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### Course: Differential Calculus > Unit 3

Lesson 5: 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?

- At3:10, 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)
- At3:03, 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.(38 votes)

- Will someone please explain how to go from y=e^x to x=e^y algebraically?(2 votes)
- 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.(26 votes)

- 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)
- 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)

- I'm not really sure how 1/e^x comes from g'(f(x)) at around4:00, can anyone explain that for me, please?(3 votes)
- 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)

- My left ear loved the video ;-)(2 votes)
- What did you have in your right ear? ;P(7 votes)

- 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?(8 votes)

- 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)
- 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)

- 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)
- 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)

- Hi, I was trying to solve this question and was wondering if anyone knew how to approach it?

d/dx f^−1(4) where f(x) = 4 + 2x^3 + sin (πx/2)

for −1 ≤ x ≤ 1.(1 vote)- So the goal is to evaluate d/dx(f^-1(x)) at x=4.

So f'(x) = 6x^2 + (pi/2)cos([pi/2]x))

Now the question is at what point should the derivative be evaluated. The key thing to note is the coordinates of x and y are swapped for the inverse.

So the x-coordinate for the inverse is 4 however the coordinate is swapped. So the for non-inverse function y=4.

So now the x-coordinate needs to be found for f(x)=4.

=> 4 = 4 + 2x^3 + sin(pi(x)/2)

=> 2x^3 + sin(pi(x)/2) = 0. By inspection x = 0 satisfies the equation.

Alternatively substitute x=4 for the inverse function then find the y-coordinate. The inverse function is

x = 4 + 2y^3 + sin((pi/2)y)

=> 0 = 2y^3 + sin((pi/2)y) since x=4.

Therefore y=0.

So the coordinate for the inverse function is (4, 0) and the non-inverse function (0, 4)

So you choose evaluate the expression using inverse or non-inverse function

Using f'(x) substituting x=0 yields pi/2 as the gradient.

=> d/dx f^-1(4) = (pi/2)^-1 = 2/pi since the coordinates of x and y are swapped.

So the gradient is given by

lim(dx->0) (f(x+dx) -f(x))/( (x+dx)-x )

However the x and y coordinates are swapped so the gradient for the inverse according differentiation by first principles is

lim(dx->0) ( (x+dx)-x ) / (f(x+dx) -f(x))

in others words the gradient of the inverse function is the reciprocal of the gradient of the non-inverse function*__________________*

Alternative solution

From the inverse function: x = 4 + 2y^3 + sin((pi/2)y)

d/dx f^-1(x) =>

1 = 6y^2(dy/dx) + (pi/2)cos([pi/2]y)(dy/dx) (1)

This dy/dx next to each y(in equation (1)) comes from implicit differentiation. This is just a result from chain rule. If you want you can replace y with u and then apply chain rule and you will get the same result.

Equation (1) => 1 = ( 6y^2 + (pi/2)cos([pi/2]y) )dy/dx

=> 1/( 6y^2 + (pi/2)cos([pi/2]y) ) = dy/dx

substituting y=0 yields 2/pi for the derivative for the inverse function.(3 votes)

- 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.(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.