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

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

### 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
- Example I: Sketch f'(x)
- Example II: Sketch f'(x)
- Example III: Find the Derivative of the Following Function sing the Definition
- Example IV: Determine f, f', and f'' on a Graph
- Example V: Find an Equation for the Tangent Line to the Graph of the Following Function at the Given x-value
- Example VI: Distance vs. Time
- Example VII: Displacement, Velocity, and Acceleration
- Example VIII: Graph the Displacement Function

- 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

### AP Calculus AB Online Prep Course

### Transcription: More Example Problems for The Derivative

*Hello, welcome back to www.educator.com, 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 problem*0338

*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 functions*1094

*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 derivative*1544

*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 www.educator.com.*1895

*We will see you next time, bye.*1897

0 answers

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.