 Hi, everyone. This lesson is on the power role for finding derivatives. So if you think back to what you've learned so far about finding derivatives, you've learned two limit definitions that we refer to as the first and second limit definition of a derivative. What we have here is the first of some shortcuts or alternatives to finding derivatives, alternatives to using those limit definitions, and this is the first of them. So what the power role states is that if f is a function such that f of x is equal to x to the n, where n is an element of the rational numbers, remember q is our mathematical symbol for the rational numbers, and remember what a rational number is, is any number that can be expressed as a fraction. Then f prime of x, the derivative, is equal to n, which was the exponent, times x now raised to the n minus 1 power. So let me give you a really quick little example. So if we had a Quintic equation, 2x to the fifth minus 6x squared plus 6x minus 12, the derivative you would have y prime is equal to, so you do 2 times 5, you do the coefficient times the exponent, that's 10, x, and you subtract 1 away from the exponent, so now it's 4 minus, now you do 6 times 2, that's 12, x, take 1 away from that exponent, so that's just to the first. If you like to write in the one exponent, please do feel free, you might find it helpful for a little while. Plus, now remember the exponent on this linear term, the x is simply 1, so you have 6 times 1, so that's simply 6. When you take 1 away from the exponent, of course, you have now x to the 0, which we know is 1, so it's really just 6, okay? And then the constant term minus 12, that really just goes away, and we're going to talk about y in a second. So what we have here is a really quick way now to get the derivative. Now think about how you would have had to do this before. This was just an equation that you were asked to find the derivative of, so we probably would have had to use that first limit definition of the derivative, the one with the h in it, and remember how that numerator goes, f of x plus h, so imagine having to figure out, you know, if you substitute x plus h in there and have to do that to the fifth power, you go ahead and do that if you really feel like it. But that's why these rules, and as I said, this power rule is just the first of a few that you're going to learn shortly. They really are shortcuts if you want to think of them that way. So let's talk a little bit more about why this 12, when we took the derivative of that, that constant term, why did that come out as 0? Now there's two different ways you can think about this, and we're going to use that 12 as our example. So that constant term was a negative 12, and a constant term we know has obviously no x attached to it. You could think of it as x to the zero power, and x to the zero power we know is 1, so really you just have minus 12. Well, apply the power rule to that. If you do negative 12 times zero, obviously that's zero. You have x to the negative one, which would be zero over x to the first. Well, zero over that's just going to be zero, assuming of course x is not zero itself. So that's really one reason why when you take the derivative of a constant term, it just comes out zero. Now let's look at it from a geometric point of view. So let me draw a set of axes, and I'm going to think of that negative 12 as an equation, y equals negative 12. So these are my axes. So let me go ahead and graph that if I think of it as y equals negative 12. Well, we know the graph of y equals negative 12 is simply a horizontal line down here at negative 12. So let me put it right about here. Well, it's a horizontal line, right? What is the slope of any horizontal line? Zero. All right, so really what you have here are two different interpretations to explain why the derivative of a constant term is zero. You can take it from an algebraic point of view, which utilizing the power rule, you can also take it from a geometric point of view in terms of thinking of the derivative as the slope of a line tangent to a curve. Well, if this is a horizontal line, the slope anywhere is going to be zero. All right, so you really do have two different explanations, if you will, as to why the derivative of any constant term is always going to be zero. So let's take a look at some more practice problems, and we'll just practice using the power rule. All right, so the first one is g of x equals cube root of x. Now one thing you might want to do, think of this as written with an exponent, an exponential form. Because if you think about how the power rule goes, you're multiplying that exponent times the coefficient. So if you can think of things in terms of having an exponent, it's much easier. So the coefficient is obviously just understood to be one, so when you find your derivative, it's simply going to be one-third x, and now you have to subtract one from the exponent, so that now gives us an exponent of two-thirds. Now let's try the next one. So we have four-fifths times two is eight-fifths, t, subtract one from your exponent, and your exponent is just one. Again, if you find it helpful to write the one exponent for a little while, while you're still learning this and it's still new to you, feel free to do that. Now the next one, we'll rewrite that also in exponential form. So that would be two x to the one-half. So our derivative, we have two times a half, which is just one. Again, if you find it helpful to write the one there, please do feel free. You have x, take away one from the exponent, and that gives you a negative one-half. You're free to leave it like that, or if you find it preferable, you can rewrite it in terms of a positive exponent. Either one is fine. So let's try a few more. So we're going to rewrite this one also. So that would give us one over two. This would be x to the two-thirds. All right, but think about how the power rule goes. It needs to look like a coefficient times x to the n. All right, so we might have to rewrite this a little bit better. So let's rewrite it as the coefficient one-half, but then x to the negative two-thirds. Because your expressions need to look a certain way in order to apply the power rule to them. You need coefficient x exponent. All right, so that's really what you need to make sure everything looks like in order to correctly apply the power rule. So this derivative, we have one-half times two-thirds. That's going to be a negative one-third x. Now we need to subtract one away from our exponent. So that's minus three-thirds. So our new exponent is a minus five-thirds. Again, it's perfectly acceptable to just leave your answer like that. Now in the next one, we might want to rewrite the first term. We could rewrite that as a coefficient of one-half and x to the fourth. Now as you get better at doing these, you very well might find you don't feel it necessary to rewrite some of these. You can just do that in your head and that's perfectly fine. But starting out, it might help you visually to just rewrite them. So this derivative, we have negative one-half times four. So that's negative two. x take away one from the exponents. So that's cubed. Plus, now we have three times three is nine x and that becomes squared. And then the derivative of the last term simply is two. Let's think about this for a second really quick. Think of it geometrically. If I asked you to graph the equation y equals negative 2x, we all know that's a line through the origin with a slope of negative two. Hence the derivative is negative two because the derivative is slope of the line tangent to the curve. And if we had an equation y equals negative 2x, the slope is always negative two. So we can also think of something like that from a very graphical point of view. Let's look at a couple of other examples. Now this might be a problem in which you would have used the second limit definition of the derivative because we have an equation x to the fourth minus two and we are asked to find the slope, the derivative, at specified x values. So this would have been a good candidate for the second limit definition of the derivative. But let's try using the power rule. We will want to go ahead and find your derivative first. So let's do that. And this is a pretty easy one. We would just have f prime would be 4x cubed. And that's really it because remember the constant two will have a derivative of zero. So all we're really doing now is evaluating this derivative at all these different x's. That's it. So the first one would be f prime of negative one that we want. So that gives us a negative four when we substitute negative one in place of our x. The next one is f prime of zero. So that's just going to be zero. Now think about for a second what that's going to imply. And we're going to look at the graph in a minute. If that has a derivative, a slope of zero, that means it's a horizontal tangent line. All right, so we'll see. Check that out in a couple minutes on the graph. And then the last one would be f prime of two. So we have to cube the two. So that's eight times four is 32. So think about what that's going to mean. That's a pretty steep slope. So that's telling us that x equals two on the graph of this original function because we are talking the graph of the original function x to the fourth minus two. That's going to be pretty steep there. So let's go check that out on the graph and see if it actually looks that way. So I have entered in y one x to the fourth minus two and I admit I already played with the graph a little bit to get a nice one for us. So these you can really check it very easily by using your calculators built-in functions. If you do second trace and that number six dy dx, let's do the first one negative one. Remember we got negative four for our derivative and there it is. Now notice again visually what's happening there at x equals negative one. Notice how the curve is decreasing. Remember you always look at curves from the left to the right and at this point at x equals negative one. That curve is decreasing right there. Hence a negative slope. So let's try x equals zero. Remember that's the one where we got a derivative of zero and there it is. And lo and behold it is a minimum point on this graph where we have a horizontal tangent line. So let's try that x equals two. Remember that was the one that had a really steep slope to it. And there it is and notice how steep it is there. It's definitely increasing. Again looking at the graph from the left to the right. It is increasing and it is pretty steep there. Now you'll notice you got a decimal 32.mini zero's eight. That's because when your calculator does its derivatives it's not quite using the power rule. It's using some other numerical approximations of the derivative. So you can almost think of it that your answer is more accurate. So that's one thing to keep in mind. So that's why the calculator's looking a little different than what we got. So let's go back and look at one final type of problem. So now we want to find an equation of the tangent line to the function at the given point. We looked at problems like this before when we did the limit definitions. But remember we'd have to go through the second limit definition in order to obtain the slope, the derivative, at that specified point. And then we can use point slope form of a line to actually write the equation. So now it's just a little bit faster because we have the power rule to use in order to get that slope. So if we apply the power rule to this we would have y prime equals, that would just be 2x minus three for our derivative. Now our slope is going to come from doing y prime at one. So if we substitute one in for our x we end up with a negative one. So remember that gives us our slope. So now we can substitute into point slope form of a line. And remember how that goes. It's y minus y one equals our slope m times the quantity x minus x one. So here we would have y minus zero equals negative one times the quantity x minus one. Y minus zero of course is y. You could easily distribute that negative one if you wish. So we have negative x plus one and that is the equation of the tangent to the curve at that specified point. So let's try one more of that type. Feel free if you'd like to pause this and try it on your own first. We'll begin by finding our derivative because that's how we'll get our slope. So in this case our derivative is 3x squared minus 4x substitute in the x value of two. So we'd have three times four minus four times two. So that gives us a slope of four. So that's our slope that we want to use. Once again we can substitute into point slope form of a line. So upon substituting we would have y minus one equals four times the quantity x minus two. You can distribute that out and put it in slope intercept form if you wish. You get four x minus seven if you do that. But remember according to the way in which the AP exam is graded it is perfectly acceptable to leave it in point slope form and stop there. So the option is yours.