 Welcome back to our lecture series, Math 1210, Calculus 1 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Missildine. Lecture 19 in our series starts Chapter 3, which is going to focus specifically on derivatives. Now, surely by now you've realized, if you've been following along with this series, that calculating derivatives can be a very cumbersome process, at least if we try to compute derivatives using the definition that is to say by using limits of difference quotients. Now, fortunately in this chapter, you'll be telling me some techniques or so-called shortcuts, as people like to call them, for calculating the derivative of differentable functions, amongst other things, of course. So it's our goal, particularly in this lecture, to provide techniques that help us compute the derivatives of polynomial functions and exponential functions. In that vein, this video will introduce us to the so-called power rule. So imagine we have a function which is a monomial function, that is, it's x to some power n, where n is a positive integer. In that case, we claim that the derivative of f of x, that is the derivative of the monomial x to the n, is going to equal n times x to the n minus 1. To try to remember this, consider the following idea. If we take x to the n and we take its derivative, what the power rule is going to do is you take the power and you bring it down as the coefficient, so you get n times x, and then you have to reduce the power by 1. In order to prove the power rule, we're going to have to use the definition of the derivative, which, although it's difficult to compute with, it's very important because of the interpretations of the derivative. It's because of the definition of the derivative as a limit of difference quotients, that we can interpret the derivative as the slope of a tangent line, or as an instantaneous rate of change. Remember that f prime of x is equal to the limit as h approaches 0 of f of x plus h minus f of x all over h. In this context, since f of x is equal to x to the n, that means f of x plus h is going to become x plus h to the nth power, and likewise f of x becomes x to the n. What we have to do to proceed forward is we have to deal with this x plus h to the nth power, which because it's a power x to the n here, it's a positive power, this will look like x plus h times x plus h times x plus h all the way down until we get an x plus h, and there's going to be n of these total inside the product. Now one would normally have to multiply this all the way out, and what are the possibilities one gets when you do that? You're going to get x times x times x all the way down to x. If you take all the x's, there's going to be one way, one product of that form, it's going to be x to the n. Now on the other hand, what if you take one of the h's, you take h times x times x times x all the way down to x, you're going to get an x to the n minus 1 times an h. That's if you take the first h, but what if you take the second h? You can still take x times x times x all the way down to x, that's another x to the n minus 1 times h. You could also take the third h, you get x times x times h times x times x times x, you're going to get another x to the n minus 1 times h. If you figure out how many possible products you could have that form, well, you could take the first h, you could take the second h, you could take the third h, you could take the fourth h, all the way down to the last one. In the end, if you take all these possibilities of x to the n minus 1 times h, you're going to get n of those inside of the product, so we see that right here in the expanded form. Now, by virtue of the binomial theorem, we could keep on going with this. If you want to count how many ways can you get x to the n minus 2 times h squared, that's going to come down to n choose 2, where n choose 2 in this context means n times n minus 1 all over 2. If you want to know how many x to the n minus 3s times x cubes you could get, well, that's going to be n choose 3, where n choose 3 has the form n times n minus 1 times n minus 2 all over 6. We don't really need to worry about the details of all of those things. It's only going to be the first two that we really care about. I mostly care about these ones in the expansion. All that's going to matter for the other ones is that they're divisible by h when you multiply these things out. Why do we care about the x to the n so much? Well, that's because there's another x to the n that's going to cancel it out, so they're going to get each other. And so now the first term in our sequence is going to be n x to the n minus 1 h. You'll notice this resemblance to the formula right here. Well, since you cancel out the x to the n, everyone else in the numerator is divisible by h. Notice you have an h. You have an h. You have an h. You have an h. And so we could factor each and every one of those h's out of the expression. So take the limit as h goes to 0. Factor out the h. You'll be left behind at n x to the n minus 1 plus this n n minus 1 over 2 x to the n minus 2 times an h. And then all these other middle terms, you'll be left with an h to the n minus 1. That's in the numerator. There's all sets above h, for which now we found a common factor of h in the top and in the bottom. So they're going to cancel out, giving us what's left right here. And so we have our first term in x to the n minus 1. Again, well, that looks similar. Everyone else is going to be a multiple of h. So when h goes to 0, we see this is going to be 0. This is going to be 0. This is going to be 0. Everyone's going to be 0 except for n times x to the n minus 1. And so this proves the power rule in the context of a monomial function. Fortunately, we can use the power rule when n is any real number. That is, if x to the n is an arbitrary power function, x is our variable, which is the base, and n is any real number as the exponent, then the derivative of x to the n with respect to x will equal n times x to the n minus 1. We'll see the proof of the general power rule later in our series after we've developed a technique called logarithmic differentiation. So what I want to proceed to do is to show you how we can use the power rule to calculate the derivative of various power functions. Consider the function y equals x to the 6th. Well, by the power rule, if we take the derivative dy dx of x to the 6th right here, we end up with 6. That's the exponent, right? So see the exponent comes out as the coefficient 6 times x to the 6 minus 1. Subtracting the powers, we get 6 times x to the 5th, which is the derivative of x to the 6th. 6 times x to the 6th is the correct derivative. And be aware, the computation we did here is dramatically simpler than the calculations we did previously using the definition of the derivative. It's important that we understand the definition, but there are certain patterns we see occurring over and over again that we can simplify the calculation by utilizing those patterns we see here. We can also take the derivative of y equals t. The name of the variable doesn't really matter too much. We want to calculate dy dt, the derivative of y with respect to t. To do that, we have to recognize that t itself is a power function. If you just have the single variable t, that can be written as t to the first. And so by the power rule, t to the first, its derivative will become 1 times t to the 1 minus 1 power, or 1 times t to the 0. And raising a variable to the 0 power is going to give you just 1 itself, 1 times 1. The derivative of t is equal to 1. Now, by fortune, if you were to write this expression dt over dt, right? Because after all y equals t, we could have written it that way too. If an algebra student were to walk in the room right now and not have any knowledge about calculus, they'd be like, dt over dt, it cancels, it's equal to 1. So it turns out that algebra student would be right for the wrong reasons. Well, sort of. I mean, after all, derivative is the limit of a difference quotient. Let me emphasize a quotient. It's a ratio. It's a fraction. If you have the same thing on top and bottom, it's going to cancel out to be 1. Derivatives inherit that property through the limit process because they themselves are fractions. So the algebra student who doesn't know any calculus is actually right. dt divided by dt is going to equal 1. How about this next one? y equals 1 itself. It's the constant function 1. Well, like we just saw, 1 is just x to the 0 power. So if we take the derivative of y with respect to x, where y is x to the 0 right here, by the power rule, we get 0 times x to the 0 minus 1. And who cares what happens to x to the 0 minus 1 power? If you times something by 0, you end up with 0, which is the derivative of this constant function 1. Now, in other situations like this example, y equals 1 over x cube, we have to sometimes be a little bit more creative. Can we write the function somehow as a power function? Well, having an x cubed in the denominator means we could use a negative exponent, y equals x to the negative 3 to write it as a power function. Therefore, the derivative of y is going to be the derivative with respect to x of x to the negative 3, which by the power rule we get negative 3 times x to the negative 3 minus 1. Subtract 1 from the exponent. This is an important thing. If I take the exponent of negative 3 and I subtract 1 from it, I'm going to get the exponent of negative 4, which it's always good practice to rewrite your function in a style resemblance of the original one. There is no negative exponent in the original expression. So we'll write the derivative as negative 3 over x to the fourth, like so. The next one, this one, we have y equals x to the fourth third. Clearly, it's a power function, so we can jump straight to calculate its derivative by the power rule, for which we're going to get four thirds times x to the fourth thirds minus 1. You see this when you have a fraction exponent? You have to subtract 1 from it. So for subtracting 1, we have to find a common denominator, rewrite this as three thirds. And so four thirds take away three thirds will be a one third. So we're going to get four thirds times x to the one third power. Since the original one had these fractional exponents, I'm going to leave it that way. You could write it as four thirds times the cube root of x, but again, I want to resemble the original style of the problem. And then with this one, y equals the square root of z. Well, again to view it as a power function, we should write the square root of z as z to the one half power. Hence the derivative of y with respect to z, that is, we have to take the derivative of z to the one half power. By the power rule, we get one half times z to the one half minus one power, which if you write that as the two halves power, this becomes one half times z to the negative one half power, which is an acceptable way of writing the derivative. But if we want to resemble the original one, we didn't have any exponents, we had a square root. So we could write this as one over two times the square root of z. That would be an acceptable way of doing it. Now let's do something a little bit different, but same basic idea. Let's find the equation of the tangent line of the curve y equals x times the square root of x at the point one comma one. So we have to find an equation for the tangent line, and remember that that tangent line is going to look like y minus f of a is equal to f prime at a times x minus a. That is, the slope of the tangent line is the derivative. And so if our function f of x is given as x times the square root of x, we didn't find its derivative in order to find this tangent line. But we don't want to do the definition of the derivative, can we shortcut it somehow? Well, only if we can recognize this function as a power function. We'll notice that the square root of x, like we saw previously, I could write that as x to the one half power, and x itself is x to the first, and adding the exponents by the usual exponent laws, we get x to the three halves power, which means that the derivative of x is going to equal three halves x to the one half power. Or, well, we don't really need to rewrite it because what we actually care about is f prime at one, which is our a value, right? Our point here, this is our a value, this is our f of a, or in this case, f of one. So we just need to figure out what f prime of one is, which when we plug one in here, we're going to get three halves times one to the square, to the one half power, which is the square root of square root of one is one. So you end up with three halves. So putting this together, our tangent line equation will look like y minus one, the y coordinate of the point of tangency, then the slope of the tangent line is the derivative, which was three halves, then we get x minus one, where one was the x coordinate of point of tangency, distribute the three halves through, we end up with three halves x minus three halves, then we need to add one to both sides to put in slope intercept form. And upon doing so, if we do that, we're going to get y equals three halves x minus one half.