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So, now that you learned the derivatives of several functions, here's an important concept that's going to help you find the derivative of more functions. It's the concept of the inverse function. Now, what is an inverse function? Well, if your function does a certain thing, then the inverse function is the one that undoes that thing. So, for example, if the function f sends the number 3 to the number 5, then the inverse function sends the number 5 back to the number 3. And now, you're going to learn that if you know the derivative of function f, then the derivative of the inverse function is actually very easy to calculate, because they are very well connected. So, the mental image I have of an inverse function is the following. Let's say that you have a person, and the function f does something to the person. It puts a hat on them. What would the function g have to do if it's the inverse? Well, it would just remove the hat so that you get the original person back. In other words, the inverse does the opposite of what f does. So it undoes what f does. In a more mathematical example, let's say you have the variable x, and f puts on a hat on it, so it squares it, let's say. So what would g have to do? Well, undo the square, so remove the hat. g is the function that takes as input x squared and returns x. Before we get to see what f and g are, let's take a look at some notation. So if g of x and f of x are inverses, we write this as g of x equals f inverse of x with an exponent of minus one, but this is not one over f of x. This is purely notation. And if you take x and apply f and then apply g, you get back the x you started with. So basically, g undoes what f does. So what are f and g in the example on the left? Well, f is easy. It's simply x squared. And g is the function that undoes the square. So it's basically square root of x. And the reason is that if you take x and square it and then take the square root, you get back x. And this is only for x positive numbers. And also, we're only taking the positive square root, not the negative. But in short, if f does something, g has to undo it to get your first input back. Now the derivatives of inverses are really nice. Let's take a look at the plot of x squared and now the plot of g of y equals the square root of y. Notice that in this plot, the units are different in the left and the right. And also in the plot in the left, the horizontal axis is x and the plot on the right, the horizontal axis is y. But in short, if a point with coordinates a, b appears on the left, then the point with coordinates b, a appears on the right. So for example, on the left, we have the point a half, a quarter. On the right, we have the point a quarter, a half. And this is because a quarter is the square of a half, and a half is the square root of a quarter. So the point, basically, f one half is equal to one quarter, and g of one quarter is equal to one half. The point one, one also appears on both, because f of one is one, as one squared is one. And g of one is one, because the square root of one is one. The point three halves, nine fours also appears on the left, and the point nine four, three halves appears on the right. And similarly, the point square root of three, three appears on the left, and the point three square root of three appears on the right. Now I want you to take a careful look at the tangents over these points in the left plot and in the right plot. And let's question yourselves. Do you think there's a connection between the slopes of the tangents on the left and the slopes on the tangents of the right? Well, consider that the plot on the right is the reflection of the plot on the left over the diagonal y equals six. So there's got to be a really nice connection between these slopes. Let's take, for example, a secant. Before we draw tangents, let's draw a secant. So let's take the point one, one, which is on both plots, and let's take the point one point five, two point twenty five on the left, because two point twenty five is the square of one point five, and the point two point twenty five, one point five on the right. So let's calculate the slope of the secant line. On the left, it's delta f divided by delta x. So the change in the vertical coordinate divided by the change in the horizontal coordinate. And on the right, this is going to be delta g over delta y, where delta g is the change in the vertical coordinate and delta y is the change in the horizontal coordinate. However, these plots are a reflection of the other. So therefore, any horizontal distance on the left is equal to the vertical distance on the right and vice versa. In other words, this delta g is the same as this delta x and this delta y is the same as this delta f. So therefore, the slope delta g over delta y is the same as delta x over delta f. Now notice that as deltas go to zero, f prime of x is delta f over delta x. And equally, as delta y goes to zero, then g prime of y is delta g over delta y. So putting this together, we get that g prime of y is one over f prime of x. And this looks complicated, but it only says that if f and g are inverse functions, then the derivative of g is one over the derivative of f. So let's look at a simple example. And this time we have two plots of x squared and square root of y, where the actual units are going to be the same on the left and on the right, so each square is one by one. So let's look at the tangents. So at the point 1, 1, we have that f of 1 is 1 and g of 1 is 1. So the point 1, 1 appears on both plots. Now take a look at the tangent on the left. Since f prime of x is 2x, because it's the derivative of x squared, then f prime of 1 is 2, which means that this orange line has a slope of 2. Now let's take a look at the slope on the right, the tangent. G prime of y is what we want to find. We want to find the slope of that tangent. Well, since g prime of 1 is 1 over f prime of 1, then this is 1 over 2, which is 1 half. Therefore, a half is the slope of the tangent on the right. So it makes sense, right? If the slope of the tangent on the left, the orange one, is 2, then the slope of the tangent on the right is 1 half, the reciprocal of 2. Now let's do another example, the point 2, 4. So on the left we have the point 2, 4, and on the right we have the point 4, 2, because f of 2 is 4 and g of 4 is 2. Now the slope on the left, well, since the derivative f prime of x is 2x, then f prime of 2 is 4, which is 2 times 2. So what do you think this slope is going to be? It's probably going to be 1 quarter, right? Because it's 1 divided by the slope on the left. And indeed it is, because g prime of 4 is 1 over f prime of 2, and that's 1 quarter. So in other words, if f and g are inverse functions, then the derivative of g is 1 over the derivative of f.