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## Calculus 1

### Course: Calculus 1>Unit 3

Lesson 5: Differentiating inverse functions

# Derivatives of inverse functions: from table

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.E (LO)
,
FUN‑3.E.1 (EK)
Given a table of values of g, its inverse h, and its derivative g', Sal evaluates the derivative of the inverse, h', at a given x-value.

## Want to join the conversation?

• if f(x) and g(x) are inverse of each other then does it mean that domain of f(x) will be the range of g(x) and range of f(x) will be domain of g(x) and vice versa?
(11 votes)
• Simply Yes. That's why they are inverse of each other.
(8 votes)
• so h'(x) is undefined when g'(h(x)) is zero ?
(3 votes)
• Yes, because g'(h(x))·h'(x)=1. If g'(h(x)) is 0, there is no value of h'(x) that makes the equation true, because it simplifies to 0=1.
(3 votes)
• What if the value for h(3) was not given and all we had is g'(3) and g(3)? What would we do to solve this then?
(4 votes)
• When will I ever be using this after school just wondering?
(3 votes)
• Depends on what you do after school. Derivatives are just functions that define the instantaneous rate of change at an input value. This is valuable when dealing with physics in any sense as far as I know, and there are plenty of physics based careers out there.
(3 votes)
• Hi! I was wondering if using Khan Academy for AP exam prep was enough even if you are not taking an actual AP Calc BC course at school. I go to a Canadian high school (so there are no AP classes) and I am taking my exam in May. Are Khan Academy and a prep book enough? Should I take another course?
(4 votes)
• You will have to be a very driven and self-motivated student to be successful at this. You certainly can be, but practice regularly and extensively if you want to have a shot. I think having the classroom experience and having a teacher/mentor is definitely an advantage though.
(1 vote)
• are proofs presented in calculus classes?
(1 vote)
• proofs can be presented depending on your teacher but can usually be found in a book
(3 votes)
• if h(g(x))=x is the inverse of the function g, does that make g(h(g(x)))=x the inverse of the inverse, and shouldn't the inverse of the inverse just be the original function (g(x))?
(1 vote)
• This isn't really a question about how to do this but I was wondering if derivatives of inverse functions would be still expected on the AP Calculus test (even if there is not a lot). I understand that Sal said that we MIGHT see this in our calculus classes but will it be expected on the AP test? If so, which one? AB or BC or both?
(1 vote)
• As far as I know, the derivatives of inverse functions will be expected on the AP Calculus AB test. You will probably be expected to know the derivatives of inverses on the BC test also since BC is just building off of AP Calculus AB.
(2 votes)
• At . Sal simply plugs in h(x) to be h(3) when he was solving for h'(3). Are you allowed to do that? Why are you able to do that? I am not sure because while he was writing that he admitted that he was guessing.
(1 vote)
• The equation for h'(x), when h and g are inverses, is 1/(g'(h(x))). So to find h'(3), he plugged in 3 for every x in the equation, giving h'(3)=1/(g'(h(3))). This would be the same if you were finding any other equation, you plug in the number you are looking for anywhere there is an x.
(1 vote)
• Is this class part of Calculus 2? or Calculus1?
(1 vote)

## 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.