Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Calculus AB
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

Start Learning Now

Our free lessons will get you started (Adobe Flash® required).
Get immediate access to our entire library.

Sign up for

Membership Overview

  • Unlimited access to our entire library of courses.
  • Search and jump to exactly what you want to learn.
  • *Ask questions and get answers from the community and our teachers!
  • Practice questions with step-by-step solutions.
  • Download lesson files for programming and software training practice.
  • Track your course viewing progress.
  • Download lecture slides for taking notes.
  • Learn at your own pace... anytime, anywhere!

Derivative I

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
  • Derivative 0:09
    • Derivative
  • Example I: Find the Derivative of f(x)=x³ 2:20
  • Notations for the Derivative 7:32
    • Notations for the Derivative
  • Derivative & Rate of Change 11:14
    • Recall the Rate of Change
    • Instantaneous Rate of Change
    • Graphing f(x) and f'(x)
  • Example II: Find the Derivative of x⁴ - x² 24:00
  • Example III: Find the Derivative of f(x)=√x 30:51

Transcription: Derivative I

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

Today, we are going to start talking about the derivative formally.0004

Let us jump right on in.0008

We finally arrived at the how.0011

When we started this course, we talked about this thing called the derivative.0014

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.0018

We also said that the how, we gave you an expression for the how.0027

It involves the limit, we have gone through this process of talking in a detailed way about the limit and0031

now we are going to come down to actually finding derivatives analytically.0038

At least, the first step of finding derivatives analytically.0043

Let us go ahead and work in blue here.0048

We have arrived at the how.0050

We have arrived at the how for derivatives.0056

Given f(x), given some function f(x), the derivative which we symbolize with the prime symbol f’(x) is =0070

the limit as h approaches 0 of f(x) + h - f(x)/ h.0093

Basically what this says at is, what this says is, when are given some function f(x), you are going to form this quotient.0103

You are going to take x + h, you are going to form that, whatever that is.0119

You are going to subtract from it f(x), you are going to divide by h.0123

You are going to simplify this expression and you are going to take the limit as h approaches 0.0126

What you are going to get is a function, that function is your derivative.0130

We symbolize it with f’.0134

It is derived from the original function.0137

Let us do an example.0140

Example 1, find the derivative of f(x) = x³.0147

f’(x) = the limit as h approaches 0 of f(x) + h – f(x).0165

I think it is a good idea when you are doing these problems to actually start each problem by writing down the definition.0178

That way it will stick in your mind, /h.0184

That = the limit as h approaches 0 of, f(x) + h, f is x³.0190

This is going to be x + h³, that is the first one, - f(x) which is x ^ / h.0199

Clearly, we cannot just plug in h as 0 because it would give us in the denominator.0212

We have to simplify this expression0218

In this case, simplification means just expanding, adding, subtracting, multiplying, dividing.0220

Doing whatever you have to do until we find a simplified expression, that we cannot do anything else to.0226

And then, we will take the limit again and see what happens.0230

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 0.0234

x³ go away, you are left with 3x².0260

3x² h + 3x h² + h³/ h.0272

We factor an h from the top, it equal the limit as h approaches 0 of h × 3x² + 3x h + h²/ h.0282

H goes away, you are left with the limit as h approaches 0 of the function 3x² + 3x h + h².0299

Now we plug in h, h was 0, you plug in 0.0318

That goes to 0, that goes to 0, you are left with 3x².0322

There you go.0328

f’(x), your derivative of x³ is equal to 3x².0328

Now recall, we have a derivative.0337

A derivative is two things, a derivative is many things actually.0341

But it is definitely these two things.0344

A derivative is the slope of the tangent line to the curve, at a given x.0348

Notice this is a function, because when you are following a curve, the tangent line, the slope of the line actually changes.0366

The derivative is also an instantaneous rate of change.0376

Remember, when we talked about rates of change,0379

we said that there was an average rate of change that is the secant line between two points.0383

We have some function.0387

If I take two points and draw a line, we call that the secant line.0389

The average rate of change, that is the slope of that line.0393

But if I have a tangent line, from that point, the slope of that is the instantaneous rate of change.0396

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?0403

It is an instantaneous rate of change.0410

What we are hoping will become a conditioned response when you see derivative,0416

is that you are thinking two things, depending on what the problem is.0420

But you will think that this is the slope of the tangent line, to the curve at a given point.0423

And it is the instantaneous rate of change of the function at that point.0428

We just want it to be a conditioned response.0434

Derivative, instantaneous rate of change, and slope of tangent line.0436

Instantaneous rate of change of the function, at a given x.0440

Let us talk about some notations for the derivative.0452

We have seen f’(x), a lot of times we will leave off the x and we will just write f’.0465

You will see y’(x), if you express it as y = x³.0474

Then, the derivative is going to be y’ = 3x².0479

We leave the x off, we do it as y’.0484

Then, there is this one, dy dx.0487

This notation, this one is derived from Δ y/ Δ x.0494

Δ y/ Δ x is a slope.0508

Dy dx is the slope of the tangent of line.0512

It is the instantaneous slope.0515

That is a symbol that we use to describe the derivative, as opposed to the average.0518

Average, instantaneous.0527

Secant line, slope of the tangent line.0531

I really have to slow myself down.0544

Whenever, we form the quotient of any change in y/ any change in x,0547

no matter what those variables are, they could be t, s, q, p, whatever.0565

Whenever we form the quotient Δ y/ Δ x and it pass to the limit,0568

which is what we did right, it is the limit of this f(x) + h - f(x)/ h.0579

This is a change in y/ a change in x.0588

When we pass the limit, that just means when we take the limit of this expression as h goes to 0.0591

When we pass the limit, the Δ's turn into d.0595

This notation reminds us that we are talking about a slope and instantaneous rate of change.0609

If x changes by a really tiny amount, y changes by that tiny amount, that is what that means.0643

If you were to see dy/ dx = 3.0651

It is the same as dy dx = 3/1.0655

If I change x by 1, I change y by 3.0657

The rate of change is the slope.0661

Recall that the slope and rate of change are synonymous.0670

You are going to hammer that point a lot, my apologies.0674

Recall that a slope and a rate of change are synonymous.0678

How do you spell synonymous, all of a sudden I forgot, are the same.0703

Recall that a rate of change is, recall what a rate of change is.0712

When we have a Δ y/ a Δ x, this is equivalent to saying,0730

when we change the x value by a unit amount, how much does y change?0741

When I change x by a unit amount, this number up on top gives me the amount by which y changes.0762

When we say that the function is x³ and when we differentiate it to get 3x², we have f’(x) = 3x².0771

We can write it as y’(x) = 3x².0788

We can write it as dy dx = 3x².0794

Let us choose a specific value.0801

Let us choose a specific value for x.0808

Let us say at x = 2.0819

Therefore, dy dx at x = 2, this is the symbolism that we use.0822

Now we can write it as f’ at 2, y’ at 2.0831

Dy/ dx, we put a line there and 2, like that, if you want.0834

I suppose it is not going to be the end of the world, if you did something like that,0838

whatever notation make sense to you.0843

As long as you understand it and the people that you are writing it for understand it.0844

dy dx at x = 2, dy dx is 3x².0849

It is going to be 3 × 2², it is going to be 12.0854

This is the same as 12/1.0859

This means, when we are at the point 2, we said that x = 2.0863

f(2) is equal to 8 because f is equal to x³.0891

When we are at the point 2,8, from that point, if we move one unit to the left or to the right,0897

then f changes by 12 units, up or down.0933

In this case, it is going to be down or up.0945

That is what this means, dy dx = 12, means dy dx is 12/1.0952

If I change x by 1 unit, I change y by 12 units.0957

Here is what it looks like.0966

Here I have my function y = x³, that is my red line.0968

We said that dy dx which is the derivative is equal to 3x², that is a function dy dx.0974

The reason it is a function is because the slope of the line, as you see, the slope along the curve changes.0983

Therefore, depends on what x is.0989

dy dx at x = 2, we said it is equal to 12, which is the same as 12/1.0992

This is the slope of the tangent line.1001

A tangent line is the one that is in the broken up black.1003

Here is our 2, here is our point 2,8.1008

At that point, the tangent in line is that line right there.1013

It has a slope of 12, that is what the derivative means.1018

The slope of the tangent line is 12 at x = 2.1028

The instantaneous rate of change of f at x = 2 is 12.1040

From this point, if I move that way or this way by one unit, my function is going to change by 12 units.1061

That is what that means.1071

The derivative gives me a function.1073

When I put a specific x value in, it gives me the slope of the tangent line at that point.1077

It also gives me the rate at which the function changes from that point.1083

Notice that the derivative is a function of x itself.1094

f(x), you take the derivative, you get another function, which we symbolize f’(x).1141

Our example x³, we took the derivative and we got 3x².1146

We often will graph both together.1153

Both the function and the derivative.1160

In fact, in your problems, sometimes given the graph of f(x), you are going to be asked to graph f’(x).1169

Also sometimes given f’(x), sometimes you will be given the graph of the derivative of the function.1195

You have to recover, sometimes given f’(x), given the graph of f’(x) not just f’(x).1207

Given the graph of f’(x), you must recover the graph of the original f(x).1220

Let us look at x³ and x².1238

I’m sorry, let us look at x³ and its derivative 3x², together on the same graph.1240

The red is the x³ and the broken up black is the 3x².1249

Here we have f(x) is equal to x³.1257

We have f’(x) which is equal to 3x².1262

This is the slope of f(x) at any, I should say, at the various values of x.1269

This is going to be very important.1289

You want to take your time when dealing with these graphs,1290

so it can take you a little bit just to sort of wrap your mind around the fact1293

that you are talking about two things that are connected, but are somehow separate.1296

Now f itself is a function of x.1302

f’(x) is also function of x.1305

Each one of those has a graph.1308

This is x³, this is 3x².1311

But the 3x² is the derivative of the x³.1315

The derivative tells me what the slope of the graph is, at that point.1321

As I move x negative to positive, again, we are always moving from negative values, working our way that way.1326

Left to right, negative to positive.1336

Notice the slope of the curve here, the slope is positive which is y, and 3x²,1338

the derivative is the slope of the curve which is why it starts above the x axis.1349

But notice as x moves that way, the slope of the function is decreasing towards 0.1354

That is what this demonstrates.1367

This shows that it starts up here, it is declining towards 0.1369

Here the slope of the function itself is 0, which is why the derivative function is 0.1378

Now the slope starts to become positive again.1390

Positive, positive, positive, positive, positive.1394

But it is not just becoming positive, it is becoming more and more positive.1397

Therefore, the black line, the derivative is telling me that.1402

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.1405

The function is the red, the derivative is that.1416

The derivative is the slope.1419

The slope is positive but it decreases towards 0.1421

It starts to increase again, that is what this line is telling me.1424

Let us do another example.1432

Let us go ahead and work in blue.1438

Let us do example number 2.1440

Find the derivative of x⁴ - x², I wish I had not chosen such a complicated function.1448

Find the derivative of x⁴ - x².1464

We know that f’(x) is equal to,1468

I always do that, after 35 years, I still forget to write down my limit.1473

It equal the limit as h approaches 0 of f(x) + h - f(x)/ h.1479

I’m not going to keep writing it over and over again.1490

I’m just going to work with the function.1493

I will just simplify it and then we will take the limit at the end.1497

f(x) + h, this is going to be x + h⁴ – x + h².1501

That is the f(x) + h – f(x).1513

f(x) is x⁴ - x²/ h.1517

Again, you plug in 0, you are going to have 0 in the denominator.1525

You have to simplify it out.1529

Let us multiply all of this out.1531

We are going to expand that.1533

This is going to equal x⁴ + 4x³ h + 6x² h²1536

+ 4x h³ + h⁴ - x² + 2x h + h².1548

And then, we are going to do this - this – that.1569

Remember we have to distribute.1571

It is going to be –x⁴ + x²/ h.1573

We are going to get, x⁴ cancels x⁴.1586

We are going to get 4x³ h + 6x² h² + 4x h³ + h⁴ - x² - 2x h - h² + x²/ h.1595

-x² and +x² cancel, I’m going to factor out an h.1625

It is going to be h × 4x³ + 6x² h + 4x h² + h³ - 2x - h/ h.1630

The h is canceled, I'm left with my final 4x³ + 6x² h + 4x h² + h³ - 2x – h.1652

Now I take the limit.1669

Now we take the limit as h approaches 0 of this thing.1674

It is going to be 4x³ + 6x h + 4x h² + h³ - 2x – h.1681

h goes to 0, that goes to 0, that goes to 0, that goes to 0.1700

You are left with that.1705

f’(x)is equal to 4x³ - 2x.1711

We have y’(x) = 4x³ - 2x.1722

You will see dy dx = 4x³ - 2x.1729

Let us take a look at f(x) and f’(x) in the same graph.1735

The red is your f(x), this was your x⁴ – that.1741

That is your derivative which was 4x³ - 2x.1751

The derivative graph tells you what the slope is of the graph.1758

The slope is negative but it is increasing.1763

Notice here the slope is 0.1772

Here the slope is 0, here the slope of the function is 0.1774

Therefore, the derivative, this derivative graph, the black, 0, 0, 0.1779

Here it is negative, so it is below the axis.1789

Here the slope of the graph is positive.1792

It is above the x axis.1795

Here the slope of the graph is negative.1799

This is negative, it is below the x axis, it hits 0.1800

After this point, the slope starts to become positive again.1804

It arises and it is above the x axis.1809

The derivative is the description of how the slope is changing, as you move along the graph.1811

This is a description of how the slope of the original f(x) is changing.1823

It is a function of x in its own right.1828

That is all it is saying.1833

For example, at x = that value, the slope is this.1834

What is the slope, the slope is that number.1840

That is what is happening here.1845

Let us do example 3, find the derivative of f(x) = √x.1851

We actually did this in one of the example problems from a previous lesson, but let us do it again formally.1871

We have f’(x) = f(x) + h – f(x).1877

Again, I forgot my limit.1885

I think it is unbelievable, I never remember to write down my limit.1888

I remember to take the limit, but I never remember the write it down.1893

The limit as h goes to 0 of f(x) + h - f(x)/ h.1896

Again, I’m not going to write the limit over and over again.1908

I’m just going to go ahead and work with the function.1910

√x, f(x) + h is √x + h – √x, that is the f(x) divided by h.1916

I have to manipulate it.1927

I’m going to go ahead and multiply by the conjugate of the numerator.1928

It is going to be √x + h + √x/ √x + h + √x.1933

I end up with x + h - x/ h × √x + h + √x.1944

That and that go away, leaving me just h on top.1956

h and h cancel, I’m left with 1/ √x + h + √x.1958

Now I take the limit.1969

The limit as h goes to 0 of 1/ √x + h + √x.1971

h goes to 0, this turns to 0.1980

I'm left with 1/ √x + √x.1982

I get 1/ √x, that is my f’.1985

f’(x) = 1/ 2√x.1991

I have got f’(x) = 1/ 2√x.2010

y’(x), same thing, different notation, 1/ 2√x.2017

And the last notation, dy/ dx = 1/ 2√x.2023

Now I’m going to ask the question, what is dy dx at x = 3?2030

Very simple, dy dx at x = 3 is equal to, just plug in 3, 1/ 2√3.2041

What is the f(3)?2060

f, we said that f(x), the original function is just √x.2069

f(3) is equal to √3.2076

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.2085

f of that which is √3.2127

It has a slope of 1/ 2√3.2138

Notice how I did that.2143

The curve itself, there is an x value.2146

That x value is going to give me y value.2150

The point is going to be your xy or your x f(x).2153

The derivative at that point is the slope of the line.2157

Now you have a slope and you have a point that it passes through.2161

You can find the equation of that line.2164

The equation of the tangent line is, we know that it is y - y1 = m × x - x1.2170

Here it is going to be y - y1 is √3 = the slope which is 1/ 2√3 × x – 3.2184

There we go.2203

The red is the graph, the black is the tangent line to the graph at x = 3.2206

At the value x = 3, I go up here.2215

This point right here, this is my 3√3.2219

This is y = √x.2229

This line, the equation of this line, we just found it.2235

This is y - √3 = 1/ 2√3 × x – 3.2240

y - y1 = m × x – x1.2250

This point is easy to find.2255

Just plug in the x value and you get the y value.2256

That is going to be the 3 and √3.2258

The slope, that is from the derivative.2261

f(x) = √x, f’(x) 1/ 2√x, it is that simple.2267

I hope that made sense.2283

Let me write out everything here.2287

f(3) = 3, the tangent line touches 3√3.2290

Now f’ at 3 = 1/ 2√3.2303

The slope of this tangent line is 1/ 2√3.2311

Let us go ahead and take a look at the graph and its derivative as functions of x.2323

Here is the function y = √x, this is the derivative.2332

It is y prime, it is equal to 1/ 2√x.2337

The graph, the original graph, the derivative, the black, describes how the slope changes as x gets bigger and bigger.2343

Notice the slope is almost straight up infinite.2353

It is positive, positive, positive, positive, positive.2359

Always a positive slope along the curve but the positive slope is decreasing.2362

The tangent line is actually dropping down getting close to a slope of 0.2374

That is what this be describes.2379

High, positive, positive, positive, positive, but the slope is getting closer to 0.2381

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.2387

That is what that tells me.2396

Thank you so much for joining us here at

We will see you next time, bye.2401