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Calculus: Derivative of the Square Root Function


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About this Lesson

  • Type: Video Tutorial
  • Length: 15:19
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 166 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Intro to Derivatives (10 lessons, $17.82)
Calculus: Some Special Derivatives (2 lessons, $4.95)

In this lesson, we will calculate the derivative of the square root of x using the definition of the derivative. As part of this problem, we will go through how to calculate the instantaneous rate of change at a given point in time. Further, we'll examine how to use a position function to find the point in time when an object will be moving at a particular instantaneous rate of change (or velocity) by setting the derivative of the position function to that rate of speed and then solving for the variable.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Recent Reviews

awesome instructional!
~ nachan

This is great lesson if you need to understand using the square root of x in calculus. I love the props he uses to help explain each problem! He also explains how to find the instantaneous rate of change by finding the derivative. He also explains the definition of the derivative and explains a step by step process of finding the problem.

awesome instructional!
~ nachan

This is great lesson if you need to understand using the square root of x in calculus. I love the props he uses to help explain each problem! He also explains how to find the instantaneous rate of change by finding the derivative. He also explains the definition of the derivative and explains a step by step process of finding the problem.

An Introduction to Derivatives
Some Special Derivatives
The Derivative of the Square Root Function Page [1 of 4]
Okay. Well, this is the last opportunity to try one of these questions out, and I think you’ll enjoy this one. The last one is the old favorite. Let f of x equal the square root of x—square root of x. Find the instantaneous rate of change of that function when x equals four. I’m going to try to give you a physical sense of this, if you think of a physical sense of this. Let’s pretend that this actually describes the rate—I’m sorry—describes the position, rather—of the little bear bicyclist that we looked at earlier.
For the bear bicyclist, we want to take a look at the position of this thing and suppose it’s given by that. I actually want to show you—the reason why I picked the bear bicyclist is because I can actually capture the live movement in terms of the velocity. I want you to see this and watch him, and then we’re going to study it a little bit more. But just for fun, you can watch this, and here’s how it’s going to look.
Now, maybe you noticed that it started off sort of fast, and then it sort of tapered down and slowed down. And let’s see if that really is what’s being described here. Okay. Well anyway, to find the instantaneous rate of change, I just remind you that—well, how do you find instantaneous rate of change? It’s just an application of the derivative. So, this is sort of a no-frills, no-bells, no-whistles kind of question. All we have to do here is find the derivative and then evaluate it at four and that will give us the instantaneous rate of change.
Okay. Well, let’s try that. Let’s try one right now on the fly. Well, you have to remember, of course, the definition of the derivative. Well, there it is. And here’s our function. So, let’s see how we proceed in finding the derivative of this square root function. So, again I start off, as always, just writing in f prime of x equals the limit as delta x approaches zero of—well—f of x plus delta x.
Now, f of x equals the square root of x. So, what does f of x plus delta x? Well, wherever I see this x here, I going to replace it by all that stuff right there. So, I’m going to replace it by x plus delta x. And so, I see the square root of x plus delta x. That’s just this piece, right here. See how that follows looking at this? Let’s pick out the x and put an x plus delta x.
Now, I have to subtract off f of x. Well, that’s just subtracting off the square root of x. Divide all that by delta x. There’s the limit. Let’s take it. If delta x goes to zero, then that term right there goes to zero, and so I see the square root of x minus the square root of x. Well, that’s zero. Okay. We’re going to add zero on the bottom too. Mmm. Indeterminate form. Needs more work.
What would you do here? Well, this is not really a problem where factoring is going to sort of be a good idea. However, you might recall some of those cool tricks we learned about finding limits that when you see square roots like this, a good idea sometimes is to multiply by a clever choice of one—a clever choice of one. By multiplying something by one, you’re certainly not going to change the value of it.
So, I’m going to write this. This equals the limit as delta x goes to zero. I’m going to write all that stuff out there—square root of x plus delta x minus the square root of x all over delta x. So now, I’m going to multiply that by one, and what’s the choice of one that I’m going to use here? Well, so there will be something divided by itself, and I’m just going to write in all this square root stuff. The thing that scares me is the square root. So, maybe I should just multiply everything through by the square root and hope all those radicals will just go away.
Now, a good choice would be to put in a minus sign there, but if you try that, you’ll see square roots will still remain when you FOIL and untangle this. However, if you change the sign of whatever that is, that was the trick we saw when we were looking at limits earlier. If you multiplied by the exact same object as that square root, but just change the sign, then the inside term and the outside term will drop out and there’ll be no square roots on the top. That was the trick. Now along with that trick, I think we’re home free. Let’s take a look and see.
The limit as delta x goes to zero—notice, I keep writing that out, by the way. Boy, your prof would be so impressed with you if you keep writing that out every time. Okay, well now, let’s multiply the top. We multiplied the bottom. So, I multiply the top. So, that’s going to be a little bit FOILly here. Let me do that for you right now. It’s the first times the first. Well, those two things are the same, and since it’s a square root, what do we see? Well, when you square a square root, the radical just lifts. So, all I see here is x plus delta x. That’s just that product right there.
Then, I’m going to have an inside term, which notice is minus the square root of x times the square root of x plus delta x. However, again, I immediately add the square root of x times the square root of x plus delta x. So, these two terms will actually add up to give zero. They cancel out. I could write that down, but I won’t. They just cancel. And I’m left with the last term and a minus square root of x times a plus square root of an x, gives me a minus x. Because, again, the radical lifts, but there’s a negative sign there.
So, the top actually cleans off quite nicely. In fact, you might already see those two terms actually cancel. I won’t cancel them right now, and it’s all divided by the bottom. And remember my philosophy about bottoms. I don’t distribute. In fact, I try to avoid touching the bottom as much as possible, because the more you touch it, the more complicated I think you can make it. So, I just keep everything there, but I do remember that it’s all that times delta x. I didn’t make that usual mistake that I tend to make, forgetting to multiply everything through. I keep it nestled in the parentheses right there.
Okay, well now, I noticed the happy fact that this x and this minus x actually annihilate themselves, and all I’m left with on the top now is a delta x, and then I have a delta x to factor on the bottom. Well there’s the zero over zero. It’s sort of, almost, like a hunting game. Can you hunt down the zero over zero that makes this thing an indeterminate form.
I think this is great fun. Don’t you think this is fun? You have to admit, it’s sort of fun. The idea is to hunt down. It’s like a mystery. Can you find it? It’s like a Sherlock Holmes kind of thing. Okay. Well, there it is. We can cancel them out. So, I’m going to cancel them out. Now, by the way, when you cancel this out, you are left with an invisible one on top. Don’t forget that. I don’t want you to think there’s a zero there. There’s an invisible one, because it’s one times delta x.
Okay. And now, what do we have? Well now, what we see is the limit as delta x approaches zero of—and on the top, I just have a nice clean one. Look how nice that is. And on the bottom, I just have the square root of x plus delta x plus the square root of x. So, notice how I migrated the square roots of x that were originally on top—where we can subtract it, which is bad—to square roots on the bottom where they’re being added. That’s happy.
I can now actually take the limit. Let delta x get really, really small. Let it approach zero. Let it head to zero. As it gets smaller, smaller and smaller, what happens? Well, this term right in here shoots down to zero. And so, what are we left with? So this equals—now I’m actually taking the limit, so I’m not going to write the limit anymore, because I’m actually taking the limit. This equals one divided by—well, the square root of x plus zero plus the square root of x.
So, what is this square root of x plus the square root of x. Is that just x? Does the radical lift because I have two of them? No. I’m adding, not multiplying. So, if I have one square root of x and another square root of x, how many square roots of x do I have? I’ve got two of them. So, in fact, this equals one over two square roots of x and that’s the derivative.
So now, we see the derivative of the function square root of x is one divided by two square root of x. The question, I remind you—or you can just read it over there, if you don’t want to be reminded of it—is to find the instantaneous rate of change when x equals four. Or, if you think about it in terms of the little bear travelling, how fast is the bear travelling at the fourth second?
Now, let’s pick that up. That’s just now evaluating the derivative, which gives us the velocity or instantaneous rate of change—f prime at four. So, I now plug in four into this formula. I see one over two square root of four. And the square root of four I know is two; and two times two is four. So, this is one-fourth. Oh, I guess I never specifi—oh, I did specify the feet. I think it’s feet—let’s say feet per second. So, it’s going one-fourth feet per second. That’s how fast it was going at the fourth second. Okay?
Let me ask you—actually ask you two little follow-up questions on this before we leave it—before we leave this entire topic, in fact, for a while. The first question is—well, this was the first question. I must note here that the derivative—let me just write the derivative here—is still equal to one over two square root of x. And we just saw that the derivative evaluated at the fourth second, or x equals four is equal to one-half.
Let me now ask the—I’m going to—I’m sorry, there was a typo. We saw it actually one fourth. It’s one-half times one-half. Look how—look—watch how I’m going to say this, folks. I’m going to say it right here live. This equals a fourth.
Okay. Now, how—um—here’s the next question I wanted to ask you now. When was the bear travelling at a rate of two feet per second? The bear is traveling, and I want you to watch and tell me when it is traveling at a certain speed. When is it traveling two feet per second? Slowing down. Slowing down.
When was it traveling two feet per second? How would you find that? Well, this gives me velocity. This gives me rate. I want to know when is that equal to two. So, I would set this equal to two. You might think that I should plug in two, but if you think about it, plugging in two is not the way to go here, because plugging in two would be saying, “I want to know what’s happening at the second second.” What I’m interested in, actually, is when is it going at a particular speed? So, what I’m going to do here is take two divided by the square root of x , and I’m going to set it equal to two. So, I’m setting this equal—set it equal to two. And that will tell me for what x was the velocity equal to two.
And now, I solve this by multiplying—how would you solve this, by the way? I’d multiply both sides, first, just by two to get rid of that two. And if I do that, I would see one over the square root of x would equal—and then I’d have two times two is four. Then, you might want to multiply—maybe multiply everything through by the square root of x to get it out of the bottom, and then it would be here as a one equals four times square root of x. You might want to divide through by the four, there. And I would see that one-fourth equals the square root of x.
And if one-fourth equals the square root of x, what does x have to equal? You can either think about it, or just square both sides, and you’d see that tells us that x must be one over sixteen. So one-sixteenth of a second is the answer to the question. At one-sixteenth of a second, the bear was traveling at a rate of two feet per second.
Okay. And now, my last little question to you is when is the bear traveling at a rate of a half a foot per second? It’s going very, very slowly—half a foot per second. Well, I would do the exact same thing. What I would do is I would take the derivative and set it equal no longer to two, but set it equal to a half. So, if I take the derivative and set it equal to a half, that would tell me when the bear is traveling at a rate of one-half feet per second.
How would you solve that? You could multiply everything through by two. Notice then, they cancel. And then you could multiply everything through by the square root of x. I think you would see very quickly that we have the square root of x has to equal one. If the square root of x has to equal one that means that x equals one.
So, let me summarize all the stuff here. There’s a lot of equations here. In fact, no words—sort of interesting—no words, just equations here. So, let’s try to capture the spirit of what’s going on here. What we’ve just discovered is the following. At—uh—at—when x equals one—so—so, act—let’s say. Let’s start here. So, at the first sixteenth of a second, the bear is traveling at a rate of two feet per second. Then, at the first second—so a little bit later on—the bear is traveling at a rate of a half a feet per second. So, it’s actually slowing down. This is a slower rate than this.
And then, we already saw that after four seconds, the bear is traveling at one-fourth feet per second. So, what’s the bear’s velocity doing? Well, it’s actually decreasing. The bear is slowing down. The bear, in other words, is decelerating. Because at the beginning, the bear is going very, very fast. Then, sort of in the middle, the bear is going a little slower. And then, toward the end, the bear is going much slower. And that was why I was able to show you this and say watch how it starts off fast, but then slows down. See? So, that’s another question. In this example, the bear doesn’t stop, but it slows down.
So, there you can get a sense of what these numbers actually mean, and also, how to take the derivative of the square root of x, which is a much more elaborate function.
Okay, well now, we see all the different ways of taking these derivatives by definition. Notice that each and every one of the examples that we looked at together all have the feature that I used the very definition of the derivative in order to solve them. And now, you know what? Well, it turns out that I really am a huge fan of this definition. I love it. I think it’s really powerful. I think it’s a really beautiful idea. In fact, I think it’s one of the greatest ideas of human kind, really. I think it should be in the top group of them.
But, you can only take the derivative by definition for so long. And even myself, as a huge fan of this, I have to admit, after a while, I get tired of going through all the algebraic steps and then taking that limit and "eeeee" and see indeterminate form. If I see an indeterminate form one more time, I may just explode right on camera.
So, one has to wonder is there an easy way of taking the derivative that doesn’t require us to actually us the definition of the derivative? Maybe there’s some easy techniques that will give us the derivative without actually going through all the toil of taking this limit. The answer, happily, is yes. And I’m very delighted to report, we’re finally in a position to actually take a look at that. That’s what coming up next, and I’ll see you in a bit. Thanks.

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