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INSTRUCTORS Raffi Hovasapian John Zhu

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For more information, please see full course syllabus of AP Calculus AB
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Lecture Comments (3)

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Post by Peter Fraser on April 11 at 07:13:27 PM

Hi Raffi

At 8:35 I think the divisor for the complex fraction should not have changed to a negative sign for the x^-1/2 term; also at 8:58 I think the x^1/2 - (x + h)^1/2 expression should be x^1/2 + (x + h)^1/2.

Sorry in advance if I'm wrong in this.

Peter :)

1 answer

Last reply by: Professor Hovasapian
Fri Apr 7, 2017 6:44 PM

Post by Peter Fraser on April 6 at 03:47:34 PM

3:33: That derivative shape's really interesting because I think it's suggesting that an antiderivative of a rational function such as 1/x ought to have that sort of red coloured graph.

More Example Problems for The Derivative

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Example I: Sketch f'(x) 0:10
  • Example II: Sketch f'(x) 2:14
  • Example III: Find the Derivative of the Following Function sing the Definition 3:49
  • Example IV: Determine f, f', and f'' on a Graph 12:43
  • Example V: Find an Equation for the Tangent Line to the Graph of the Following Function at the Given x-value 13:40
  • Example VI: Distance vs. Time 20:15
  • Example VII: Displacement, Velocity, and Acceleration 23:56
  • Example VIII: Graph the Displacement Function 28:20

Transcription: More Example Problems for The Derivative

Hello, welcome back to, and welcome back to AP Calculus.0000

Today, I thought we would do some more examples on the derivative.0004

Let us jump right on in.0009

Typical example, graphs are going to be very huge.0013

You are going to be given a function, find the derivative graph.0015

Or the derivative graph, find the function, work your way back.0018

The following is a graph of f(x), the original function.0022

On the same graph, sketch f’(x).0024

They just want a sketch of the derivative.0028

One of the things that we know -- when you are working with graphs and their derivatives, the most important thing is slope, where slope is 0.0033

We know that a slope of 0 means a horizontal tangent line.0042

On this graph, we have a horizontal tangent line here, here, and here.0046

At these x values, namely, let us say somewhere around there.0056

Here it looks like this is 0.0060

Here, those are the places where the derivative, because the derivative is the slope of the graph.0062

It is where the derivative graph hits 0.0068

Now to the left of this point, we notice that the slope is negative.0071

It is below the axis.0076

As x moves this way, as we hit this point, that is where we are going to hit 0.0080

From this point to this point, the slope is positive.0086

It changes and it increases a little bit, and it decreases towards 0.0090

Basically, it goes like this and then it passes 0 again.0097

Now you notice the slope is pass this point, the slope is negative.0102

It gets more negative and it starts to go positive again until it hits 0 again.0106

It gets more negative, it starts to go positive until it hits 0 again.0113

Then, it keeps going up.0117

Look for horizontal tangents, when you are working with graphs and their derivatives.0120

Whether it is first derivative or second derivative, or whatever.0125

There you go, that is the derivative.0128

This is black, this is your f’(x).0129

Same thing, the following is a graph of f(x), sketch f’(x).0136

Here, we notice that f(x) is not differentiable here.0142

Anytime you remember when you have a cusp, 0147

that is one of the ways that a function cannot be differentiable at that point.0148

Everything else is fine.0152

We look at this and we see that the slope moving from of course left to right, 0154

we are moving this way from negative values to positive values, the slope is negative.0159

It is actually becoming more negative.0165

At that point, it is not differentiable.0169

The derivative is not defined there.0174

It is a negative slope, it means it starts below the x axis, becomes more negative.0176

The derivative graph looks like that.0183

You are just graphing what the slope does.0185

That is what the f’ graph is, that is what the derivative graph is.0187

What is the slope doing?0191

Starting negative becoming more negative and not touching 0.0193

Here, the derivative just passes 0.0201

It starts really highly positive, it stays positive but it becomes less positive.0205

Highly positive becomes less positive, and that is it.0210

This hyperbola looking thing, this is your f’ graph.0217

That is all it is, what is the slope doing, that is all you have to ask yourself.0221

Find the derivative of the following function using the definition.0232

I think I’m going to work in blue here.0237

Again, it is always a great idea to write the definition.0239

Just get accustomed to writing it in a nice, just to be systematic in your approach.0242

We know that the definition f’(x), the derivative = the limit as h approaches 0 of the following quotient, f(x) + h.0247

It must be nice if I make my + signs legible.0257

x + h - f(x)/ h.0262

We form this quotient first, simplify as much as possible, and then we take the limit.0266

Let us do it, I'm going to go ahead and just work with the difference quotient first.0278

And then, I will do the limit at the end.0285

I do not want to keep writing limit as h goes to 0.0287

I will just work with the quotient first.0291

f(x) + h is going to be 1/ √x + h – f(x).0293

f(x) is 1/ √x/ h.0302

I take the common denominator for the top.0309

It becomes √x – √x + h/ √x × √x + h/ h.0312

I’m going to keep simplifying this here.0334

Actually you know what, it looks like this particular one going to cause a little bit of a problem0338

because we are still going to end up with, if we take h to 0,0349

it looks like here the denominator is 0 as h goes to 0, because when we put 0 in there for h.0353

I’m going to try an alternate procedure here.0367

I’m going to start again.0369

I’m going to go 1/ √x + h - 1/ √x/ h.0372

This one, the reason I stop is because at this point, there is really not much you can do.0382

I mean, this is h/1.0389

When you flip it, you are going to end up with an h at the bottom.0391

When you take h to 0, you are going to end up with 0 in the denominator.0394

That is what is happening here.0398

When that happens, because you really cannot simplify this much more than that this.0399

Because we have 0 in the denominator, we cannot really do anything with that.0405

We have to try an alternate procedure, an alternate manipulation.0408

That is why we are going to try an alternate manipulation here.0412

I go back to the beginning, this thing.0415

I think what I’m going to do them is multiply it by the conjugate of the numerator.0417

It is going to be × 1/ √x + h + 1/ √x/ 1/ √x + h + 1/ √x.0422

This is going to end up equaling 1/ x + h - 1/ x/ h × 1/ √x + h + 1/ √x, when you simplify this.0441

Let us see what we have got here.0462

I might as well continue on.0465

This is going to equal x - x + h/ x × x + h.0468

I just found the common denominator for the numerator, and I leave the bottom as is.0478

1/ √x + h + 1/ √x.0484

This is going to end up equaling, x - x cancels, you are going to end up with -h/ x × x + h/ h × 1/ √x + h -1/ √x.0490

Now the h cancel and you are left with,0519

Now we have -1/ x × x + h/, do a common denominator here, √x - √x + h/ √ x × √x + h.0526

This = -1/ x × x + h × √x × √x + h.0548

Again, this is just algebraic manipulation, that is all it is, /√x + √x + h.0558

That is fine, I will go ahead and write out the whole thing, not a problem.0577

This is going to be -1/ x × x + h ×, this is √ × √, I can combine them.0579

It is just going to be x × x + h, under the radical together, / √x + √x + h = -1/, 0589

This can cancel with that, one of these.0605

I’m left with √x × x + h.0612

On the bottom, this one, this cancels one of the radicals here.0618

This just leaves me with that.0622

This is going to be × √x + √x + h.0624

Now I take the limit of this.0633

I take the limit as h goes to 0.0637

This is just the same as f(x), I have just manipulated it.0640

The limit as h goes to 0.0644

Now I take h to 0 and I end up with -1/ √x² × √x + √x.0647

Let me go to the next page, it is not a problem.0669

It = -1/ √x × 2√x = -1/ 2, just √x, x².0671

x² × x is x³.0686

There you go, that is your derivative.0688

Or we can write it as 2x³/2, that is another way that we can write it.0693

Again, that is one possibility, or a third possibility is you can bring this up to the top.0703

You can write -, this x³/2 is going to be x⁻³/2 /2.0710

Any one of these is absolutely fine, it is not a problem.0717

Notice that f’(x), we have this thing right here.0721

If it is 0, if x is 0, we are going to end up with 0 in the denominator.0736

The original function f(x) is not differentiable at 0.0740

If you try to graph it, you can see that it is not going to be differentiable at 0.0749

Analytically, you can see that it is not going to be differentiable at 0.0752

Let us try something else here.0764

The following is a graph of f(x), f’, and f”.0765

Our first and second derivatives along with the original function.0770

Decide which is which and be able to explain why you made the choices that you made.0773

Again, when you are dealing with situations like this, you work with 0 slopes.0777

You look for a function, you look for the graph that has a horizontal tangent.0782

You see where it hits 0 on the x axis, that is going to be the derivative of that function.0788

Again, work with horizontal tangent which is 0 slopes.0793

Let me work in black actually because one of my graphs is blue.0814

I have a horizontal tangent here.0818

I find that the blue graph is where it hits 0.0822

The blue graph is the derivative of the red graph.0826

Now the blue graph has a horizontal tangent here.0829

As I go straight up from there, I notice the green graph has crosses 0 at the x value 0834

where the blue graph has a horizontal tangent, which means that the green graph is the derivative of the blue graph.0843

What you get is f, the original function is the red graph.0851

The blue is f’, the first derivative.0856

The green is going to be the f”.0861

That is it, just work with horizontal tangents and see where it is.0866

The graph has a horizontal tangent, see where the x value crosses 0, that is the derivative of that function.0869

Find an equation for the tangent line to the graph of the following function at the given x value.0881

A tangent line, we need to find the derivative.0887

We need to find the y value, whatever that is going to be, for the original function.0891

And then, we are going to do y - y1 = the slope.0896

We are going to put the x value into the derivative function to get the slope, × x – x.0909

x1 is 4, y1 we are going to find.0915

Let us get started.0921

We know that f’(x) is equal to the limit as h approaches 0 of f(x) + h - f(x)/ h.0923

Again, I'm going to work with just the difference quotient.0940

I'm going to get the f(x) + h is 5 × x + h/ x + h + 3² - f(x) which is 5x/ x + 3², and all of that is going to be /h.0946

I'm going to end up with, I’m going to take a common denominator here.0975

I’m going to get 5 × x + h × x + 3² - 5x × x + h + 3²/ x + h + 3 × x + 3²/ h.0978

I’m not going to do the expansion.1026

I’m going to leave the algebra to you.1027

When you expand, when you multiply everything out, when you expand, multiply, and simplify, you get the following.1029

Once you simplify it, you take the limit as h goes to 0.1052

In other words, you put 0 in wherever h shows up in the expression.1055

You are going to get your derivative.1058

You get f’(x) is equal to -5x + 15/ x + 3³, that is the derivative of the function.1061

Now the original f(x), that was equal to 5x/ x + 3².1073

We need to find the y value for where the original function = 4.1088

This is going to be the hardest part, when dealing with derivatives and functions1094

is making sure you keep the original function separate from the derivative function.1098

They are not the same.1105

Here we want to find f(4), the original function.1107

You are going to end up with 20/49, when you put it in.1110

The tangent line touches the graph of the original function at the point 4, 20/49.1114

Now the slope, that is the derivative.1136

We want f’ at 4.1138

That is going to equal -5 × 4, we are using this one, + 15/ 4 + 3³.1142

It is going to end up equaling -5/ 343, if I done my arithmetic correctly which I often do not.1156

Please check that.1161

My slope is -5.1163

The equation of the tangent line to the original function at x = 4 is, we said it is y - y1 = the slope × x - x1.1169

It is going to be y - 20/ 49, that is our y value = our slope -5/ 343 × x – 4.1188

That is it, nice and simple.1202

You find the derivative, you find the slope.1203

You find the x and y values of the point that it passes through.1205

Now you got a slope and you got a point, you are done.1208

The following graph shows a distance vs. time graph for two runners.1219

One runner is the blue.1223

One runner is our blue graph and runner two is our red.1225

Answer the following questions.1229

Describe the race from beginning to end.1231

Distance vs. time.1232

Distant on the y axis, time on the x axis.1234

The slope of the graph is the y axis/ the x axis.1240

y/x, Δ y/ Δ x.1245

Distance/ time, we already know what distance/ time is.1248

Distance/ time = velocity.1256

The slope at any point along the graph tells me how fast the runner is going.1259

It is always going to be that way.1264

The derivative is always going to be the slope, you know that already.1266

In this case, it is a physical application.1269

Here the y value is distance, the x value is time.1271

Distance/ time, velocity.1274

In this case, the slope gives me the velocity of the person at any given moment.1276

We notice that the red runner, it is constant velocity.1281

It is one speed.1284

The blue runner starts slowly, his slope is almost horizontal but he speeds up.1285

At some point, his slope is actually steeper than the slope of the red.1293

It is actually going faster than the other person.1299

What is happening here is very simple.1304

They start running, the person in red is running faster initially.1306

At some point, where the slopes are the same, they are the same speed but the red is further ahead.1311

However, as the blue starts to speed up his velocity, eventually, at this value,1323

it looks like maybe 14 or 15 seconds, he catches up with the first one.1329

They are actually at the same distance, that is what is happening.1334

Red, blue catches up eventually, that is what is happening.1338

That takes care of part A, when do the runners have the same velocity?1344

Velocity is a slope, they have the same velocity when the slope of the blue graph is the same as,1347

We look for something like that when they are parallel.1353

Maybe somewhere around 8.2 seconds or something like that.1356

When is the distance between the runners greatest, the part c.1363

The distance is greatest when the gap between them, when that distance is the greatest.1367

It looks like it is somewhere around 8 seconds.1376

You can argue, maybe it is 7 or 9, something like that.1379

But basically you are looking for the gap between the two.1384

The distance, that is the y axis, when is that distance the greatest.1388

That is it, nice and simple.1395

Again, slope is the y axis/ the x axis.1397

In any given real world situation, it might be temperature time, it might be distance time, whatever it is.1401

In this case, it was distance/ time.1408

Distance/ time, we know it is velocity.1411

The displacement of a particle from the origin.1420

As it moves along the x axis it is given by x(t) = t³ – 7t² + 12t – 3.1422

What this means is that as time goes from 0 forward, 1, 2, 3, 4, 5, 6, 7 seconds, 1430

when I put both t values in here, the number that I get, the x tells me where on the x axis I am.1437

It gives the x coordinate, that is what this means.1444

Find the displacement, the velocity, and the acceleration of the particle at t = 5 seconds.1449

It is asking me, at 5 seconds, where is it, how fast is it going, in what direction, and is it speeding up or slowing down.1453

Displacement, where is it.1461

Velocity, how fast is it moving and in what direction.1463

Positive velocity to the right.1468

Negative velocity to the left.1469

We are on the x axis, we are moving this way and this way.1471

Positive velocity is that way, negative velocity is that way.1475

Acceleration, if it is positive, it is speeding up at that point.1479

If it is negative, it is slowing down.1483

You can have a particle with a positive velocity and a negative acceleration.1487

It is moving to the right but it is slowing down.1490

It is getting ready to stop.1494

f(t), displacement.1500

It is t³ – 7t² + 12t – 3.1504

We want to know what x of 5 is.1512

When we put it in, we get 7.1514

At 5 seconds, on the x axis, that point, my point is there.1517

Now velocity, the velocity is a function of time, it is equal to the first derivative of the displacement function.1527

The derivative of this function, you can either do it by doing the definition of derivative1544

or you can do that thing that we told you about.1551

3t² - 14t + 12, remember, I think we discussed that real quickly at the end of one lesson 1554

or at the end of the problem lessons.1564

You take this number, you bring it down here.1566

You multiply it by that and you reduce the degree by 1.1568

This is the velocity function.1572

At any time t, this function tells me how fast I’m moving.1573

The velocity at 5 = 17.1579

This is positive which means that I'm here and I’m actually moving to the right at 17 m/s, ft/s, mph, whatever my unit is.1583

Now the acceleration function.1596

The acceleration at any given time t is equal to the derivative of the velocity, 1598

which is equal to the second derivative of the displacement.1604

Now I take the derivative of that.1610

2 × 3 is 6, 6t, drop the degree by 1.1612

1 × 14 is 14, -14, drop it by 1, it is just t⁰ which is 1.1617

It just stays 14, that is my acceleration function.1626

My acceleration at 5 is equal to 16.1631

This is positive.1638

I’m at 7, I'm traveling at 7, let us say m/s.1642

I’m accelerating, I’m speeding up going to the right.1646

Not only am I going to the right, but I’m actually getting faster going to the right, which means 16 m/s² acceleration.1650

At t = 5, I’m at 17.1659

At 6 seconds, 1 second later, I’m at 17 + 16.1662

I’m 33 m/s.1666

One second after that, I’m at 33 + 16, I’m 49 m/s.1669

I'm here, I’m traveling this fast, and I’m speeding up that fast per second.1675

Let us say this is meters, this is going to be meters per second, this is going to be meters per square second.1683

Displacement, when you take the first derivative, you get the velocity.1691

When you take the derivative of velocity, you get the acceleration.1694

Now graph the displacement function from the previous problem.1701

Then use the graph to describe the motion of the particle during the first 5 seconds.1705

The function was x(t) = t³ – 7t² + 12t – 3.1711

Here is the graph of the function.1716

We want you to describe what is happening to this particle.1718

Describe its motion during the first 5 seconds.1722

5 seconds is here.1727

Between 0 seconds and 5 seconds, what is the particle doing?1729

At t = 0, I'm at, it looks like -3.1735

Let me draw it up here.1742

Here is my 0, let us say this is -3.1746

I start at -3, that is what this graph is telling time.1753

At time = 0, it is this way.1756

Now my slope is positive which means that the particle is actually going to be moving to the right.1758

It is positive but it is slowing down.1769

At 1 second, the velocity hits 0.1773

At 1 second, the particle actually stops momentarily.1776

The velocity which is the slope becomes negative and becomes more negative.1782

Now the particle is moving this way.1787

Again, but it starts to slow down and at just about 3.5 seconds, it hits 0 again.1792

The slope is 0, the velocity is 0 again.1800

Now the velocity becomes positive.1803

It turns around again and starts moving to the right.1805

This time the slope starts accelerating to the right.1809

That is what is happening.1812

At 5 seconds, it actually hits 7.1814

It starts at -3, it starts moving to the right.1820

Stops momentarily then starts moving to the left.1829

Stops again and starts moving to the right.1832

At 5 seconds, I’m at 7, that is what is happening.1835

Once again, if this is my 0, let us say this is my 3.1840

Let us say this is my 7.1845

My particle starts here, moves to the right, moves to the left, moves to the right.1847

At 5 seconds, I'm there, that is what is happening.1854

That is what the graph is telling me.1857

Again, distance vs. time.1858

Time is on this axis, x is the displacement.1861

It is the distance, it is how far I am along the x axis.1864

The graph is not describing the path that the particle is following.1868

It is not moving in two dimensions.1875

It is moving in one dimension.1876

The path is telling you what is happening in that one dimension.1878

The slope is the velocity.1882

Positive velocity moving to the right.1884

0 velocity, it stops momentarily.1887

Negative velocity, it is heading the other direction, that is what is happening.1889

Thank you so much for joining us here at

We will see you next time, bye.1897