Welcome to Calculus. I'm Professor Greist. We're about to begin, lecture 16, on the differentiation operator. You may have come into this class thinking of derivatives, in terms of numbers, or slopes. But we've broadened our perspective. By thinking in more abstract and conceptual terms. We're going to culminate that process in this lesson by thinking of differentiation as an operator. This more global perspective is going to lead us not only to better comprehension, but better computational abilities. Some derivatives do not follow from the basic rules that we've covered thus far. For example, what's the derivative of 2 to the x? If you don't recall, you may be tempted to write x times 2 to the x minus 1. That would be a mistake. That is not the derivative. What about some other functions? Arcsin of x. Do you recall the derivative of that? What about x to the x, or x to the x to the x, or more complicated functions still? These can be differentiated. But we cannot use the basic rules that we've learned thus far. What we will use, is the idea that differentiation is an operator and that it acts on equations as well as functions. And more importantly it works well with other operators, such as exponentiation or integration, taking a limit or in particular, the differentiation operator works well with the natural algorithm operator. And that is the basis for that method known as logarithmic differentiation. Let's begin with a simple example. Compute the derivative of e to the x. Now, I know that you know what that derivative is, but let us compute this via logarithmic differentiation. The key step will be to write either the x as y, and then to simplify this equation, we'll apply the natural logarithm to both sides. On the left, we obtain log of y. On the right, log of e to the x, that is Nx, and now we can differentiate this equation. On the left we'll obtain dy over y, on the right we'll obtain dx. And by manipulating these differentials, we can easily compute dy dx to be y. That is e to the x. Now, this process is simple enough. Let's step back and think of what you are doing. What we're really doing is taking our initial equation and applying the natural logarithm operator, and feeding the results of that into the differentiation operator. That is what allows us to compute this derivative. Now you can compute other things similarly. For example, you could show that the derivative of a to the x, where a is a constant, is a to the x times log of a. That's a fun exercise. Odd, there's more to logarithmic differentiation than just this. In fact applying operators gives lots of examples of things that are not quite logarithmic differentiation, but follow a similar pattern. For example, let's look at the derivative of log of x. Now, we both know the answer to that. But let's apply a similar methodology. We'll let y be equal to log of x. And now, how do we simplify this? Well, we can apply the exponentiation operator. And then apply the differentiation operator. So, exponentiating this equation gives e to the y equals e to the log of x. That is x. Now, assuming that we know how to differentiate exponentials, we can apply the differentiation operator. And on the left we'll obtain E to the Y, DY, on the right DX. Manipulating differentials to solve gives 1 over X as the dirivative. And so we see, that if you know the derivative of log, you can compute the derivative of the exponential. If you know the derivative of the exponential, you could likewise compute the derivative of the log. But that's not all we can do. If asked to compute the derivative of arcsin, how would we proceed? What would the appropriate operator be? It's pretty clear that sin is what we want to use to feed into the derivative. For example, if we let y be equal to arcsin of x, and apply the sin operator to both sides. What do we get? Well, on the left sin of y, on the right because arcsin is the inverse of sin, we obtain simply x. And now because we know the derivative of sin this is going to be pretty easy. We got cosin of y dy equals dx. And manipulating to obtain the dy dx. We get 1 over the cosin of y. And that is the secant of y. Now, that is not exactly the form of the answer. That we want. So we need to do a little bit of trigometric manipulaation. Given the right triangle with angle Y. Since the sin of Y is equal to X, we could assume that the opposite side length is X, and the hypotenuse is one. Meaning that the adjacent side length is root 1 minus x squared. This allows us to compute the secant as 1 over the square root of 1 minus x squared, and that indeed is the derivative of ARCSIN. You can derive many other similar results using this same principal. You can compute the derivative of our arch cosine, of arch cotangent. You can even compute the derivatives of the inverses to the hyperbolic trig functions. Arcsinh, arcosh, and arctanh, and it's not only derivatives that one can computer using these methods. Operators are very versatile and helpful across mathematics. Let's do an example involving a limit. Compute the limit as x goes to infinity of quantity 1 plus A over x to the x power. Here, a is a constant, and note the difficulty implicit here. X is going to infinity, so the term within the parenthesis is going to 1, but you're raising this to higher and higher powers. You might be tempted to say, but this limit is 1 to the infinity, which seems like it ought to be 1, but that is not applicable let's see what that limit is. We are going to, as before let y be defined as quantity 1 plus a over x to the x power. And before applying the limit operator. We're going to apply something else in this case, the natural logarithm, to pull down that exponent of x. On the left we obtain the log of y. On the right, using what we know about logarithms, we obtain x times the log of one plus a over x. Now we have something that we can work with. We're not differentiating. We are taking a limit. On the left is, close to something we want to get, the limit of log of y. On the right, we hit the limit as x goes to infinity of x times log, 1 plus a over x. Now, how is this helping us at all? well, the logarithm is in the form, log of one plus something that becomes very small, as X goes to infinity, so we can use our Taylor expansion. And rewrite that as a over x plus something in big o of one over x squared. That limit is going to be easy to compute. The leading order term is in fact a, and that is the limit not of y but of log of y. And so exponentiating this equation, and exchanging with the limit gives the limit, x goes to infinity of e to the log of y. That is y is e to the a and that is indeed what the limit of 1 plus a over x to the x. Is, let's step back for a moment. Look at what we've done. We had a challenging limit to compute. A limit that we could not figure out directly. And so we took an indirect route, by applying the natural logarithm operator. Then take the element then applying the exponentiation operator. The wonderful thing is that we can exchange these with limits, in this case, and take the long way around to get to where we want to go. Let's look at another example. This one seems intimidating. What is the derivative of x to the x? We're going to follow the same path as before. Let y be equal to x to the x. How do we simplify that exponent? Clearly, the natural logarithm is what we're going to want to use. When we apply the natural log, we obtain on the left, log of y. And on the right, x times log of x. Now we can differentiate. On the left we'll obtain dy, over y. On the right, using the product rule, we get x over x, plus log of x, dx. That simplifies a little bit doing some algebra gives us that dy, dx equals y times quantity 1, plus log of x. Replacing that y with x to the x shows us. But the derivative of x to the x, is x to the x times quantity 1 plus log of x. And that was easy enough to compute, but let's think. This function, x to the x, is super exponential. It's bigger than the exponential function. And when we compute it's derivative, it's even larger than x to the x. It is of order that big o, x to the x times log of x. Now if we think about what happens when we differentiate functions and look at their growth rates. If I take something that is super exponential in its growth rate and differentiate it, I'm losing control over the growth rate. The growth rate is potentially much larger, as we saw in the example. Of x to the x, going to something that was in big O of x to the x times log of x. The moral of this lesson is that no matter how difficult the differentiation problem looks, taking advantage of operators is going to help you, with a tricky computation. And so ends Chapter two on differentiation. We've come a long way from thinking about a derivative as a slope. Indeed, we've considered derivatives from the point of view of Taylor Series and asymptotics. We've considered differentials, and ultimately, differentiation as an operator. We'll begin our next chapter, Chapter three, by inverting that operator, and looking at integration.