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### Course: AP®︎/College Calculus BC>Unit 3

Lesson 4: Differentiating inverse functions

# 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!).

## Want to join the conversation?

• 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.
• 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.
• Will someone please explain how to go from y=e^x to x=e^y algebraically?
• You can't, those two equations aren't equivalent.

If you meant go from y=e^x to x=ln(y), you just take the natural log of both sides and simplify ln(e^x) as x.
• 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
• 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
• I'm not really sure how 1/e^x comes from g'(f(x)) at around , can anyone explain that for me, please?
• 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).
• My left ear loved the video ;-)
• What did you have in your right ear? ;P
• 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)
• 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?
• 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?
• 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))
• 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!
• 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))]
• 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)
• If 𝑔'(𝑓(𝑥)) = 0, then 𝑔'(𝑓(𝑥))⋅𝑓 '(𝑥) = 0, which means that 𝑔(𝑓(𝑥)) is a constant, and 𝑓 and 𝑔 are not each other's inverses.