Question 1

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.

Accepted Answer

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.

Question 2

At 1:05, 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.

Accepted Answer

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))

Question 3

Correct me if I'm mistaken, but can't you switch x and y and then solve for y to get the inverse derivative?

Accepted Answer

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.

Question 4

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?

Accepted Answer

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.

Question 5

my calculation is:
f'(x)=1/h'(f(x)) ==> f'(-2)=1/h'(f(-2)) ==> f'(-2)=1/h'(-14), and then solve for h'(-14).

It's still correct, right

Accepted Answer

Yep. Seems correct to me.

Question 6

I cant understand how f(h(x))=x, can you explain this?

Accepted Answer

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

Question 7

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.

Accepted Answer

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!

Question 8

I think we get the same result from g(h(x)) and h(g(x)), correct me if I'm mistaken. How is this possible? I see they yield the same result but I can't get the intuitive sense of it.

Accepted Answer

If g and h are inverse functions, then g(h(x))=h(g(x)). So of course we get the same result, they're the same function.

AP®︎ Calculus AB (2017 edition)

Course: AP®︎ Calculus AB (2017 edition) > Unit 4

Derivatives of inverse functions: from equation

Want to join the conversation?

Video transcript