https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html

A theorem is something that needs to be proved.
A definition is something that captures a useful concept.

Your counterpoints definition may be useful in grasping or memorizing both the idea behind the relation between the two derivatives and the procedure for computing one of them in terms of the other. Still, it does not replace the theorem. I, therefore, do not see to which it is an alternative.

Yes, I realized this after posting. I wrongly inserted the word definition in the post when I meant to say theorem.

"Your counterpoints definition may be useful in grasping or memorizing both the idea behind the relation between the two derivatives and the procedure for computing one of them in terms of the other. Still, it does not replace the theorem. I, therefore, do not see to which it is an alternative."

I suppose what I am asking is if my explanation is sufficient to show the relationship without mentioning the form:

(d/dx) = 1 / <(d/dx)[f^(-1)'(x)>]

...which for me is hard enough to type, let alone explain to high school students...

It was not my intention to "rewrite" the theorem, only to give some clarity to the idea presented in the theorem without having to mention the rather involved notation....

so not an alternative to the theorem itself, but rather an alternative way to present the theorem....

I hope I am making sense...

Still, if the idea is that, by the end of the day, the students know how to compute the derivative of the inverse function, then sometine around the lunch you'll have to produce the formula. You may use the chain rule to derive the formula without any resort to coordinates.

But whatever works for you - just do it. I am always apprehensive of general preferences. I do not believe there is always the best approach that works for all students and for all teachers.

If f and g are inverse functions, then f having the point (x,y) means that g has the point (y,x). Since the tangents as well as the functions are reflected over the line y = x,

f'(x) = 1 / g'(y)

Perhaps, as an extra contrivance, one may distinguish between "variable" x and y and "constant" x₀ and y₀. The reason is that in y = x and f'(x) = 1 / g'(y), x and y play different roles. In

f'(x) = 1 / g'(y)

y is a value for x, so that I'd rather write

f'(x₀) = 1 / g'(y₀).

It may work even better.