If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Calculus 1

### Unit 3: Lesson 5

Differentiating inverse functions

# Derivatives of inverse functions: from equation

Given that f(x)=½x³+3x-4, Sal evaluates the derivative of the inverse of f at x=-14.

## Want to join the conversation?

• 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. •  ``  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                dxThe 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.
• 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. • 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))
• Correct me if I'm mistaken, but can't you switch x and y and then solve for y to get the inverse derivative? • 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.
• • 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? • I cant understand how f(h(x))=x, can you explain this? • 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
• 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)) • 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. • 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!
• is it possible to start by getting the inverse of the function (without solving completely) by essentially switching the x's and y's, then do implicit differentiation. • The lecture confused me because of the notation used. For example, does h’(x) refer to dh/dx, dh/dy, df⁻¹/df or ? and derivative at what point?

Here’s an alternative explanation:
Given (x, y) in the coordinate plane such that y = f(x), and given the inverse of f exists and also is continuous at (x , y), such that x = f⁻¹ (y) , or f⁻¹ ( f(x) ) = x:

Then by Leibniz' notation:
dy/dx @ (x) = df (x)/dx …………and……….. dx/dy @ (y) = d f⁻¹(y)/dy ……(1)

The fundamental relationship described in lecture (with elaborate proof not shown) is:
dx/dy @ (y) = 1 / [ dy/dx @ (x) ]

Which now is more intuitive – although do note how I tracked the variables, y and x. Substituting from (1) above:

df⁻¹(y)/dy = 1 / [df (x)/dx ] ….. with handy note that on the right side of the equation, x = f⁻¹ (y) .
So slopes of f and f⁻¹ at (x, y) are reciprocals.

The upcoming table problems illustrate my confusion with the notation used in the lecture. Let's say you have function f(x) and its inverse, h [ shown in the lecture and most of the problems as "h(x)" ]. Coordinate pairs listed in the table as (x, h(x) ) actually could be interpreted in 2 ways:

Say x=a and b=f(a) represents a specific point (a, b) at which to calculate f' or h'. Does a table value pair (x, h(x)) refer to the point (a, h(a)), or does it refer to (x=h(b), b)? Well, actually it refers to the latter.

To avoid making substitution errors when I do these problems, I start out by copying and rearranging the table so that all points and functions map to the same x-y coordinate system. Simply put, I swap each pair (x, h(x)) so they map to the same Cartesian x-y plane as (x, y=f(x)). Thus, I map pairs (x, h(x)) to => (h(y), y), because in the specific x-y plane, x = h(y) and y = f(x).

In my opinion, the K.A. convention of using the same symbol for the independent variable of both a function and its inverse is prone to error, especially when their functions are being used in the same formula, or when their values are being mapped to the same Cartesian plane. 