# Derivatives of inverse functions: from table

## Video transcript

- [Voiceover] 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, let's say one set right over there, that's another set right over there, and 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. 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 and so 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 and I did all of that so 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, 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. Alright, so they want us to
evaluate H prime of three. They don't even give us H prime of three. 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 and
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 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. So let's start with the expression, well let's start with,
let's do this one over here. So if we have G of H of X is equal to X, so we 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 X, and on the left-hand side well
we just apply the chain rule. This would be G prime of H of X, 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
G prime of H of three is and 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 one over G prime of H of X. Now in some circles,
they might encourage you to memorize this and 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, 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 three is. H prime of three is going to be equal to one over G prime of H of three, 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 four,
G prime is equal to 1/2. So G prime of four is equal to 1/2. So H prime of three is
equal to one over 1/2. So one over 1/2, one divided by 1/2 is the
same thing as one times two. So this is all equal
to two and we are done.