
Now f’ at 3 = 1/ 2√3.
00:38:30.900 --> 00:38:42.900
The slope of this tangent line is 1/ 2√3.
00:38:42.900 --> 00:38:51.900
Let us go ahead and take a look at the graph and its derivative as functions of x.
00:38:51.900 --> 00:38:57.000
Here is the function y = √x, this is the derivative.
00:38:57.000 --> 00:39:03.300
It is y prime, it is equal to 1/ 2√x.
00:39:03.300 --> 00:39:13.500
The graph, the original graph, the derivative, the black, describes how the slope changes as x gets bigger and bigger.
00:39:13.500 --> 00:39:19.400
Notice the slope is almost straight up infinite.
00:39:19.400 --> 00:39:21.800
It is positive, positive, positive, positive, positive.
00:39:21.800 --> 00:39:34.600
Always a positive slope along the curve but the positive slope is decreasing.
00:39:34.600 --> 00:39:39.100
The tangent line is actually dropping down getting close to a slope of 0.
00:39:39.100 --> 00:39:40.900
That is what this be describes.
00:39:40.900 --> 00:39:47.400
High, positive, positive, positive, positive, but the slope is getting closer to 0.
00:39:47.400 --> 00:39:55.800
If I want to know what the slope of the tangent line at that point is, it is going to be that number right there.
00:39:55.800 --> 00:39:58.800
That is what that tells me.
00:39:58.800 --> 00:40:01.200
Thank you so much for joining us here at www.educator.com.
00:40:01.200 --> 00:40:02.000
We will see you next time, bye.