# Help with a proof!

1. Jun 28, 2012

### danerape

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

Thanks

2. Jun 28, 2012

### danerape

and vice-versa of course

3. Jun 28, 2012

### 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)

4. Jun 28, 2012

### danerape

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

5. Jun 28, 2012

### danerape

the above is in regard to the link provided

6. Jun 28, 2012

### vela

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

7. Jun 28, 2012

### 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.....

Dosen'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, dosen't that mean that the proof above is not general, in other
words, what if I cant solve y=f(x) for x as a function of y explicitly? Then I cant 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?

Last edited: Jun 28, 2012
8. Jun 28, 2012

### christoff

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.

Last edited: Jun 28, 2012
9. Jun 28, 2012

### 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)

Last edited: Jun 28, 2012