"inverse function and its derivative"

[martis880]

(1) why are we using y as the horizontal axis

In the lecture “inverse function and its derivative” it’s not clear to me why we decided to plot the inverse using y as the horizontal axis.

We say g(f(x)) = x and so g(x) = f^-1(x);

it was hard to follow why he started to use g(y)

(2) I didn’t follow the logic here.

(A) It makes sense to me that in the graphs as labeled delta g / delta y = delta x / delta f.

(B) It also makes sense to me that the limit as delta x goes to zero of delta f / delta x is the definition of f’(x) and the limit as delta y goes to zero of delta g / delta y is the definition of g’(y).

However… delta y and delta x in (A) are not the same distance, so the jump from (A) and (B) to (C) wasn’t clear to me.

[paulinpaloalto]

The fundamental point is that you get from the graph of the function to its inverse by reflecting the graph about the 45 degree line (y = x). That is the fundamental “symmetry” and they are then showing the implications of that for the derivatives of the two functions.

As to the labels on the axes, those are just conventions. You could have called them Fred and Barney if you wanted to be perverse. It’s just a label, so it doesn’t matter. But I think they chose y for the horizontal axis in the inverse graph precisely to emphasize the “symmetry” point that I was making above. That is the reflection of what was the y axis in the first graph.

For the last point, I think you’re just missing the symmetry: the point is that it is \Delta g that is equal to \Delta x, right?

[martis880]

@paulinpaloalto

The fundamental point is that you get from the graph of the function to its inverse by reflecting the graph about the 45 degree line (y = xy=xy = x). That is the fundamental “symmetry” and they are then showing the implications of that for the derivatives of the two functions.

As to the labels on the axes, those are just conventions. You could have called them Fred and Barney if you wanted to be perverse. It’s just a label, so it doesn’t matter. But I think they chose yyy for the horizontal axis in the inverse graph precisely to emphasize the “symmetry” point that I was making above. That is the reflection of what was the yyy axis in the first graph.

This is more of a critique of the presentation style than me saying it was wrong, bc yes the labels are arbitrary at the end of the day.

If you have a curve, and you want to plot it’s inverse, you draw a new curve that is the first curve reflected about the x=y line. But then if you also switch the axes labels (you make y the horizontal and x the vertical), you are essentially reflecting twice. So imo it’s confusing to do things this way. I felt like he didn’t really explain why he choose y to be the horizontal axis he just did it.

@paulinpaloalto

For the last point, I think you’re just missing the symmetry

I see how delta g and delta x are the same in (A). But still not clear to me why this proves (C). You can construct two curves that are not related at all and go through the same points, and you’d have the same claims as (A). But then (C) would not hold. For example, in the second graph, if instead of g(y) = y^(1/2) you draw a line through the two points, and use that as your curve. In that case you’d have the same claim as (A) but then (C) wouldn’t necessarily follow. Something is missing to prove (C).

[paulinpaloalto]

@martis880

But then if you also switch the axes labels (you make y the horizontal and x the vertical), you are essentially reflecting twice.

You would be reflecting twice if the vertical axis were labelled x, but it’s not, right? It’s labelled g(y), which is the point.

@martis880

I felt like he didn’t really explain why he choose y to be the horizontal axis he just did it.

You just have to read the labels on both axes and then it all makes sense.

@martis880

You can construct two curves that are not related at all and go through the same points, and you’d have the same claims as (A). But then (C) would not hold.

But the whole point is that what we’re discussing here is not some arbitrary case of a different function in the second graph. It’s the inverse function, so all those relationships and how you get from one to the other in that instance should be obvious.

[paulinpaloalto]

If you agree that A is true:

\displaystyle \frac {\Delta g}{\Delta y} = \frac {\Delta x}{\Delta f}

Then the next step is:

\displaystyle \frac {\Delta x}{\Delta f} = \frac {1} {\frac {\Delta f}{\Delta x}}

Substitute the values in B and you’re done, right?

[martis880]

no, but I apologize, if I get some more time I will try to explain my concerns more clearly

[farmerinatechstack]

I’m still trying to fully grasp the concept too. One resource I’ve preferred so far is the following from Khan Academy: https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-3/v/derivatives-of-inverse-functions

[Hartej_Dhiman]

There are a lot of times when Luis Serrano’s material is way more confusing and Khan Academy’s comes to the rescue…a paid course charging $50 a month, per month should not have content that is more difficult to interpret than what is for free out there.

In the end, the emerging pattern is that a paid course (this Coursera course) is worse off at educating than a free course.