 There's one more important derivative result that we need to investigate, which is the derivative of a transcendental function. So transcendental functions have nothing to do with Ralph Waldo Emerson. Instead, this refers to an important classification. Functions fall into two broad categories. First, they're algebraic functions, which can be written using powers and roots of a variable. For example, this rather modest looking thing is an algebraic function, because the only thing we have here are powers and roots of a variable. On the other hand, transcendental functions are much more terrifying because they cannot be written using powers and roots. So one of the terrifying transcendental functions is this one, y equals sine of x. And it might not be quite so obvious why this is such a frightening function, but here's one reason why you can't evaluate this function exactly. In particular, if you use the arithmetic functions of addition, subtraction, multiplication, and division, you can evaluate any algebraic function as accurately as you want to. On the other hand, a transcendental function can't be evaluated as precisely as you want to without invoking some calculus. Fortunately, you're learning calculus. Now with a little bit of work, it's possible to prove that the derivative of sine is cosine and the derivative of cosine is minus sine. And once we have these two derivatives, then we have the derivatives of all of the trigonometric functions because all of the trigonometric functions can be defined in terms of sine and cosine. Well actually, that's not quite true. We don't need both of these derivatives. We actually only need this one because even cosine can be defined in terms of sine. And it's important if you want to work with trigonometric functions that you have a thorough understanding of the relationships between the primary trigonometric functions. So now is a good time to brush up on your trigonometry. But we won't do that here because this is a calculus course. So let's see if we can find the derivative of some simple function like sine of x squared. So we'll apply the chain rule. The last thing that we do is to take the sine of something. So we're finding the derivative of sine something. And the chain rule tells us the derivative of sine something is cosine something times the derivative of our something. Putting things back where we found them. We have an unresolved derivative. And we have our derivative of sine of x squared. As with the derivative of sine or cosine, we can with some effort find the derivative of log from first principles. But the proof is rather involved. And so we'll simply claim without proof that the derivative of log of x is 1 over x. And for example, say we want to find the derivative of log of x squared plus 1. So for variety's sake, we'll use our prime notation, the derivative of log of x squared plus 1. Well, the function is log something. And so our derivative of log something is 1 over something times the derivative of something. Putting things back where we found them. And finding our unresolved derivative gives us our derivative of log of x squared plus 1. Now we come to the world's easiest derivative. Suppose y is e to the x. So in case it's been a while since you've seen this, the log function has a base, which is e. And so if I hit both sides with a log, I get log y equals x. We can then differentiate using implicit differentiation. Over on the left hand side, I have the derivative of log y. And over on the right hand side, I have the derivative of x, which is just 1. So now I want to find the derivative of log. And my derivative of log is 1 over times the derivative of whatever. And so I get 1 over y dy dx equals 1. And I can rearrange this to find dy dx is equal to y. But now I'll put everything back where I found it. y is e to the x, and so dy dx is equal to e to the x. And this gives us a useful theorem about derivatives. The derivative of e to the x is e to the x. So let's find the derivative of y equals e to the power cosine x. So starting off with the chain rule, we want to find the derivative of e to the something, which will be e to the something times derivative something. We'll put things back where we found them and evaluate our unresolved derivative.