Derivatives of inverse functions: from table | AP Calculus AB | Khan Academy
Let G and H be inverse functions. So let's just remind ourselves what it means for them to be inverse functions. That means that if I have two sets of numbers, so let's say one set right over there, that's another set right over there. If we view that first set as the domain of G, so if you start with some X right over here, G is going to map from that X to another value which we would call G of x, x, G of X. That's what the function G does.
Now, if H is the inverse of G and frankly vice versa, then H could go from that point G of X back to X. So H would do this; H would get us back to our original value, so that's what the function H would do. We could view this point right over here; we could view it as X. So that is X, but we could also view it as H of G of X. So we could also view this as H of G of X. I did all of that so that we can really feel good about this idea.
If someone tells you that G and H are inverse functions, that means that H of G of X is X. H of G of x, H of G of X is equal to X. Or you could have gone the other way around; you could have started with a... well, you could have done it multiple different ways, but also G of H of X. I could have just swapped these letters here. The letters H and G are somewhat arbitrary, so you could have also said that G of H of X is equal to X.
So G of H of X is equal to X, and then they give us some information. The following table lists a few values of G, H, and G Prime. All right, so they want us to evaluate H Prime of 3, but they don't even give us H Prime of 3. How do we figure it out? They gave us G Prime and H and G; how do we figure this out? Well, here we're going to actually derive something based on the Chain Rule.
This isn't the type of problem that you'll see a lot of, but it is interesting. So we're going to work through it, and you... there's a chance that you might see it in your calculus class. So let's start with either one of these expressions up here. Let's start with the expression... well, let's start with the... let's do this one over here.
So if we have G of H of X is equal to X, so let me put that H of X back there, which is by definition true if G and H are inverses. Well now, let's take the derivative of both sides of this. So let's take the derivative with respect to X of both sides. Derivative with respect to 2X.
On the left-hand side, well, we just apply the chain rule. This would be G Prime of H of X times H Prime of X. That's just the chain rule right over there. And then that would be equal to... what's the derivative with respect to X of X? Well, that's just going to be equal to one.
So now it's interesting; we need to figure out what H Prime of three is. We can figure out what H of three is, and then we can use that to figure out what G Prime of whatever H... G Prime of H of 3 is. So we should be able to figure out H Prime of X, or we could just rewrite it this way. We could rewrite that H Prime of X is equal to... is equal to 1 over G Prime of H of X.
Now, in some circles, they might encourage you to memorize this. Maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it. But I'll tell you, 20 years after I took Calculus, almost 25 years after I took Calculus, this is not something that I retain in my long-term memory. But I did retain that you can derive this from just what the definition of inverse functions actually are.
But we can use this now if we want to figure out what H Prime of 3 is. H Prime of three is going to be equal to 1 over G Prime of H of 3, which I'm guessing that they have given us. So H of three, when X is three, H is four. So that is H of three there, so H of three is four.
So now we just have to figure out G Prime of four. Well, lucky for us, they have given us when X is equal to 4, G Prime is equal to 12. So G Prime of 4 is equal to 12. So H Prime of 3 is equal to 1 over 12.
So 1 over 1/2... 1 divided by 1/2 is the same thing as 1 times 2. So this is all equal to two, and we are done.