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Derivatives of inverse functions: from equation

Let's explore the intriguing relationship between a function and its inverse, focusing on the function f(x)=½x³+3x-4. We delve into the derivative of the inverse of f, applying the chain rule and the power rule to evaluate it at x=-14. Join us as we unravel this complex calculus concept.

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  • leaf green style avatar for user choonklog
    Is it also possible that dy/dx[x]=dy/dx[1/2y^3+3y-3]?
    I've got the solution of dy/dx=1/(3y^2/2+3). But I don't know how to apply h'(-14)=-2 to that.
    (6 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user John
      You had written:
        dy         d [  1               ]
      ---- x = ----[ --- y^3 + 3y -3 ]
      dx dx [ 2 ]

      Where did you get "y" from? The problem in this video is written in terms of just x and the functions f(x) and h(x).

      I would encourage you to go back and try it again without using "y", just keep everything in terms of "x". Kind of like this..


                f( h(x) ) = x

      d d
      ---- f( h(x) ) = ---- x
      dx dx

      The right (d/dx x) is just 1.
      Using chain rule on the left gives us:

      f'( h(x) ) h'(x) = 1

      Multiplying each side by 1/f'( h(x) ) :
      1 1
      ----------- * f'( h(x) ) h'(x) = 1 * ------------
      f'( h(x) ) f'( h(x) )


      1
      h'(x) = ------------
      f'( h(x) )


      To solve this for h'(-14), e.g. x = -14, we need
      to know h(x). Which is a pain because they didn't give
      us the definition of h(x).

      We did get that additional "trick question" hint
      that f(-2) = -14.
      Which doesn't help us with h(x). Not directly, anyway.
      But we can use the inverse function definition to get h(x).

      Because f(-2) = -14 at x = -2,
      and because f(x) and h(x) are inverse functions, h( f(x) ) = x.
      This means that h(-14) = -2
      Because h( f(-2) ) = -2.

      That is enough to start solving the expression on the right:

           1
      -------------
      f'( h(-14) )

      1
      -------------
      f'( -2 )


      Now you still need to work out f'(x) but that is
      straight forward polynomial differentiation.
      (42 votes)
  • blobby green style avatar for user D. Ashley Nelson
    At , Sal says if h (x) is the inverse of f(x), then h' (x) = 1 / f ' (h(x)). Is the derivative of a an inverse its reciprocal? I thought reciprocals and inverses were distinct and separate functions. Thank you for taking the time to ex[plain. I am teaching self and appreciate the help.
    (5 votes)
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    • male robot hal style avatar for user Palash Goiporia
      Reciprocals and functions are different functions. This is what Sal did -
      According to the Chain rule
      (f(h(x)))' = f'(h(x)) * h'(x)
      if f(x) and h(x) are inverses, then f(h(x))=x
      so (f(h(x)))' is just equal to dx/dx which is 1
      This means that f'(h(x)) * h'(x) = 1
      If you divide both sides by f'(h(x)) you get
      h'(x) = 1/(f '(h(x))
      (15 votes)
  • duskpin ultimate style avatar for user Brianna Seepersad
    Correct me if I'm mistaken, but can't you switch x and y and then solve for y to get the inverse derivative?
    (5 votes)
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    • leaf green style avatar for user Abdullah
      Yes, however, finding the inverse of a cubic function is very difficult. You can find the inverse of a quadratic function by completing the square.

      Finding the inverse of a simple cubic function, for example, f(x) = x^3 is easy.

      But finding the inverse of f(x) = x^3 + 5x^2 + 2x - 6 is very difficult, if not impossible.
      (11 votes)
  • blobby green style avatar for user hrubycha
    What is the inverse function of f(x) = 2x^3-5
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leafers sapling style avatar for user Naozi
    I still don't understand the first step of implicit differentiation. Why the derivative with respect to x on each side of the equation should be equal? What does it mean?
    (3 votes)
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  • aqualine ultimate style avatar for user Liang
    my calculation is:
    f'(x)=1/h'(f(x)) ==> f'(-2)=1/h'(f(-2)) ==> f'(-2)=1/h'(-14), and then solve for h'(-14).

    It's still correct, right
    (2 votes)
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  • leaf orange style avatar for user Prasid Sarkar
    I cant understand how f(h(x))=x, can you explain this?
    (2 votes)
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    • orange juice squid orange style avatar for user Tyler
      This is a rule that we learned from regular inverses. Since they are inverses, this must be true. For example, if
      f(x) = sqrt(3x - 4)
      the inverse of that is
      h(x) = (x^2 + 4)/3
      Test it out, it works! Now, if you plug in h(x) to f(x), you get x. Try plugging in f(x) to h(x). You get x. Note that h(x) is basically the opposite of f(x):
      In f(x) you multiply x by 3, subtract 4, and square root it
      In h(x) you square it, add 4, and divide x by 3
      (5 votes)
  • blobby green style avatar for user Li Hans
    In the previous video, it showed that g(f(x))=x. Why is it that in this video, everything is swapped and that f(h(x))=x instead? The formula comes out different too. In the first video, formula is f'(x)=1/g'(f(x)) while here, it is h'(x)=1/f'(h(x))
    (3 votes)
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  • blobby green style avatar for user Miray Atar
    I think we get the same result from g(h(x)) and h(g(x)), correct me if I'm mistaken. How is this possible? I see they yield the same result but I can't get the intuitive sense of it.
    (1 vote)
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  • blobby green style avatar for user Trần Quốc Đạt
    Does the function need to be bijective in order to have an inverse function?
    Also, how can we prove that the inverse of a continuous (and differentiable) function is also continuous (and differentiable)? I'm stuck with the proof.
    (2 votes)
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    • mr pink red style avatar for user andrewp18
      A function must be bijective in order to be invertible. If you are unsure why this is true, I suggest you look into the inverse function videos in the algebra section.

      First note that differentiability implies continuity because it is a stronger condition. That being said, the inverse of a function is indeed differentiable if the original function is differentiable, as graphically, the inverse of a function has a graph that is the reflection of the graph of the original function across the line 𝑦 = 𝑥.

      While this an intuitive explanation, it is not rigorous. We can show this rigorously as follows:
      By the definition of inverse functions, we have:
      𝑓⁻¹(𝑓(𝑥)) = 𝑥
      Then we differentiate both sides:
      𝑓'(𝑥) • 𝑓⁻¹'(𝑓(𝑥)) = 1
      Thus we have:
      𝑓⁻¹'(𝑓(𝑥)) = 1/[𝑓'(𝑥)]
      𝑓⁻¹'(𝑥) = 1/[𝑓'(𝑓⁻¹(𝑥))]
      So 𝑓⁻¹ is differentiable as we can find its derivative. Comment if you have questions!
      (2 votes)

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.