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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

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 that if if F and H are inverses then H prime of X H prime of X is going to be equal to it's going to be equal to 1 over F prime of H of X 1 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 funk funk ssin and it's inverse we know that if we have a function in 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 equal to x this literally this is comes out of them being each other's 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 but then if you took the derivative of both sides of this what would you get let me do that so if we take the derivative of both sides of this d DX on the left hand side d DX on the right hand side I think you see where this is going you're essentially going to get a version of that the left-hand side use 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 1 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 14-4 sorry H prime of negative 14 is going to be equal to 1 over F prime of H of negative 14 H of negative 14 now have they given us H of negative 14 well they didn't give it to us explicitly but we have to remember that F and H are inverses of each other so if F of negative 2 is negative 14 well H is going to go from the other way around if you input negative 14 into H you're going to get negative 2 so H of negative 14 well this is going to be equal to negative 2 once again they are inverses of each other so H of negative 14 is equal to negative negative 2 and once again I just swapped these two around that's what the inverse function will do if you're mapping from if f goes from negative 2 to negative 14 H is going to go from negative 14 back to negative 2 so now we want to evaluate F prime of negative 2 well let's let's figure out what f prime of X is so f prime of X is equal to C which can leverage the power rule so 3 times 1/2 is 3 halves times X to the 3 minus 1 power which is just the second power plus the derivative of 3x with respect to X well that's just going to be 3 and you could view that it's just the power rule if this was X to the first power 1 times 3x to the zeroth power well X to the 0 is just 1 so you just left with 3 and derivative a constant that's just going to be 0 so that's F prime of X so f prime time of f prime of negative two is going to be three-halves times negative two squared is four positive 4 so plus three so this is going to be equal to 2 times 3 plus 3 so 6 plus 3 is equal to 9 so this denominator right here is going to be equal to 9 so this whole thing is equal to 1 over 9 so this involved this was a just you know this isn't something you're you're going to see every day this isn't a typical problem in your calculus class but it's it's interesting