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 continue our discussion of the derivative.
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Let us jump right on in.
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Let us start with an example, example 1.
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I guess we will stick with black today.
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It is not a problem, we will see how that goes.
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Find the derivative of 2 + x/ 3 – x.
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We know how to do this, nice straightforward.
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f’(x) is equal to the limit as h approaches 0 of f(x) + h - f(x)/ h.
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Again, we are going to work with just a function.
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f(x) + h, it is going to be 2 + x + h/ 3 - x + h - 2 + x/ 3 – x.
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That is that one and that one, /h.
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It is like a little bit of a simplification to do.
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This is just 2 + x + h/ 3 - x – h - 2 + x/ 3 - x/ h.
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We are going to find a common denominator on top.
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This is going to be 3 - x × 2 + x + h -, this × that,
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-2 + x × 3 - x - h/ 3 - x - h/ 3 - x/ h.
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I’m going to multiply it out.
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I end up with something that looks like this, I hope.
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+ 3x + 3h - 2x - x² - xh – 6 - 2x - 2h + 3x - x² - xh/ x.
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I'm sorry, this is 3 - x - h × 3 – x, and all of that /h.
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We get 6 - 6 because the negative distributes.
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We get + 3x - the 3x, we get -2x.
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A - and - 2x - x², - and - x² – xh.
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What we are left with is just the 3h and the 2h.
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3h - -2h, that is going to give us, it is going to equal 5h/ 3 - x - h × 3 - x/ h
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which is going to equal 5h/ h × 3 - x - h × 3 – x.
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The h cancel, you are just left with our function 5/ this thing which, now we are going to take the limit of that.
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We get the limit as h goes to 0 of 5/ 3 - x - h × 3 – x.
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h goes to 0 and we are left with 5/ 3 - x².
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That is that, our f’(x) is equal to 5/3 – (x)².
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Let us take a look at the graph and its derivative.
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Again, the graph is in red, the derivative is in black.
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As x increases, notice as x increases from left to right,
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notice the slope of f, as it increases, the slope of the graph increases.
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It is always positive which is why the black is positive above the x axis, but it is actually increasing which is why the black goes up.
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3, of course.
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What was our original f(x)?
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Our original f(x) was 2 + x/ 3 – x, it is not defined at 3, there is an asymptote.
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Our f’ was 5/ 3 - x², also not defined.
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Here pass 3, the slope is always positive but notice now the slope is actually,
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from your perspective, the slope is changing, it is decreasing.
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Always going to stays positive but it is decreasing, getting close to 0.
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That is why the derivative graph looks the way that it does.
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The black graph.
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As x increases from left to right, the slope of f, let me go back to black.
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The slope of f is positive and increases at 3.
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Neither f nor f’ exist, we said it is not differentiable there.
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It is not defined there and it is not differentiable there.
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The function is not defined, the derivative is not defined.
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It is not differentiable.
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We say f is not differentiable, it means something is going on there.
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It is not smooth there.
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It is not differentiable at x = 3.
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Past 3, the slope is always positive above the x axis, that totally decreases.
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Let us talk about differentiability.
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We know the definition of the derivative.
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If the limit that we take, when we take the derivative, the limit as h approaches 0 f(x) + h - f(x)/ h.
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If the limit exists, the function is differentiable, in other words, it has a derivative.
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If the limit does not exist, because we have seen situations where the limit does not exist,
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we say function is not differentiable.
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It does not have a derivative there at that point.
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That is all that is going on.
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We say that a function is differentiable at, a, if f’ at a exists.
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In other words, f’ at a specific point is equal to limit as h approaches 0 of f of the point.
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Now it is not just x + h, it is the actual point + h - the actual point/ h.
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If f’ of a exists, in other words, if this limit actually exist as a finite number.
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If this limit exists as a finite number.
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Again, notice that we actually put a specific value in for this normal f(x) + h - f(x)/ h.
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If we speak about differentially at certain point, we put the value in.
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Actually, we did not have to do that.
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The fact of the matter is we can just stick with the normal f(x) + h - f(x)/ h.
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See if we actually get a derivative, some function of x, some f’(x).
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And then, take the a and put it into the f’(x) function and see if it actually works out.
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In the previous problem, we had f was 2 + x/ 3 – x.
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And then, when we took the derivative of that, we ended up with f’ being equal to 5/ x - 3².
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The question was, is it differentiable at 3?
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It did not matter, you can go ahead and just use the normal general expression f(x) + h - f(x)/ h, take the limit.
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When we took the limit, we ended up with an actual function.
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Now we can put a in.
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If you put 3 in here, you notice that 3 is not going to work, it is not defined.
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Therefore, it is not differentiable.
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It is up to you, you can put the a in, form the expression and take the limit.
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Or you can just use the general expression as a function of x.
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Find the limit if it exists, and then plug the a in and see if the derivative is actually defined, either one is fine.
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I hope that made sense.
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Let me just write this all out.
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Notice we actually put the specific a value into the limit expression.
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Or we can work more generally and say f(x) is differentiable if f’(x) exists.
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That is it, the same thing, f’(x) = the limit as h approaches 0 of f(x) + h - f(x)/ h.
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If you take the limit and you get a function.
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It is going to exist for some values of a function, for certain values.
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It is not going to exist so it is not going to be differentiable at those particular points,
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where the derivative f’(x) is not defined or if there are some other problem there.
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Recall, for a limit to exist both left hand and right hand limits have to be the same.
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Right hand limits must equal each other.
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Graphically it means this.
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That is our a, we have a secant line.
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This point is a + h.
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The limit as h goes to 0 of f(a) + h – f(a)/ h.
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That is the slope of that line.
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As I get closer and closer, I’m going to get, this is the secant line.
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I’m going to have another secant line, another secant line.
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At some point, it is going to get so close, I’m going to end up with a tangent line.
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The same thing from this side, if I have a secant line and if I get closer and closer,
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I’m going to have a bunch of other secant lines that pass through that point.
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You see the secant line slope is approaching a certain value.
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In the other end, the secant line slope is approaching a certain value.
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If the two lines match the left hand and the right hand, that is what it means for the slope to exist.
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The left hand slope and the right hand slope have to equal.
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The left hand limit and the right hand limit have to equal.
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If they equal each other, we end up with the derivative.
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Now if I plug in a value of a and calculate it, the two numbers have to equal each other.
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It has to exist first of all, but they have to equal each other.
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If I do it with general function, find the function and that function is going to be undefined somewhere.
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I can do it either way.
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I can either use the expression with a specific value of a or I can use just the general expression with x and put a in afterward.
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But this is what it means for a limit to exist.
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The slopes have to match from left and right.
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Geometrically, this means, as you pass to the limit as h approaches 0,
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the secant line slopes from the left and the right, they approach the same numerical value.
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In other words, the secant lines become the same tangent line.
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Same numerical value means the same tangent line.
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Geometrically, this means, the graph is smooth.
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It does not have any kinks in it.
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Geometrically, this idea of differentiability means the graph is smooth.
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Smooth means no sudden jerks, no sudden changes of direction, or going off to infinity.
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When dealing with graphs, a differential function, any point that all of the sudden has a sudden of change of direction,
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the function is not differentiable at that point.
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Any point x where the graph goes off to infinity, the graph is not differentiable.
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Differentiable means smooth, nice transitions.
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It goes off to infinity.
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You have something like this.
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Notice, all, everything, there is no sudden change of direction.
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If you have something like this, it is not differentiable there and it is not differentiable there.
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The slope all of a sudden goes from here to here.
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The slope was from here to here.
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No cusp, any cusp on the graph means not differentiable.
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If you have a situation where it goes to infinity, it is not differentiable at that x value.
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Example 2, this one is a bit involved, but that is not a problem.
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It is a really great example.
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Example number 2, show analytically that f(x), the function f(x) = the absolute value of x, is not differentiable at x = 0.
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We know what this graph looks like.
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Here is our graph, it is just going to be that thing right there.
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It is perfectly differentiable here, perfectly smooth and perfectly smooth.
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Notice here it has a sudden change of direction.
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The slope and the slope are not equal.
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Geometrically, because we can see that it is not differentiable.
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We want to show analytically that it is not the case.
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We have to show that the left hand limit and right hand limit, even though they exist, they are not equal to each other.
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Remember, in order for there to be differentiability, the limit has to exist.
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For the limit to exist, the left hand and the right hand limits have to be equal to each other.
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I hope that make sense.
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We must show that f’ at 0 does not exist, that is non differentiability.
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In other words, we have to show that it is infinite or the left hand limit does not equal the right hand limit.
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We have our f(x), f(x) = the absolute value of x.
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0 is the dividing line, they also told us that we are concerned with 0.
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We must check x greater than 0 and x less than 0, and compare the limits.
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In other words, we are going to do the limit as x approaches 0 from below.
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We are going to take the limit as x approaches 0 from above.
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We are to see what those two limits are.
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If the limits are the same, it is differentiable.
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If they are not the same, it is not differentiable.
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This is how we do it analytically.
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We have to show the limit exists, we have to show that limits equal each other.
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For x greater than 0, we have f’(x) is equal to the limit as h approaches 0 from above of f(x) + h - f(x)/ h =
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the limit as h approaches 0 from above of the absolute value of x + h - the absolute value of x/ h.
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The absolute value of x + h, when x is greater than 0, recall the definition of absolute value.
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Let me do it another way, the absolute value of anything under the absolute value sign is that thing,
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when x is greater than 0, when a is greater than 0.
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Or it is –a, when a is less than 0.
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Since x is greater than 0, this just becomes the limit as h approaches 0.
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The absolute value signs go away.
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It is just x + h - x/ h.
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x cancel and you are left with the limit as h approaches 0 from above of 1 which = 1.
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I hope that makes sense.
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For x less than 0, f’(x) is equal to the limit as h approaches 0 from below of f(x) + h - f(x)/ h =
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the limit as h approaches 0 from below.
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This is x is less than 0.
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This thing is actually going to be –x + h - -x.
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Because the absolute value of x, when x is less than 0 is –x, that minus stays.
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I hope that makes sense.
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/h, = the limit as h approaches 0 from below - x - h + x/ h.
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= the limit as h approaches 0 from below -1.
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The left hand limit, when we approach 0 is -1.
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The limit exist, it is -1.
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The right hand limit as x approaches 0 from above was +1.
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The left hand limit exists, the right hand limit exists,
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but the limits are not equal to each other which means that it is not differentiable.
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-1 does not equal 1.
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f(x) is not differentiable but it is differentiable everywhere else.
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It is differentiable from negative infinity all the way to 0, union 0 all the way to positive infinity.
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It is differentiable everywhere else, because everywhere else is defined.
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As long as x is less than 0, the slope is always going to be -1.
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As long as x is greater than 0, the slope is always going to be +1.
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0, it cannot decide which slope to take, -1 or +1.
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Because it cannot decide, because they are not equal, it is not differentiable.
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It is not smooth, there is a sudden change of direction.
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It is not differentiable, that point causes problems.
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That is all that is going on here.
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Let us go ahead and draw these out.
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We know what f looks like, this is f(x) = the absolute value of x.
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It is not differentiable here, it is not smooth there.
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The graph of the derivative f’(x).
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When it is bigger than 0, we said the slope is 1.
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When it is less than 0, the slope is -1.
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Notice it is not defined here.
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The f’(x) is not defined here.
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This is a discontinuity, it is not differentiable, it is not smooth.
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The function is continuous but the function is not differentiable.
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It is okay, you can have a function that is continuous.
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Notice change direction, I do not have to lift my pencil off.
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This is not the f graph, this is the f’ graph.
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Be very careful, this is going to be probably the single biggest problem for a couple of weeks
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until you just learn to separate the fact that you can have a function and
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you can have the derivative of the function also be a function.
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You are going to be graphing both, keeping the graphs separate.
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Now in calculus, my best advice is slowdown, be very careful.
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There is going to be a lot of stuff going on in the page.
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We will have symbolism.
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This symbolism is going to be very subtle.
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The difference between f and f’.
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If your eye tends to move quickly, it is going to miss it.
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The function itself, the original function is continuous.
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I do not have to lift my pencil up.
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The graph of the function shows that it is not differentiable.
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It is not smooth there.
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It does not go like this.
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Let us formalize that notion.
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Let me draw the graph again.
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This us our coordinate axis, this is our f(x) = the absolute value of x.
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What is f(0), f(0) is just 0.
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The absolute value of 0 is 0.
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What is the limit as x approaches 0 of the absolute value of x?
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It is also 0.
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Let me break this one up actually.
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The limit is x approaches 0, we have to do the left and right hand limits.
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The limit as x approaches 0 from above of the absolute value of x = the limit as x approaches 0 from above of x.
00:31:50.600 --> 00:31:56.000
Because when x is positive, x approaches 0 from the right, all positive numbers.
00:31:56.000 --> 00:32:01.100
The absolute value of x is positive = 0.
00:32:01.100 --> 00:32:11.600
The limit as x approaches 0 from below of absolute value of x = the limit as x approaches 0 of –x.
00:32:11.600 --> 00:32:16.700
When we are dealing with negative numbers, it is approaching 0 there.
00:32:16.700 --> 00:32:18.500
The absolute values of x is –x.
00:32:18.500 --> 00:32:22.800
Here I still get 0.
00:32:22.800 --> 00:32:30.000
All three are equal, the left hand limit, the right hand limit, and the value of the function.
00:32:30.000 --> 00:32:32.100
Therefore, the function is continuous.
00:32:32.100 --> 00:32:39.900
What I just demonstrated is the analytical way of showing that the absolute value function is continuous.
00:32:39.900 --> 00:32:45.900
All three are equal.
00:32:45.900 --> 00:32:55.200
f(x), the absolute value function is continuous at x = 0.
00:32:55.200 --> 00:33:04.800
Notice what I have done, what I have done here, the limit of the function as x approaches 0, that is not what I just did.
00:33:04.800 --> 00:33:20.700
What I just did is I found the limit as h approaches 0 of f(x) + h - f(x)/ h.
00:33:20.700 --> 00:33:23.400
I took the derivative.
00:33:23.400 --> 00:33:27.700
The derivative means finding the limit of this quotient.
00:33:27.700 --> 00:33:34.900
Here, as h approaches 0, here I’m taking the limit as x approaches 0 of the actual function itself.
00:33:34.900 --> 00:33:39.400
When I’m taking the limit of the function itself, I’m checking for continuity.
00:33:39.400 --> 00:33:44.600
When I’m finding the limit of this thing called the Newton quotient, I'm finding the derivative.
00:33:44.600 --> 00:33:49.400
We see that analytically, we have demonstrated that the function is continuous.
00:33:49.400 --> 00:34:02.300
And previously, we demonstrated analytically but is not differentiable.
00:34:02.300 --> 00:34:16.100
You can have a function, let me say that again.
00:34:16.100 --> 00:34:20.900
I think it is very important to say that again.
00:34:20.900 --> 00:34:29.600
The limits that we just evaluated, those were the limits as x approaches 0 of f(x).
00:34:29.600 --> 00:34:36.000
The limits that we evaluated prior to that to show differentiability or non differentiability,
00:34:36.000 --> 00:34:45.900
those were the limit as h approaches 0 of f(x) + h - f(x)/ h.
00:34:45.900 --> 00:34:48.900
This concerns continuity.
00:34:48.900 --> 00:34:51.000
This concerns differentiability.
00:34:51.000 --> 00:34:54.300
Do not mistake the two.
00:34:54.300 --> 00:35:01.200
I hope the last two things did not confuse you.
00:35:01.200 --> 00:35:37.200
You can have a function be continuous at a point a, but not differentiable at that point at a, as we just saw.
00:35:37.200 --> 00:36:04.200
However, if f(x) is differentiable at point a, then it is continuous at a.
00:36:04.200 --> 00:36:07.800
Differentiability implies continuity.
00:36:07.800 --> 00:36:11.700
But continuity does not imply differentiability.
00:36:11.700 --> 00:36:15.400
If I know a function is differentiable at 5, I know that it is continuous at 5.
00:36:15.400 --> 00:36:19.600
I’m given continuity automatically, as a free gift.
00:36:19.600 --> 00:36:25.000
But if I'm only continuous at 5, that tells me nothing about whether it is differentiable at 5.
00:36:25.000 --> 00:36:27.700
Continuity does not give me differentiability for free.
00:36:27.700 --> 00:36:31.000
Different ability gives me continuity for free.
00:36:31.000 --> 00:36:34.700
Because sometimes, we are going to know some function is continuous, we are not going to know it is differentiable.
00:36:34.700 --> 00:36:42.500
Other times, we are going to know a function is differentiable, we automatically going to know that it is continuous.
00:36:42.500 --> 00:36:57.600
Differentiability implies, remember this double headed arrow means implies continuity.
00:36:57.600 --> 00:37:03.600
In other words, if it is differentiable then it is continuous but not the other way around.
00:37:03.600 --> 00:37:06.300
Notice it is not a double headed arrow.
00:37:06.300 --> 00:37:26.700
If it were a double headed arrow, it would mean it works this way and it works that way.
00:37:26.700 --> 00:37:33.300
Let us see what we have got, what is next.
00:37:33.300 --> 00:37:40.500
Differentiability implies continuity but not the other way around.
00:37:40.500 --> 00:37:51.600
In case this idea of implication, differentiability implies continuity but it does not go the other way.
00:37:51.600 --> 00:37:54.000
If it did, we have a double headed arrow.
00:37:54.000 --> 00:37:57.700
Now let us get some real world examples of what this idea of implication means.
00:37:57.700 --> 00:38:01.300
When we say if something is true then something else is true.
00:38:01.300 --> 00:38:06.400
If I have, if a then b, that does not mean if b then a.
00:38:06.400 --> 00:38:08.800
It means only if a then b.
00:38:08.800 --> 00:38:12.400
If it were true the other way around, let us call it converse.
00:38:12.400 --> 00:38:15.700
I would specify if b then a.
00:38:15.700 --> 00:38:17.500
They are definitely different.
00:38:17.500 --> 00:38:25.000
Let me give you a couple of real world examples, in order to help with this intuitive notion of what implication actually means.
00:38:25.000 --> 00:38:38.800
Real world example of this thing which is the logical implication.
00:38:38.800 --> 00:38:42.400
It is huge in mathematics because everything is about if then.
00:38:42.400 --> 00:38:49.400
Now I can write, if rain then clouds.
00:38:49.400 --> 00:38:51.500
In other words, rain implies clouds.
00:38:51.500 --> 00:38:56.900
What that means is, if it is raining, I know automatically it is cloudy.
00:38:56.900 --> 00:38:58.400
There is no other way it is going to happen.
00:38:58.400 --> 00:39:04.400
However, just because it is cloudy, it does not mean that is raining, that is what this means.
00:39:04.400 --> 00:39:06.200
It works one way, it does not work the other way.
00:39:06.200 --> 00:39:11.400
We are very specific in mathematics about this.
00:39:11.400 --> 00:39:25.100
Clouds do not imply rain, but rain always implies clouds.
00:39:25.100 --> 00:39:26.600
Differentiability implies continuity.
00:39:26.600 --> 00:39:29.900
In other words, if it is differentiable, we automatically know it is continuous.
00:39:29.900 --> 00:39:36.800
But just because something is continuous, it does not mean that it is differentiable.
00:39:36.800 --> 00:39:38.700
How can a function fail to be differentiable?
00:39:38.700 --> 00:39:41.700
What we just saw.
00:39:41.700 --> 00:39:48.000
How can a function fail to be differentiable?
00:39:48.000 --> 00:40:03.000
How can a function not be differentiable at a point?
00:40:03.000 --> 00:40:06.000
The first way is a sharp kink.
00:40:06.000 --> 00:40:10.800
I should say a sharp change of direction.
00:40:10.800 --> 00:40:15.000
Sharp change of direction, that is one way.
00:40:15.000 --> 00:40:19.200
If we have a sharp change of direction like the absolute value function, it is not differentiable there.
00:40:19.200 --> 00:40:37.900
A discontinuity in the original function of the derivative, the f(x).
00:40:37.900 --> 00:40:44.800
Three, an infinite slope.
00:40:44.800 --> 00:40:51.700
If you have an infinite slope, in other words, the slope is all of a sudden goes vertical.
00:40:51.700 --> 00:40:55.400
Change in y/ change x, change in 0, it is infinite.
00:40:55.400 --> 00:40:57.500
Vertical line, whatever vertical line, there is no slope.
00:40:57.500 --> 00:41:02.900
The slope is undefined, it is not differentiable there.
00:41:02.900 --> 00:41:11.600
If you have something like rapid change in direction, it is not differentiable at that point.
00:41:11.600 --> 00:41:15.500
It is not differentiable at whatever x value is.
00:41:15.500 --> 00:41:25.700
A discontinuity function is not differentiable at that point.
00:41:25.700 --> 00:41:32.000
The last one, if you have a function that goes vertical and it was like that,
00:41:32.000 --> 00:41:36.800
the slope goes vertical, it is not differentiable at that point.
00:41:36.800 --> 00:41:39.400
Three ways you can fail differentiability.
00:41:39.400 --> 00:41:46.000
Sharp change in direction, a discontinuity, or an infinite slope.
00:41:46.000 --> 00:41:49.600
Let us talk about the final topic here.
00:41:49.600 --> 00:41:56.200
Let us see where we are, maybe higher derivatives.
00:41:56.200 --> 00:42:02.500
I think I will go to blue, just a happy color and makes me happy.
00:42:02.500 --> 00:42:09.800
Higher derivatives.
00:42:09.800 --> 00:42:17.000
I know that if I have a function f(x), I can take the derivative of it and I get some f’(x) which is also function of x.
00:42:17.000 --> 00:42:19.400
I can take the derivative of that.
00:42:19.400 --> 00:42:22.400
As a function of x, I can take the derivative again.
00:42:22.400 --> 00:42:25.400
I can take the derivative as many times as I want.
00:42:25.400 --> 00:42:27.800
As long as I end up with something that is actually meaningful.
00:42:27.800 --> 00:42:33.500
I get the second derivative which we symbolize with a double prime and a triple prime,
00:42:33.500 --> 00:42:37.700
and a quadruple prime, however many derivatives you can actually take.
00:42:37.700 --> 00:42:43.100
y, take the derivative, you get y’.
00:42:43.100 --> 00:42:50.400
Let us do it this way. If I have a function at y, if I take the derivative, dy dx, if we are using that symbolism.
00:42:50.400 --> 00:42:55.500
Now when we take the second derivative, when we take the derivative again,
00:42:55.500 --> 00:43:00.300
the symbol is d² y dx².
00:43:00.300 --> 00:43:06.600
If it was the 4th derivative, it would be the d⁴ y/ dy⁴.
00:43:06.600 --> 00:43:11.200
This is a symbol telling me that I have taken the derivative twice.
00:43:11.200 --> 00:43:19.300
Let us talk about this.
00:43:19.300 --> 00:43:26.300
Let us talk about this peculiar symbolism.
00:43:26.300 --> 00:43:37.400
What the heck is d² y/ dx² mean, why do we symbolize it that way.
00:43:37.400 --> 00:43:43.100
If you do not want to do this, yes I do.
00:43:43.100 --> 00:43:52.600
In mathematics, we have something called an operator.
00:43:52.600 --> 00:44:01.600
An operator is a symbol that tells you do something to a function.
00:44:01.600 --> 00:44:13.700
A symbol that tells you to perform an operation on the symbol.
00:44:13.700 --> 00:44:23.900
It is a fancy word for, it is a symbol to tell you to do something to a function, to perform an operation on a function.
00:44:23.900 --> 00:44:50.000
Given f, taking the derivative of f is an operation that you are performing on the function f.
00:44:50.000 --> 00:45:05.300
It is an operation you perform on f, in order to get f’.
00:45:05.300 --> 00:45:08.600
You have a function, you operate on that function with the differential operator,
00:45:08.600 --> 00:45:15.200
with the derivative operator, and you get another function.
00:45:15.200 --> 00:45:31.700
The derivative operator is symbolized ddx.
00:45:31.700 --> 00:45:44.600
Whenever you see d/ dx and there is something that follows it, that means take the derivative of what is following it.
00:45:44.600 --> 00:46:11.000
When you see ddx of x², this says take the derivative of x².
00:46:11.000 --> 00:46:15.300
It is that simple, it is just take the derivative of x².
00:46:15.300 --> 00:46:32.700
Now we said y is also a symbol used for functions.
00:46:32.700 --> 00:46:46.200
We have seen it for years now, ever since probably 6th or 7th grade.
00:46:46.200 --> 00:46:49.500
y = x², we see that all the time.
00:46:49.500 --> 00:46:57.300
It can be f(x) = x² or we just write y = x².
00:46:57.300 --> 00:47:05.200
Dy dx is equivalent to ddx of y.
00:47:05.200 --> 00:47:11.600
We just decide to get rid of the parentheses and put the y there, up on top with the numerator.
00:47:11.600 --> 00:47:25.100
Dy dx which is equivalent to ddx of y, which is equal to ddx of the function x² says, take the derivative of y.
00:47:25.100 --> 00:47:30.500
That is it, take the derivative of y.
00:47:30.500 --> 00:47:41.000
When you see dv dt, it is telling you z is some function of some variable.
00:47:41.000 --> 00:47:45.200
Take the derivative with respect to t of that.
00:47:45.200 --> 00:47:48.800
It is telling you that z is actually some function of t.
00:47:48.800 --> 00:47:52.100
It is telling you take the derivative of z.
00:47:52.100 --> 00:47:55.400
Dy dx, take the derivative of y.
00:47:55.400 --> 00:48:02.000
That is all it is, find f’, find y’, whatever it is.
00:48:02.000 --> 00:48:17.300
Given y, apply the differential operator that gives you dy dx.
00:48:17.300 --> 00:48:21.800
Apply the differential operator again, in other words take the derivative again,
00:48:21.800 --> 00:48:27.800
that gives you, you are taking ddx of this thing which is a function.
00:48:27.800 --> 00:48:37.900
You are doing ddx of dy dx, multiplied out symbolically.
00:48:37.900 --> 00:48:46.600
d × d is d², dx dx we just put the 2 on top of that.
00:48:46.600 --> 00:48:52.000
d² dx, this is actually the whole thing dx.
00:48:52.000 --> 00:48:53.000
I hope that makes sense.
00:48:53.000 --> 00:48:57.200
That is where the symbolism comes from.
00:48:57.200 --> 00:49:00.800
It is just the idea of applying this thing called an operator.
00:49:00.800 --> 00:49:03.200
In this case, it is the derivative operator.
00:49:03.200 --> 00:49:06.300
Later in the course, you are going to learn something called the integral operator
00:49:06.300 --> 00:49:15.000
which means take the integral of the function, instead of take the derivative of the function.
00:49:15.000 --> 00:49:16.500
Symbolically, that is all it means.
00:49:16.500 --> 00:49:22.200
It is all based on this symbol, an operator which says do something to a function.
00:49:22.200 --> 00:49:30.000
When you have done something to a function, you are going to get another function.
00:49:30.000 --> 00:49:53.400
Let us do example 3, find dy dx and d² y dx² for y = x³.
00:49:53.400 --> 00:50:01.800
The first derivative, we need to apply the differential operator once to get something.
00:50:01.800 --> 00:50:05.400
And then, we need to apply the differential operator again to get something else.
00:50:05.400 --> 00:50:09.400
Again, applying the differential operator just means taking the derivative.
00:50:09.400 --> 00:50:15.700
Finding the limit as h approaches 0 of f(x) + h - f(x)/ h².
00:50:15.700 --> 00:50:21.100
In this case, when we do it, this is our f(x).
00:50:21.100 --> 00:50:25.700
Once we get the first derivative, now we are going to treat this as a function of x and
00:50:25.700 --> 00:50:29.200
we are going to ignore this and treat this as an f(x).
00:50:29.200 --> 00:50:36.700
We are going to be taking the limit of f of this, the derivative.
00:50:36.700 --> 00:50:56.800
The first derivative, the first derivative is just f’(x) is equal to the limit as h approaches 0 of f(x) + h - f(x)/ h.
00:50:56.800 --> 00:50:58.300
We did this already in the previous problem.
00:50:58.300 --> 00:51:00.700
I do not want to go through this process again.
00:51:00.700 --> 00:51:10.300
Let us just say because we are dealing with y = x³, we found that f’(x) is equal to 3x².
00:51:10.300 --> 00:51:12.400
That is our first derivative.
00:51:12.400 --> 00:51:15.400
Now we want to take the second derivative.
00:51:15.400 --> 00:51:24.400
The second derivative means take the derivative now of 3x².
00:51:24.400 --> 00:51:42.100
f”(x) is equal to the limit as h approaches 0 of f’(x) + h - f’(x)/ h.
00:51:42.100 --> 00:51:45.800
This is f, the derivative give me a prime.
00:51:45.800 --> 00:51:53.600
Here is f’, this is f’’.
00:51:53.600 --> 00:51:57.200
It is a prime on top of that prime, that is what is happening here.
00:51:57.200 --> 00:52:16.400
This is going to be the limit as h approaches 0 of this function 3x + h² - 3x²/ h.
00:52:16.400 --> 00:52:24.600
When I expand this and simplify it, I end up with 6x.
00:52:24.600 --> 00:52:29.700
You should do it for yourself to actually make sure that it works.
00:52:29.700 --> 00:52:39.000
What you have got is x³.
00:52:39.000 --> 00:52:42.000
I can go to the next page.
00:52:42.000 --> 00:52:47.700
x³, the first derivative.
00:52:47.700 --> 00:52:51.900
When we take the derivative of that, we get 3x².
00:52:51.900 --> 00:52:56.400
We take the derivative again and we get 6x.
00:52:56.400 --> 00:52:58.900
If I wanted to, I can take the derivative again.
00:52:58.900 --> 00:53:00.400
I would get 6.
00:53:00.400 --> 00:53:04.300
If I wanted to, I can take the derivative one more time and I would end up with 0.
00:53:04.300 --> 00:53:08.500
In this case, this is differentiable four times.
00:53:08.500 --> 00:53:13.600
You will actually see how we go from this to this to this to this.
00:53:13.600 --> 00:53:20.500
For the time being, we have been using the limit as finding the derivative.
00:53:20.500 --> 00:53:28.000
Our process of finding the derivative so far is taking the limit as h approaches 0 of f(x) + h.
00:53:28.000 --> 00:53:32.000
We actually go through this tedious algebraic process.
00:53:32.000 --> 00:53:38.600
In the subsequent lessons, we are going to teach you quick ways to go from here to here,
00:53:38.600 --> 00:53:42.000
without having to go through this process.
00:53:42.000 --> 00:53:43.800
Thank you so much for joining us here at www.educator.com.
00:53:43.800 --> 00:53:45.000
We will see you next time, bye.