Calculus: Uses of the Power Rule

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Computational Techniques
The Power Rule
Using The Power Rule Page [1 of 4]
Well, now we have this really easy way of finding derivatives. We spent so much time working through this really, really hard, difficult way of using limits. In fact, maybe, you’re even sort of mad at me, mad at your prof, mad at your book, in fact, mad at everybody—mad at life. Because we’ve worked through all these problems using that long limit definition, when all along we knew—and I have to admit; I have to ‘fess up—I did know the easy way when I was telling you about the hard way, knew everything. Well, why did they do that? Why are they so mean? Why is Berger so mean? Well, the answer is—first of all, it’s fun to be mean. You should try it some time. It’s really—no—that’s a joke. Don’t be mean to anyone. But the truth is to really appreciate what the derivative means, I think it’s important to really work through a few of those and really get a sense of what you’re accomplishing.
You know, I don’t want you to think that just because something you can do fast, or something is easy, it means that it’s sort of worthless. It turns out that there is really a conservation—in some sense—of energy, always. If you put a lot of work into something that turns out, it probably is something important. And the derivative is extremely important. And I think it’s important for you to have an appreciation for the value of that and for the subtlety of it. And by working through a few of those examples, as you have done, I hope, you really get a sense that this is an important idea. It’s an important thought.
Now that you have that sense and you have a sense of what the subject is, and you’ve made it your own, now we can move on to actually looking at the power of that idea by looking at applications. Well, we don’t want to, then, keep taking the derivative by definition. So, I hope you’re not too upset. Take it on as an opportunity to really get a sense of what these objects mean.
Okay. Anyway, so much waxing philosophy. What really has happened—by the way—what we’ve discovered is a really powerful formula for finding derivatives of certain functions. I want to write that formula down for you right now. If I have a function f of x equals x to some number power—call it the n—then the derivative of it is extremely easy to find. I just bring down the exponent in front, multiply it by x raised to the power and minus one.
You know, all the stuff we’ve been doing, it’s sort of like—let me try to give you a visual example and show you what this has really been like. It’s like we’ve been having this thing that we’ve been carrying around—a heavy, heavy thing—a heavy burden, almost. You see, all this stuff, and it’s all in here and it’s messing around. And we were taking the limits and doing all this stuff. And we understand it all now, but it was a lot.
And now, what we’ve just about to do—what we’re just doing—is realizing that we could just dump it all out—dump it all out, and, in fact, get rid of it all—just clean that off. And what we’re left with—which was, at first, was complicated, we now see turns out to actually be quite beautiful—quite simple. Really neat, really neat.
Okay. Let’s take a look at some examples now and see how to use this stuff and get really fast at taking derivatives. Well, the first example that we’ll take a look at is—well, how about this one. F of x equals, well, x squared. Well, that’s pretty easy, because remember the recipe. The recipe is just bring that down out in front. We bring the two out in front, multiply it by x raised to the power two minus one, which is one. Or you could just write that as two x. And that’s the derivative. Look how fast we can do these things.
Let’s try another one. Let’s try f of x equals x to the two thousand power. A lot bigger exponent, but no more work, because the work is the exact same. The answer would be—I bring the two-thousand out in front – doesn’t make a difference if the number’s big or small—multiply that by x raised to the one thousand nine-hundred and ninety-ninth.
Let’s try this one. F of x equals x. Okay, now this one, even though it’s just x, it’s going to be a little tricky. We have to use the formula and be very careful. What would this look like? F prime of x equals — well, there’s an invisible one here. Remember, there’s an invisible one. If you can’t see anything, it’s x to the one power. And if you bring that one down in front, then we just have a one here, and then I have x raised to the power one minus one, which is zero.
Now, remember, that x to the zero power is one. So this is actually equal to one times one, which equals one. So, in fact, the derivative of x is the number one. Okay? Just follow the formula. You go to be a little careful, because we have x to the zero power here. So you just got to be a little bit careful. X to the zero power is one.
And let me try one last example here. It’s funny, with this really easy formula, it turns out that the really simple cases are sometimes that ones that are really, really hard, because they get confusing. How about taking the derivative of the function fifteen. Now, you might be saying, “Wait a minute, there’s no x’s in there. I don’t see any x’s in there.” There aren’t any. This is a constant function.
So, if I say to you, “What is f of three?” wherever you see an x, you plug in three. But there are no x’s, so the answer’s fifteen. What’s f of seven? Fifteen. What’s f of minus twelve? Fifteen. What’s f of sixty-four? Fifteen. It’s the constant function fifteen. This function just always spits out fifteens. It’s an easy machine. You put anything in. Fifteen comes out, always.
What’s the derivative of it? Well, how can I think about that? Well, fifteen is actually fifteen times x to the zero power, if you think about it that way. Use x to the zero—I remind you—is just one. Now, if you bring that zero out in front as we are to do, what do you notice? I have zero times all this stuff. Well, zero times anything is zero. And so, I discover is that the derivative of this is zero.
Moral? Moral number one is that the derivative of a constant function is zero. The derivative of just x alone—the derivative of sort of lines—they’re just going to be numbers, like one. And then, if you’ve got the derivative of x squared, or other things, then you just use this formula directly.
In fact, right now, let me give you an opportunity to try to make a guess. So, here’s a chance for you to make a guess. F of x equals sixteen x cubed. And I want you, right now, to take a guess as to what you think the derivative of this is. So, right now, go off and take a guess. We’ll be right back.
All right. Well, let’s see if you guessed the right answer. Well, how can we think about this? This is something we haven’t seen before, because this is actually something we know about, but then I have to multiply it by sixteen. And it turns out that there is a great fact—and this is a fact that I want to tell you about now—that if you have a function, and you multiply it by a number—let’s say you take a number, any number. It could be like sixteen. And then you multiply that number by a function. And then, you take the derivative of that. Remember, that little prime there means take the derivative. Then, you know what you get? It’s actually pretty easy. You just take that number that you got and you multiply it by the derivative of the function. And that’s the formula.
So, it’s actually a very easy procedure to generate even more complicated functions and take the derivative. All you have to do, if you see a constant number, like sixteen, you basically just hold it out in front like a holding pattern. Just keep it steady, holding it there, holding it there. Don’t touch it. Now, just take the derivative of that, and keep that sixteen right there. That’s the idea.
So, for example, in this problem, what would be the derivative? And maybe you guessed this. By the way, I didn’t tell you this function and this fact before, so if you didn’t guess the right answer, don’t worry about it. It’s an opportunity for you to guess. Maybe you actually guessed on it. If you guessed the right answer, by the way, terrific! That was great!
All right, now what would I do? Well, what I’d do is I’d just keep that sixteen out in front—so, I’m taking the derivative of that—keep the sixteen out in front, and now, put in the derivative of this. So, I just take the derivative of that. Well, the derivative of x cubed—what do you do? You bring down the three and multiply it by x to the three minus one. So, I see three x squared. And so, now we have sixteen times three, which I believe is forty-eight. You know, I am so bad at doing multiplication, by the way.
In fact, you know—I think actually, I think the first time I actually had to multiply in this entire collection of lectures—the first time I had to multiply on the fly. So I would always ask my staff. Is that right? Is sixteen times three, forty-eight? Good. It’s always good to check your answer.
Okay. And so, that is the derivative of this function, and notice that I used this basic property, which is that if you have just a constant number multiplied by some function, take the derivative of that whole thing. All you do is keep the constant out in, like, sort of a holding pattern, take the derivative of the rest of the thing, and then multiply that by the constant.
So, for example, if the function—let me use a different letter just for fun—g of x equals—let’s see—ten x to the ninth, then what would the derivative be? Well, the derivative g prime would be what? You try right now. Now, you do it, and hopefully you’ll get the right answer.
Well, let’s see. I think about this constant ten in front here as sort of being in a holding pattern. Hold on to that, and I bring the nine out in front. And, in fact, the way I do it—I could write the ten here—but the way I do it, I actually visually bring this nine out in front and multiply it by that ten, and I see a ninety right there out of the get-go. X raised to the power, nine minus one, which is eight. So look, I could even do this whole problem in just one step. I don’t even have to write down the ten times the nine x to the eighth. I just bring it down visually and multiply it.
Let’s try one last one. H of x equals, let’s say, minus one-half x to the minus three. It’s a little bit more complicated, but I’m not going to panic. I’m going to use the same procedure as before. What I do here is I take that minus three and bring it out in front. So, that becomes, now, a minus a half times minus three becomes a plus three halves. So, h prime would equal three halves x to what power? The minus two. Because you take minus three and subtract one, and you get—oh, no—minus three minus one isn’t minus two, which is a good false start. But, minus three minus one is actually minus four. All right?
So, if you were stumped by that little tomfoolery there, great. Be careful. When you’re subtracting minus one from minus three you actually get—I’m sorry—if you subtract one from minus three, you get minus four.
By the way, how could you write this without that negative sign there? Let me just remind you, this is going to start to be important for us. I can get that negative sign downstairs and write it as three divided by two x and now it’s the fourth power. These two things are identical. It might look a little different, but let me just show you everything’s okay. I’ve got a three on the top here. There’s the three on the top. I’ve got a two on the bottom. There’s the two on the bottom. And that x to the minus four is the same thing as one over x to the fourth, and there’s the x to the fourth in the denominator. So, this, in fact, is the same as this. So, there’s the derivative.
Okay. Well, now I’d like to do is even build up more elaborate functions that we can now take the derivative off. So let’s ask you—let me pose this as a question. Here’s the warm-up question for you—the warm-up question. F of x equals two x cubed plus six x. And now I want you to find the derivative of that. Try it right now. See if you can do it. Guess an answer.
Okay. Let’s see if you guessed the right answer. It turns out the principle that we’re going to need here is the following fact. If you want to take the derivative of the sum of two functions, what do you do? Well, I bet that you guessed the right answer. If you want to take the derivative of that, you do what seems natural. Well, take the derivative of that, take the derivative of that, and add them up. So, I take the derivative of f, and I’d add to it the derivative of g. And it turns out that is the correct answer.
So, in this question that I posed here as a warm-up—what would the derivative of this be? Well, it’s actually pretty simple. All I’m going to do is I’ll take the derivative of this piece and then add it to the derivative of that piece. I sort of work separately. I work on there, then I work there, and I just put them together. Fairly easy.
Well, what about this one. Well, this has a little coefficient on front. But not a problem, I know how to deal with that. It’s a constant number. I bring the three out in front, so that makes this three times two, which is six x, and I raise that to the three minus one power, which is going to be two. I take the derivative of this, which is just, what? Well, this is now x to the one power. You bring the one out in front. That’s going to make this six times one times x to the zero power, because I take one minus one, x to the zero is just one. I’m just left the six, and I add those two things up, since that’s what I’m supposed to do is add them.
So, let’s take the derivative, sort of, of them individually, and then just link them together. And, in fact, that exact same formula holds if you’re subtracting. All you do is just subtract them. So, for example, let me give you a real scary-looking thing, which really isn’t that big of a deal. Let’s find the derivative of this function. Two x to the ninth minus one half x cubed plus four-thirds square root of x minus one point three seven two.
Okay. So, now, let’s find the derivative of this. By the way, maybe it’s not looking scary to you at all, but let me just point out, moments ago, we were taking derivatives using the definition. Could you imagine taking the derivative of this using the definition—putting in x plus delta x in all these places and doing all that stuff and untangling this and multiplying by conjugates and wooohooo?
Turns out, it would all work, but one big job. You’d better not make any plans for about a month if you untangle it. But now, piece of cake. What do I notice? I notice that I have all these little pieces that are either being added or subtracted together. So, I’m going to go through one at a time, take the derivative of each of those pieces, and just tag them together.
So, the derivative of this—that’s actually pretty easy. I’d bring the nine out in front. Nine times two equals eighteen, x nine minus one is eight. Then, I subtract. I bring the three out in front of this constant the half and I see three halves, x to the power three minus one, which is two. Isn’t it great! Look at this, we’re cruising through this thing. No limits, no dada. It just bloooop.
Okay. Now here, I guess I spoke a little too soon, because here, now, we’ve got to pause a moment because, remember, that doesn’t look like an exponent. What’s the power there? So, you might have to go up an do a little conversion. In fact, let me do this for you. I’m going to do a little conversion right here. This is a little side thing. I’m going to take a look at the square root of x and remind you that that equals x to the one-half power. That’s going to be useful here for what we’re doing.
And, in fact, let me just tell you the general fact that if I have x to, like, the one over—let’s say—a power, that’s the same thing as taking the eighth root of x. So, if I have x to the one-third power, that means the cube root of x, x to the one-fourth power means the fourth root of x, and so on. But, with the square root, that means one-half power.
Okay. So, when you see this, the first thing I suggest you do—and to be truthful, what I do in my mind is I first convert it to this, and then I use the formula. I bring the one-half out in front. Now, I’m going to actually do this here, so visualize in your mind’s eye, that’s x to the one-half sitting there. There it is. I bring that one-half out in front, and it becomes what? Well, I have a one-half times four over three. You can write this down if you want, but I want you to do this in your head with me.
See the four over three and visualize the one-half here. There’s a little cancellation. What am I left with? Two-thirds. And then, what do I have? I have an x to what power? Well, I had the one-half power originally. Now, I subtract one. So, I take a half minus one and I get a half. No, a half minus one gives me minus a half. So, the exponent here is minus a half. And then, I’ve got to subtract off the derivative of one point three seven two. Well that sounds like a sort of weird number. What’s the derivative of that? Well, that’s a constant. There are no x’s there. And remember what we said. The derivative of a constant is zero.
A way of thinking about that is remembering that I could write in the invisible x to the zero power here. So, when I bring that zero out in front, it multiplies everything through by zero. So, the derivative of any constant number is zero. So, that whole number there—zero. If I put in a different number—still zero. The derivative of a constant is zero.
So, that’s the derivative of this really, really long expression. By the way, you might want to write this a little differently. You can write x to the minus one-half how? In fact right now, let me give you an opportunity to try it. How would you express x to the minus one-half? I’ll give you a whole bunch of choices. Pick the one you think is correct.
Okay? Well, maybe you picked the fact that, well, that negative sign means I push underneath. So, it’s one over x to the one-half. And one-half means square root, and so x to the minus one-half power is the same thing as one over the square root of x. So, there’s another way of writing that little piece of it right there. But still, look at that. A long, long thing. We took the derivative almost effortlessly. So, the important facts here is that if you’re taking the derivative of a sum or a difference function, just take the derivative of each piece and string them together the right way.
If you have little constant coefficient in front, put it in a holding pattern, and then take the derivative of what remains.
Okay. Here’s a few examples for you to try. And up next, we’ll take a look at some more techniques, which will allow us to differentiate even more exciting and elaborate functions. See you in a bit.