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## Differentiating inverse functions

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# Derivatives of inverse functions: from equation

AP Calc: FUN‑3 (EU), FUN‑3.E (LO), FUN‑3.E.1 (EK)

## Video transcript

- [Voiceover] Let F of X be
equal to one half X to the third plus three X minus four. Let H be the inverse of F. Notice that F of negative
two is equal to negative 14. And then they're asking us
what is H prime of negative 14? And if you're not familiar
with the how functions and their derivatives
relate to their inverses and the derivatives of the
inverse, well this will seem like a very hard thing to do. Because if you're attempting
to take the inverse of F to figure out what H is well, it's tough to find, to take to figure out the
inverse of a third degree a third degree polynomial
defined function like this. So, the key the key, I guess property to realize, or the key truth to realize if F and H are inverses then H prime of X H prime of X is going to be equal to is going to be equal to one over F prime of H of X. One over F prime of H of X. And you could now use this in order to figure out what H prime of negative 14 is. Now I know what some of you are thinking, because it's exactly
what I would be thinking if someone just sprung this on me is where does this come from? And I would tell you,
this comes straight out of the chain rule. We know that if a function and its inverse we know that if we have a
function and its inverse that F of F of the
inverse of our function. So F of H of X. F of H of X. We know that this is going to be to X. This literally, this is comes out of them being
each others inverses. We could have also said H of F of X will also be equal to X. Remember, F is going to map or H is going to map
from some X to H of X. And then F is going to map
back to that original X. That's what inverses do. So that's because they are inverses. This is by definition,
this is what inverses do to each other. And then if you took the derivative of both sides of this what would you get? Let me do that. So if we took the derivative of the both sides of this D D X on the left hand side. D D X on the right hand side. I think you see where this is going. You are essentially gonna
get a version of that. The left hand side used the chain rule. You're going to get F prime of H of X. F prime of H of X times H prime of X comes straight out of the chain rule is equal to, is equal
to the derivative of X is just going to be equal to one. And then you derive,
you divide both sides by F prime of H of X and you get our original property there. So now with that out of the way let's just actually apply this. So, we want to evaluate H prime of 14. Or sorry, H prime of negative 14. Is going to be equal to one over F prime of H of negative 14. H of negative 14. Now had they given us H of negative 14. But they didn't give it to us explicitly, we have to remember that
F and H are inverses of each other. So F of negative two is negative 14. Well, H is gonna go from
the other way around. If you input negative 14 into H you're going to get negative two. So H of negative 14 well, this is going to be equal to negative two. Once again, they are
inverses of each other. So H of negative 14 is equal to negative negative two. And once again, I just
swapped these two around. That's what the inverse function will do. If you're wrapping from if F goes from negative two to negative 14 H is going to go from negative 14 back to negative two. So now we want to evaluate
F prime of negative two. Well, let's figure out
what F prime of X is. So, F prime of X is equal to remember the power rule,
so three times one half is three halves times X to the three minus one power which is the second power. Plus the derivative of three X with respect to X which
that's just going to be three. And you could do that,
it's just the power rule. But this was X to the first power, one times three, X to the zero power, but X to the zero is just one so you're just left with three. And derivative of a concept
that's just gonna be zero. So that's F prime of X. So F prime of, F prime of negative two is going to be three halves times negative two squared
is four, positive four. So plus three. So, this is going to be equal to two times three plus three. So, six plus three is equal to nine. So this denominator here is
going to be equal to nine. So this whole thing is
equal to one over nine. So this involved, this
was something you're not going to see every day. This isn't that typical
problem in your calculus class. But it's interesting.

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