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## Differential Calculus

### 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?

• 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? • 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? • 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!
(1 vote) • My left ear loved the video ;-) • 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?
• 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.
• 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) • So f^-1(x) should equal x, yes? if g(x) is then defined as the inverse of f(x), shouldn't g(x)=x and g(f(x))=f(x)? Can someone explain?
(1 vote) 