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You may have come into this class thinking of derivatives, in terms of

Â numbers, or slopes. But we've broadened our perspective.

Â By thinking in more abstract and conceptual terms.

Â We're going to culminate that process in this lesson by thinking of

Â differentiation as an operator. This more global perspective is going to

Â lead us not only to better comprehension, but better computational abilities.

Â Some derivatives do not follow from the basic rules that we've covered thus far.

Â For example, what's the derivative of 2 to the x?

Â If you don't recall, you may be tempted to write x times 2 to the x minus 1.

Â That would be a mistake. That is not the derivative.

Â What about some other functions? Arcsin of x.

Â Do you recall the derivative of that? What about x to the x, or x to the x to

Â the x, or more complicated functions still?

Â These can be differentiated. But we cannot use the basic rules that

Â we've learned thus far. What we will use, is the idea that

Â differentiation is an operator and that it acts on equations as well as

Â functions. And more importantly it works well with

Â other operators, such as exponentiation or integration, taking a limit or in

Â particular, the differentiation operator works well with the natural algorithm

Â operator. And that is the basis for that method

Â known as logarithmic differentiation. Let's begin with a simple example.

Â Compute the derivative of e to the x. Now, I know that you know what that

Â derivative is, but let us compute this via logarithmic differentiation.

Â The key step will be to write either the x as y, and then to simplify this

Â equation, we'll apply the natural logarithm to both sides.

Â On the left, we obtain log of y. On the right, log of e to the x, that is

Â Nx, and now we can differentiate this equation.

Â On the left we'll obtain dy over y, on the right we'll obtain dx.

Â And by manipulating these differentials, we can easily compute dy dx to be y.

Â That is e to the x. Now, this process is simple enough.

Â Let's step back and think of what you are doing.

Â What we're really doing is taking our initial equation and applying the natural

Â logarithm operator, and feeding the results of that into the differentiation

Â operator. That is what allows us to compute this

Â derivative. Now you can compute other things

Â similarly. For example, you could show that the

Â derivative of a to the x, where a is a constant, is a to the x times log of a.

Â That's a fun exercise. Odd, there's more to logarithmic

Â differentiation than just this. In fact applying operators gives lots of

Â examples of things that are not quite logarithmic differentiation, but follow a

Â similar pattern. For example, let's look at the derivative

Â of log of x. Now, we both know the answer to that.

Â But let's apply a similar methodology. We'll let y be equal to log of x.

Â And now, how do we simplify this? Well, we can apply the exponentiation

Â operator. And then apply the differentiation

Â operator. So, exponentiating this equation gives e

Â to the y equals e to the log of x. That is x.

Â Now, assuming that we know how to differentiate exponentials, we can apply

Â the differentiation operator. And on the left we'll obtain E to the Y,

Â DY, on the right DX. Manipulating differentials to solve gives

Â 1 over X as the dirivative. And so we see, that if you know the

Â derivative of log, you can compute the derivative of the exponential.

Â If you know the derivative of the exponential, you could likewise compute

Â the derivative of the log. But that's not all we can do.

Â If asked to compute the derivative of arcsin, how would we proceed?

Â What would the appropriate operator be? It's pretty clear that sin is what we

Â want to use to feed into the derivative. For example, if we let y be equal to

Â arcsin of x, and apply the sin operator to both sides.

Â What do we get? Well, on the left sin of y, on the right

Â because arcsin is the inverse of sin, we obtain simply x.

Â And now because we know the derivative of sin this is going to be pretty easy.

Â We got cosin of y dy equals dx. And manipulating to obtain the dy dx.

Â We get 1 over the cosin of y. And that is the secant of y.

Â Now, that is not exactly the form of the answer.

Â That we want. So we need to do a little bit of

Â trigometric manipulaation. Given the right triangle with angle Y.

Â Since the sin of Y is equal to X, we could assume that the opposite side

Â length is X, and the hypotenuse is one. Meaning that the adjacent side length is

Â root 1 minus x squared. This allows us to compute the secant as 1

Â over the square root of 1 minus x squared, and that indeed is the

Â derivative of ARCSIN. You can derive many other similar results

Â using this same principal. You can compute the derivative of our

Â arch cosine, of arch cotangent. You can even compute the derivatives of

Â the inverses to the hyperbolic trig functions.

Â Arcsinh, arcosh, and arctanh, and it's not only derivatives that one can

Â computer using these methods. Operators are very versatile and helpful

Â across mathematics. Let's do an example involving a limit.

Â Compute the limit as x goes to infinity of quantity 1 plus A over x to the x

Â power. Here, a is a constant, and note the

Â difficulty implicit here. X is going to infinity, so the term

Â within the parenthesis is going to 1, but you're raising this to higher and higher

Â powers. You might be tempted to say, but this

Â limit is 1 to the infinity, which seems like it ought to be 1, but that is not

Â applicable let's see what that limit is. We are going to, as before let y be

Â defined as quantity 1 plus a over x to the x power.

Â And before applying the limit operator. We're going to apply something else in

Â this case, the natural logarithm, to pull down that exponent of x.

Â On the left we obtain the log of y. On the right, using what we know about

Â logarithms, we obtain x times the log of one plus a over x.

Â Now we have something that we can work with.

Â We're not differentiating. We are taking a limit.

Â On the left is, close to something we want to get, the limit of log of y.

Â On the right, we hit the limit as x goes to infinity of x times log, 1 plus a over

Â x. Now, how is this helping us at all?

Â well, the logarithm is in the form, log of one plus something that becomes very

Â small, as X goes to infinity, so we can use our Taylor expansion.

Â And rewrite that as a over x plus something in big o of one over x squared.

Â That limit is going to be easy to compute.

Â The leading order term is in fact a, and that is the limit not of y but of log of

Â y. And so exponentiating this equation, and

Â exchanging with the limit gives the limit, x goes to infinity of e to the log

Â of y. That is y is e to the a and that is

Â indeed what the limit of 1 plus a over x to the x.

Â Is, let's step back for a moment. Look at what we've done.

Â We had a challenging limit to compute. A limit that we could not figure out

Â directly. And so we took an indirect route, by

Â applying the natural logarithm operator. Then take the element then applying the

Â exponentiation operator. The wonderful thing is that we can

Â exchange these with limits, in this case, and take the long way around to get to

Â where we want to go. Let's look at another example.

Â This one seems intimidating. What is the derivative of x to the x?

Â We're going to follow the same path as before.

Â Let y be equal to x to the x. How do we simplify that exponent?

Â Clearly, the natural logarithm is what we're going to want to use.

Â When we apply the natural log, we obtain on the left, log of y.

Â And on the right, x times log of x. Now we can differentiate.

Â On the left we'll obtain dy, over y. On the right, using the product rule, we

Â get x over x, plus log of x, dx. That simplifies a little bit doing some

Â algebra gives us that dy, dx equals y times quantity 1, plus log of x.

Â