Help Proving (dy/dx)=1/(dx/dy)

[danerape]

How does one go about rigorously proving that (dy/dx)=1/(dx/dy)?

Thanks

 

[danerape]

and vice-versa of course

 

[christoff]

Good question. To answer that, first ask yourself, what do you mean by \frac{dy}{dx}? What do you mean by \frac{dx}{dy}?

I recommend looking at this post. https://www.physicsforums.com/showthread.php?t=63886 (it's the same as your question)

 

[danerape]

how can y=f(x) define y implicitly? that seems explicit.

 

[danerape]

the above is in regard to the link provided

 

[vela]

You need to show some more effort here, per the forum rules. We're not here to do your homework for you.

 

[danerape]

This isn't for homework. Here is my attempt at a proof.

1.Say we are given a function y=f(x)

2.Differentiate implicit w.r.t. y and we see...

1=(df/dx)(dx/dy)

3. Now we can solve for the derivative of our choice!

Now, my question is...

doesn't the result of 2(implicit differentiation) imply that we thought of y=f(x) as y=f(x(y))? In other words,
implicit diff stems from the chain rule right? So, doesn't that mean that the proof above is not general, in other
words, what if I can't solve y=f(x) for x as a function of y explicitly? Then I can't say y=f(x) is the same as y=f(x(y))
and obtain the right side of 2 which results from the chain rule.

Any ideas?

 

[christoff]

...what if I can't solve y=f(x) for x as a function of y explicitly?

Have you ever heard of the inverse function theorem? I will summarize it for a function of one variable, since the formulation is a bit easier. The theorem effectively says that if you have a function f:A\subset \mathbb{R}\rightarrow \mathbb{R} which is differentiable and has a continuous derivative on some open set A, and f'(a)\neq 0 at some point a\in A, then f^{-1} exists and is defined in some neighbourhood of a.

In the context of your problem, this means that as long as f is C^1, and its derivative isn't zero one the whole domain, then SOMEWHERE in its domain, f is invertible. Suppose that f is invertible on U\subset A as above. Then we have a function f^{-1} : f(U)\rightarrow A which satisfies f^{-1}(f(x))=x for all x \in U.

In conclusion, if f(x)=y, then so long as y is in f(U), we have x=f^{-1}(y).

In short... you can always solve y=f(x) for x as a function of y somewhere, as long as f is continuously differentiable and isn't a constant function.

 

[tt2348]

Well.. A neat lil trick you can do is (given f^-1=g), f(g(x))=x, derivative of bother sides is f'(g(x))*g'(x)=1, proceed from there :)
(assuming invertibility)