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Derivatives of inverse functions

Functions f and g are inverses if f(g(x))=x=g(f(x)). For every pair of such functions, the derivatives f' and g' have a special relationship. Learn about this relationship and see how it applies to 𝑒ˣ and ln(x) (which are inverse functions!).

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  • duskpin ultimate style avatar for user Tuan Anh Dang
    At , Sal says that y=e^x and x=e^y. This doesn't seem to make sense, I tried to plug some numbers into the calculator and the results don't match. If I am understanding this wrong, then please tell me where I went wrong, thanks.
    (4 votes)
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    • aqualine ultimate style avatar for user Mark Geary
      At , Sal says that, to get the inverse of y = eˣ, you swap the variables to get x = eʸ. If you graph the two functions y = eˣ and x = eʸ, you'll see they are symmetric around the function y = x, so they are indeed the inverse functions of each other.
      (40 votes)
  • leaf green style avatar for user Haak
    Will someone please explain how to go from y=e^x to x=e^y algebraically?
    (3 votes)
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  • female robot amelia style avatar for user Fourth Dimension
    Is there a neat graphical or geometrical proof to this formula or is it only derived algebraically. I really want to see some beautiful math behind it
    (6 votes)
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    • starky seedling style avatar for user deka
      1. definitions
      1) functions
      a. math way: a function maps a value x to y
      b. computer science way: x ---> a function ---> y
      c. graphically: give me a horizontal value (x), then i'll tell you a vertical value for it (y), and let's put a dot on our two values (x,y)

      2) inverse functions
      a. norm: when we talk about a function, the input is x (or a horizontal value, length) and the output is f(x) or y (or a vertical value, height)
      b. inverse: let's flip the graph around on the axis of x=y (which is 45 degree angle away from both axes). thus, x_axis becomes y_axis and y_axis becomes x_axis, on our new graph
      c. naming: call this function, which gets an input of pre-y-but-present-x-axis and then spits out an output of pre-x-but-present-y-axis, g(x)
      d. pitfall: x in g(x) here is, in fact, f(x) of our original graph. so we better keep it g(f(x)), or g(y) if you will, not to confuse ourselves

      3) derivatives
      a. (only) graphically: a slope of our dot. yes it's weird that a dot has a slope. but let us use our imagination that there is a bar right on top of it, thus this bar represents the tinest difference of the lengths and heights between this dot's and its neighbor's



      2. connections
      1) derivatives of a function f(x)
      : slope of our dot on the original graph = change_y / change_x

      2) derivatives of a function g(f(x))
      : slope of our dot on the flipped graph = change_x / change_y
      = 1 / change_y/change_x
      = 1 / slope of our dot on the original graph
      but with an input of f(x) or y than x itself (don't be confused with names as i warned above!)



      3. conclusions (in a mathy sense)
      g'(f(x)) = g'(y) = 1 / f'(x)
      and
      f'(g(x)) = f'(y) = 1 / g'(x)

      with a catch:
      the names like x, y, f(x), g(x), inverse, and d/dx are just names for human conveneince. thus, if you want to really understand a concept like this one to the bottom, better not give a heavy weight on a specific name. thus, your asking of a graphical understanding is quite reasonable and helpful not just for you but for the others here

      with another catch for the catch above:
      but if you have a strong sense of handling the naming conventions, it's indeed convenient to talk with others and manipulate equations quite freely like Sal did in this video

      in short, grasp the concept (maybe) graphically or whatever in your hand first. second, digest how the namings and equations are related to the concept. and then use them to dive deeper or derive another equation with your interest



      hope this to give you a little smell of how fun math could be
      (5 votes)
  • leaf red style avatar for user euan
    I'm not really sure how 1/e^x comes from g'(f(x)) at around , can anyone explain that for me, please?
    (3 votes)
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    • leaf red style avatar for user Cavan P
      When plugging the value 'x' into g'(x), we get 1/x. Now, instead of 'x', we are plugging f(x) into the function (i.e 1/f(x)). You can then replace 'f(x)' with the actual function (in this case was 1/e^x), resulting in 1/(1/e^x).
      (6 votes)
  • blobby green style avatar for user Din Shpati
    My left ear loved the video ;-)
    (2 votes)
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  • blobby green style avatar for user mahektalreja012
    why is the inverse of y=e^x is equal to x=e^y? i'm not able to understand this. i checked a video of finding inverse of rational functions on khanacademy and sal just switched the positions of x and y in the function.
    (1 vote)
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    • male robot donald style avatar for user Venkata
      This comes from the process of finding inverses. Basically, if a function takes in x and outputs y, the inverse will take in y and output x. So, building upon this idea, if we consider y = e^(x), this takes in a value of x and outputs a value of y. So, the inverse function must do the opposite, which x = e^(y) does! It takes in a value of y and outputs x. Does that make sense now?
      (8 votes)
  • male robot hal style avatar for user Zev Oster
    I really wish we had come up with some symbolization of the inverse other f^-1(x), as it makes it seem like it is the recipricole of the function rather than the inverse. Why is it like that?
    (4 votes)
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    • male robot donald style avatar for user Venkata
      A good point. I've seen a lot of people mention this and this is really some of those "bad notation" instances we see in Math. Just to avoid confusion, we use parenthesis (f^(-1)(x) is the inverse of f(x) while (f(x))^(-1) is the reciprocal of f(x))
      (2 votes)
  • blobby green style avatar for user chris.freilich
    I'm still stuck on the chain rule concept, shown again here. We've learned previously that the notation d/dx [g(x)] is the same as g'(x). So here, we see that d/dx [g(f(x)] is actually g'(f(x))*f'(x). So, d/dx [g(x)] should simultaneously be g'(f(x)) AND g'(f(x))*f'(x). I get that somehow the phrase 'with respect to...' plays a part here, but in this video, there's no indication that g'(f(x)) is 'with respect to' anything in particular. If someone could help me with this, I feel like it's all downhill from here!
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      Yes, one of the issues with Lagrange notation ("prime notation") is that it doesn't state what variable we are differentiating with respect to.

      However, it is commonly accepted that
      𝑔'(𝑥) = 𝑑∕𝑑𝑥 [𝑔(𝑥)]
      𝑔'(𝑓(𝑥)) = 𝑑∕𝑑𝑓 [𝑔(𝑓(𝑥))]
      𝑔(𝑓(𝑥))' = 𝑑∕𝑑𝑥 [𝑔(𝑓(𝑥))]
      (5 votes)
  • boggle blue style avatar for user Bryan
    What happens if g'(f(x)) = 0? Then we can't go from

    g'(f(x)) * f'(x) = 1
    to
    f'(x) = 1/g'(f(x))

    since we can't divide by 0 right? So what would f'(x) be in that case?
    (1 vote)
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  • blobby green style avatar for user tejaskravi
    what does defferentiating mean if anyone could explain in simpler terms
    (1 vote)
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    • male robot donald style avatar for user Venkata
      It's the process of finding the rate of change of something, but over a very small interval. Like, finding the distance covered over a very small interval of time gives the instantaneous speed. If we had the distance given as a function of time, differentiating it would give us the instantaneous speed at any instant of time.

      All in all, it's like finding the slope of a graph (∆y/∆x) but we let ∆x tend to zero
      (2 votes)

Video transcript

- [Instructor] So let's say I have two functions that are the inverse of each other. So I have f of x, and then I also have g of x, which is equal to the inverse of f of x. And f of x would be the inverse of g of x as well. If the notion of an inverse function is completely unfamiliar to you, I encourage you to review inverse functions on Khan Academy. Now, one of the properties of inverse functions are that if I were to take g of f of x, g of f of x, or I could say the f inverse of f of x, that this is just going to be equal to x. And it comes straight out of what an inverse of a function is. If this is x right over here, the function f would map to some value f of x. So that's f of x right over there. And then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it, using the chain rule. And we're gonna get a pretty interesting result. What I want to do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d/dx on the left-hand side, d/dx on the right-hand side. And what are we going to get? Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x. So that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x. And then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we'd use the chain rule to take the derivative of the left-hand side. But if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f. So f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is, if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. All right, if that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be? Well, this one of those amazing results in calculus. One of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out. So we should get a result, f prime of x, e to the x should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true? Yes, it is. One over, one over e to the x is just going to be e to the x. So it all checks out. And you could do the other way because these are inverses of each other. You could say g prime of x is going to be equal to one over f prime of g of x because they're inverses of each other. And actually, what's really neat about this, is that you could actually use this to get a sense of what the derivative of an inverse function is even going to be.