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## Calculus 1

# Derivatives of inverse functions: from equation

AP.CALC:

FUN‑3 (EU)

, FUN‑3.E (LO)

, FUN‑3.E.1 (EK)

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.(6 votes)- You had written:
`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 dx

The 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.(37 votes)

- At1:05, 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.(5 votes)
- 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))(15 votes)

- Correct me if I'm mistaken, but can't you switch x and y and then solve for y to get the inverse derivative?(5 votes)
- 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.(11 votes)

- What is the inverse function of f(x) = 2x^3-5(2 votes)
- 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?(3 votes)
- A function has only one derivative. So if two expressions are equal, i.e. the functions are one and the same, their derivatives must be equal as well, since there is no second value they could take on.(3 votes)

- I cant understand how f(h(x))=x, can you explain this?(2 votes)
- This is a rule that we learned from regular inverses. Since they are inverses, this must be true. For example, if

the inverse of that is`f(x) = sqrt(3x - 4)`

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):`h(x) = (x^2 + 4)/3`

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(5 votes)

- 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))(3 votes)
- Both are equivalent. Apply the chain rule differently and you get a different equation with the same meaning.(1 vote)

- 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.(2 votes)- 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!(2 votes)

- 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.(2 votes)
- Given the equation above the switching might become messy....First off try to work on simple inverse equations (f(x) =2x and its inverse g(x) = x/2) to get the intuition/understanding of how he arrived at those equations....(1 vote)

- 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.(2 votes)- The notation used is Leibniz' d/dx form and Lagrange's f'(x) form.

h'(x) refers to the derivative of the function h, at point x.

The derivative of h at x is commonly expressed as h'(x) or dh/dx.

Beyond that, you lost me.(1 vote)

## Video transcript

- [Voiceover] Let F of X be
equal to one half X to the third plus three X minus four. Let H be the inverse of F. Notice that F of negative
two is equal to negative 14. And then they're asking us
what is H prime of negative 14? And if you're not familiar
with the how functions and their derivatives
relate to their inverses and the derivatives of the
inverse, well this will seem like a very hard thing to do. Because if you're attempting
to take the inverse of F to figure out what H is well, it's tough to find, to take to figure out the
inverse of a third degree a third degree polynomial
defined function like this. So, the key the key, I guess property to realize, or the key truth to realize if F and H are inverses then H prime of X H prime of X is going to be equal to is going to be equal to one over F prime of H of X. One over F prime of H of X. And you could now use this in order to figure out what H prime of negative 14 is. Now I know what some of you are thinking, because it's exactly
what I would be thinking if someone just sprung this on me is where does this come from? And I would tell you,
this comes straight out of the chain rule. We know that if a function and its inverse we know that if we have a
function and its inverse that F of F of the
inverse of our function. So F of H of X. F of H of X. We know that this is going to be to X. This literally, this is comes out of them being
each others inverses. We could have also said H of F of X will also be equal to X. Remember, F is going to map or H is going to map
from some X to H of X. And then F is going to map
back to that original X. That's what inverses do. So that's because they are inverses. This is by definition,
this is what inverses do to each other. And then if you took the derivative of both sides of this what would you get? Let me do that. So if we took the derivative of the both sides of this D D X on the left hand side. D D X on the right hand side. I think you see where this is going. You are essentially gonna
get a version of that. The left hand side used the chain rule. You're going to get F prime of H of X. F prime of H of X times H prime of X comes straight out of the chain rule is equal to, is equal
to the derivative of X is just going to be equal to one. And then you derive,
you divide both sides by F prime of H of X and you get our original property there. So now with that out of the way let's just actually apply this. So, we want to evaluate H prime of 14. Or sorry, H prime of negative 14. Is going to be equal to one over F prime of H of negative 14. H of negative 14. Now had they given us H of negative 14. But they didn't give it to us explicitly, we have to remember that
F and H are inverses of each other. So F of negative two is negative 14. Well, H is gonna go from
the other way around. If you input negative 14 into H you're going to get negative two. So H of negative 14 well, this is going to be equal to negative two. Once again, they are
inverses of each other. So H of negative 14 is equal to negative negative two. And once again, I just
swapped these two around. That's what the inverse function will do. If you're wrapping from if F goes from negative two to negative 14 H is going to go from negative 14 back to negative two. So now we want to evaluate
F prime of negative two. Well, let's figure out
what F prime of X is. So, F prime of X is equal to remember the power rule,
so three times one half is three halves times X to the three minus one power which is the second power. Plus the derivative of three X with respect to X which
that's just going to be three. And you could do that,
it's just the power rule. But this was X to the first power, one times three, X to the zero power, but X to the zero is just one so you're just left with three. And derivative of a concept
that's just gonna be zero. So that's F prime of X. So F prime of, F prime of negative two is going to be three halves times negative two squared
is four, positive four. So plus three. So, this is going to be equal to two times three plus three. So, six plus three is equal to nine. So this denominator here is
going to be equal to nine. So this whole thing is
equal to one over nine. So this involved, this
was something you're not going to see every day. This isn't that typical
problem in your calculus class. But it's interesting.