 Hi, everyone. This is a supplemental lesson in finding the derivatives of algebraic inverse functions. We are not talking inverse trig functions, nor are we talking about logarithms versus exponential functions. We are talking purely algebraic inverse functions. So we're going to start out with a review of some things that hopefully sound familiar to you from Algebra 2. Basically a review of inverse functions starting off with the definition of a function. Remember that a function is a relation, a set of ordered pairs, in which each element of the domain is paired with one and only one element of the range. Hopefully that definition sounds familiar to you. Now as you talked about and learned about inverse functions before, perhaps you recall there are three different representations of a function. The first one is that you can represent a function as a relation itself. And I'll provide you with an example of each of these. Remember that a relation is simply a set of ordered pairs. So for example, if you had the relation 5 comma 3, 2 comma 3, and negative 1 comma 0. So this is a function because it is true that each element, each x of the domain, is paired with one and only one element of the range. Second representation would be that as a graph. And perhaps you remember a test that you learned to see if a graph is that of a function, the vertical line test. Hopefully that sounds familiar to you. And finally, it can be represented as an equation. So for example, a simple quadratic function such as x squared minus 4. You probably then learned how to find the inverse of its function given each one of these representations. So remember the first representation was that as a relation. And I'm sure you learned that if you have a function given to you as a relation, all you simply do is swap the x and the y values. So in the example I had provided to you, the inverse of that given function would have been 3 comma 5, 3 comma 2, and 0 negative 1. Alright, this brings up an issue that we will be talking about today in that whereas the original given was indeed a function. Once you interchange the x and the y, so as to find the inverse, what you have in the end is not itself a function. Second representation remember was that of a graph and here again you swap the x and the y and replot the points. On your graph and calculator you do have the draw inverse function that will draw the inverse of any function that you have entered under y equals. Finally, if the function was given to you as an equation, and this is probably the one you did the most often, in order to find the equation of the inverse, you began again by swapping the x and the y values and then you simply solve for y. So hopefully all that sounds familiar to you. Now one primary characteristic of a function and its inverse is that the domain of f, remember that will become the range of the inverse. And I'm sure you remember that f negative 1 notation. Likewise, the range of the original function f will become the domain of the inverse. Which very much makes sense given the way in which we found each of these inverses was by swapping the x and the y, the domain and the range. So this very much makes sense. But as we saw with that relation that I provided you as an example, once you flip flopped the x and the y and found the inverse, that inverse was no longer a function. And that does become an issue. You cannot just assume that the inverse of any function that you start with is going to be a function as well. Really the only type of functions that will have inverses that are functions also are one to one functions. One to one function is a function which each x is paired with only one y and each y is paired with only one x. It is a true one to one relationship between the coordinates. Now there are two theorems related to functions and they're inverses. First is if a function f is continuous on its domain, then the inverse of f will also be continuous on its domain. Secondly, if a function f is differentiable at the point c, d, and if the derivative of f at c does not equal zero, then the inverse of f will be differentiable at the point d, c. You have to remember that if you have a point on an original function that is the ordered pair c, d, remember once you find the inverse, the point on the inverse will be the point d, c. That's why they flip flopped. So let's talk about how to then find the derivative of an inverse function. Suppose that f is a function that is differentiable on some interval i. If f has an inverse f negative one, then the derivative of that inverse, and you see there are two different notations that you'll see that are pretty common. You see the d dx notation of f inverse of x. Another way you'll see this represented, and perhaps more commonly, is f inverse prime, so the derivative of f inverse at x. Simply equal to one over f prime of x. That's the basic rule you need to know. Oftentimes though, we'll be trying to find the derivative at a certain point. So suppose you have a point a, b on the original function f, then the derivative of the inverse at b. Remember if a, b is on f, then the point b, a is on the inverse. And remember this number inside the parentheses in function notation is always the x value. So the derivative of the inverse at b will equal one over the derivative of the original function at a. And that's because a is the x coordinate for that original function. Typically where people will get messed up is not so much in the rule that it's one over the derivative of the original function. It's problems in which they're dealing with ordered pairs, and you need to find the derivative at a certain point. And they mess up, you know, which coordinate is the x for which function. So as we work through the next examples, you're going to see some notations that I found are helpful that hopefully they will be for you. So let's start with a really simple example, just of a linear function, 2x plus 3. And we're going to determine the derivative of the inverse. And then we're going to actually find the inverse, like you get back in Algebra 2, and find the derivative of that sort of as a proof that this does work. Alright, so remember the basic rule that if we want the derivative of the inverse, remember that's equal to one over the original function. The derivative of the original function. Well, what is the derivative of the function 2x plus 3? It's simply 2, so your answer is one half. Alright, but let's actually go through it the long way as a proof of sorts. Let's actually find the equation of the inverse. Alright, so we're given the function, if we rewrite it as y equals 2x plus 3. Remember that to find the inverse, we're going to flip-flop the x and the y and solve for y. So when we do that, we have x minus 3 over 2 as our function. So this is our inverse. Now we want to find the derivative of that. Now to do that, we might want to rewrite it as x over 2 minus 3 over 2. It'll make it a heck of a lot easier. The derivative, then, you know is one half. QED, quadiret demonstratum. And if you future math majors out there in college, you'll find yourself putting that at the end of your proofs. It's Latin, quad erot demonstratum. It's Latin for it has been demonstrated. That's what mathematicians typically put at the end of their proofs. Alright, so I have three practice problems lined up for us to do. So hopefully as we go through these, you'll see how I have found it's a little bit easier to walk through these. If you keep your thoughts organized as you do your work, these really are not hard. The hardest part is just making sure you think about it clearly. And the more that you can organize your work as such, I think you'll find a good bit of success with these. So the first practice problem I have for us. Suppose we consider the function square root of x. And we know that function to contain the point nine comma three. And we want to find the derivative of the inverse at three. Now you have to remember, if the original function contains the point nine comma three, remember that the point on the inverse is three comma nine. That explains why this is a three in there. So we're finding the derivative of the inverse at three. So the rule is it's going to be one over the derivative. And the question is at what x value? Well you have to remember the x value of the original function was nine. So we'll need to find f prime of nine. Alright, so we need to do a little side scratch work and find the derivative. So if we think of that original function as x to the one half, our derivative is going to be one half x to the negative one half, which we could rewrite as one over two square root of x. Just to make it easier to evaluate it. F prime of nine that we need is going to be one over two, then times three because of course the square root of nine is three. So we get one six. So we're going to take this one six, put it over into our main formula for finding the derivative of an inverse. So you have one over one six, which in the end is equal to six. The mathematics isn't that hard. Again, you just have to sort of keep everything organized. So let's try another. This time we have a quadratic function x squared plus two and we want to find the derivative of the inverse at three. Now notice they don't give you the entire ordered pair. Alright, you have to remember this three. It is the x value of the inverse. It's going to therefore be the y value of the original function. Alright, so as we set up our problem for what we want to find, we know it's going to be f prime of something. The question is, what is that something? So what we really need to do is find the x value for that original function. And we do know though that the y value of that original function would be three. So we can make use of the fact that we know the y value is three and set the original function equal to it. And we find that x equals one. Now you might be asking yourself, well, y is also negative one. Remember that these have to be functions. So we can't have like two answers if you want to think of it that way. So we do go with the primary, the principal square root. And that will be pretty common when you do come across these. Alright, so now we know we're talking about f prime of one that we need to find. We do need to find our derivative. So the derivative of this function is simply 2x, that's easy enough. We need to know f prime of one, well that's just two. So we're going to take that and put it up in place of the f prime of one. And our answer is one half in the end. And we have one more. This time we have a cubic function, 2x minus x cubed. And we're asked to find the derivative of the inverse at negative four. Alright, so as we talked about before, that negative four, it's obviously the x value of the inverse. And it is therefore the y value of the original. So if we set up what it is we're trying to find, once again we don't know what that x value is. So we're going to approach it just like we did before. And we're going to make use of the fact that we know that negative four is the y value of that function. Now this is an equation that you obviously cannot solve algebraically by hand. You will have to solve it on your graphing calculator. Easiest way might be to let y one equal negative four. Let y two equal the 2x minus x cubed. And then simply find the point of intersection. So if you go ahead and give that a shot, you should be getting that your x value in the end is two. And we're something like this on the AP exam. Of course this problem, this particular problem would be on the calculator part of the test. So then we need to find our derivative. So we know our derivative to be 2 minus 3x squared. We need to find the derivative at 2. So that would be 2 minus 3 times 4 is 12. We get negative 10. So we'll put that negative 10 up in place of the f prime of 2. And our answer is negative one tenth. So hopefully you can see if you do organize your work, it does make it easier. I have found it helpful over the years to write those little notes. You know x value of f negative one is the y value of f. Maybe that will help you.