For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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## Transcription

### 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

### AP Calculus AB Online Prep Course

### Transcription: Derivative I

*Hello, welcome back to www.educator.com, 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 and*0031

*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 expression*0218

*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 fact*1293

*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 h²*1536

*+ 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 www.educator.com.*2399

*We will see you next time, bye.*2401

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