Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Inverse functions and their derivatives functions that are not algebraic are called transcendental functions. 1 to one functions pass both the vertical line test and the horizontal line test. So if F of X1 and F doesn't equal F of X2, IE my Y value of one point doesn't equal my Y value of another point, whenever X1 doesn't equal X2 and X1 and X2 are in the domain. So basically every Y has an X and every X has AY and those are all unique XS and YS inverse functions. Suppose that F is a one to one function on a domain D and range R. The inverse function F inverse of X is defined by F inverse of A equal B if A equal F of B. So the composition of those functions. If I take the function and the functions inverse and do a composition, I should end up with a equaling F of B. The domain of F inverse is the range and the range of F inverse is the domain of the original function. So the domain of the original function is the range of the inverse and the range of the original function is the domain of the inverse. Make sure they're one to one. How do you find an inverse? Check to see if it's one to one. First exchange X&Y in the original function and then solve for Y derivative rule for inverses. If F has an interval I as domain and F prime X exists or the derivative exists and is never 0 on the interval, then F inverse is differentiable at every point in its domain. The value of F inverse of X prime at a point B in the domain of F inverse is the reciprocal of the value F prime at the point A equal F prime B where F of A equal B. We just talked about that a moment ago. So the inverse functions derivative at B equal 1 divided by the original functions derivative where we're sticking in F inverse of B or 1 / F prime of A. A notation that we could use is the inverse derivative DX. So the derivative of that inverse function at the location X equal B is really one divided by the derivative of the original function when X equal F prime F inverse of B. Or that's really just our a. So if we look at an example, F of X equal 2 X squared where X is greater than or equal to 0. If X is greater than or equal to 0, this is going to be a one to one function because two X squared is really just a parabola. But once we have this parabola, we realize that we only have X greater than or equal to 0. So we're looking at half of a parabola, which would make it 1 to one. So the first thing we're going to do is we're going to find the inverse. So if F of X = 2 X squared, Oh no, the first thing we're going to do is find its derivative. Sorry. Bring down the exponent, multiply, and take it to the one less power. So F prime of X = 4 X. So now if we're wanting to find the derivative at the inverse when A equal 5, we actually need to figure out what that Y value would be for the original function. So if I stick 5N 2 * 5 ^2, 2 * 25 is 50. So we want to find the inverse function at 50. So F inverse 50 prime is going to equal 1 divided by the derivative of the original. So four times the five that was given or 1 / 20. Now, a different way to do this problem is to actually find the inverse. So if I switch my X and my Y and solve for Y, I'm going to have X equal to y ^2, divide by two, and square root it. Because we needed X greater than equal to 0 in the original, we also knew that Y would be greater than or equal to 0 in the original. So we want the positive when we do the square root. Technically, when we take a square root, we have positive and negative, but we don't want the negative case because we need the domain of the original to be the range of the inverse, but we need the range of the original to be the domain of the inverse. So if Y is greater than or equal to 0 down here, we needed X to be greater than or equal to 0. So Y equals square root X -, 2. We could think of that as X to the half divided by two to the half if we wanted to. Now, if we take the derivative, we're going to use the power rule. We're going to bring down that 1/2, so 1 / 2 here times that two to the half and take X to the 1/2 to the one less power. 1/2 - 1 is negative 1/2 or that takes it to the bottom. So 1 / 2 * 2 to the half, X to the one half, 1 / 2 times sqrt 2, X to the one half. Now we're going to actually put in that Y value we found originally. Sqrt 50. Sqrt 50 times sqrt 2 is going to be sqrt 100. Sqrt 100 is 10. So this all ends up with Y prime equaling 1 / 20, which is what we found in this other fashion. Now the power comes from we won't always have equations where we can find the inverse. We're going to look at an example in just a moment. So in this case, we could take the function, we could find the inverse derivative at a single location, but we could also find the actual inverse function. We won't always be able to do that. So looking at another example, does a function have an inverse? Well, you're going to have some problems that they're going to ask you. Does something have an inverse and they want you to look at it and just determine yes or no. That one would be no, because it doesn't pass the vertical line test and the horizontal line test both. This one would actually, well, imagine that's a little more of a curve would be yes. However, this one would be no. So be careful. Remember it's got to pass vertical and horizontal line test. The domain and range of the original function is the range and domain of the inverse function. So if they give you a piece of a graph, say they only give you this much, and they ask you to find the domain and range. You figure out the domain and range of the original, and then you flip them for the inverse. If they ask you to draw an inverse, the inverse gets drawn by putting in the line Y equal X because an inverse is just a reflection. And then we would actually take this point and reflect it. And we would take this point and reflect it. And then if we drew and if we can draw accurately, if we folded along this Y equal X line, the top and the bottom would fall right on top of each other. So if we are given a function and we just want to find the inverse, we're going to switch the X and the Y. So we're now going to have X = 6 Y -1 / 2 Y +5 going to cross multiply because our goal is going to be to get the YS all by themselves on one side to solve for Y. So I'm going to distribute out that X and then I'm going to add and subtract. So I get everything without AY on one side and everything with AY on the other going to factor out the Y and then we're going to divide. So we're going to get Y equaling five X + 1 / 6 -, 2 X. Using correct notation, we have F inverse of X equaling five X + 1 / 6 -, 2 X. Now if we recall, if we look back here at the original, this is going to be a function where we're going to have a horizontal asymptote at 3:00 because it's degree on top and bottom are the same. So we talked about the leading coefficient. We're going to have a vertical asymptote at -5 halves because we can never, ever have the denominator be 0. If we think about sticking 0 in for X, we're going to have an intercept of 0 negative 1/5, and that's going to be enough to tell me how this graph's going to look. We're going to go close to the asymptote when we're out at Infinity. We've got to go through the intercept and we got to get close to the the vertical asymptote. The vertical asymptote is an odd degrees. So at the other side of the Y values are negative on one side, they're going to be positive on the other. So it does indeed pass both the vertical and horizontal line test. So when we do our inverse, if we thought about putting in our line Y equal X and did a reflection, we would have the graph here. So that's how we're going to find that inverse. If we wanted to graph this, 1° on top and the bottom are the same, so we'd have horizontal asymptote at -5 halves. Remember this was just the vertical asymptote a minute ago. And then we're going to have a vertical asymptote when the denominator doesn't equal 0. And in this case, it's going to be at 3:00. If I stick in 0 for my X, I'm going to get 1/6 right here. So this piece here tells me we're going to come like so. Whoops. I'll pretend that went through that point. Broad point theorem there. This is an odd degree, so it's going to come down here. So looking at the next one, if we have F of X equaling X to the 4th -3 X -1278 X greater than .75. Actually, let's go back to this last one for just a second. Did we realize that we can find the derivative here? It's a little more complex to find the inverse, but we actually have the ability. We can do the quotient rule. The derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared. We go plug, plug and simplify that out to whatever it is, but we have the ability to find the actual inverse and then find the inverse's derivative. Actually, it would have been better if I had done F inverse prime of X because we have multiple functions going on here, so it's always good to identify which one. So if I have 30 -, a negative is +2, so 32 on the top and then -10 X and positive 10X so -10 positive 10 those go away. So 32 / 6 -, 2 X quantity squared. So now if I have any point on anywhere I could find the derivative of the original and or the derivative of the inverse. The next example's going to be 1 where we can't easily find the inverse. So if we have F of X equaling X to the 4th -3 X -1278 X is greater than .75, we want to find the inverse derivative at the point X = 0 which equals F of six. So this is really saying the original functions point is an X value of 6 and AY value of 0. So the inverses point is 0, 6. So F prime of X We're going to take the derivative here and we're going to get four X ^3 - 3. So F inverse prime of B equal 1 / F prime of F inverse of B. So we want to find F prime inverse of 0, IE the value that was going in in the inverse function would equal 1 / F prime of F inverse of 0. Well, F inverse of 0 is really really, really just saying the six. What point goes with the zero? The six matched with the zero. So we're going to have 1 / F prime of six, and we're just going to put 6 into that first derivative of the original function. So 1 / 4 * 6 ^3 - 3. So if we do that, we get 1 / 861. Remember that derivative is really just telling us the slope at a given point on a function. Thank you and have a wonderful day.