# Inverting differentials

1. Dec 5, 2009

### Allday

This is a simple dummy question. What are the conditions under which the following relationship holds,

dx/dt = Inverse(dt/dx) = 1/(dt/dx)

meaning if I want to do a derivative and I know t(x) but not x(t) when can I just calculate dt/dx and put it over 1 to get dx/dt. I see this in derivations a lot, but always wonder what the realm of applicability is.

thanks.

2. Dec 5, 2009

### HallsofIvy

Staff Emeritus
First those are not "differentials", they are "derivatives". I mention that because I don't believe the "inverse" of a differential exists.

1/(dx/dt)= dt/dx (it would be better to say "reciprocal" rather than "inverse") as long as x(t) has a differentiable inverse function- and dx/dt is not 0. And it can be shown that a differentiable function has a differentiable inverse on some neighborhood of $(t_0, x(t_0))$, dx/dt is not 0 so you really only need the condition that dx/dt is not 0.

3. Dec 5, 2009

### arildno

ALLDAY:

Suppose you have an invertible function X(t), i.e, there exists a function T(x), so that
the composite function h(x)=X(T(x))=x, and the composite function H(t)=T(X(t))=t, for all values of x and t.

Then we have, for example:
$$\frac{dh}{dx}=\frac{dX}{dt}\mid_{t=T(x)}*\frac{dT}{dx}=1$$

Thus, we get:
$$\frac{dT}{dx}=\frac{1}{\frac{dX}{dt}\mid_{t=T(x)}}$$

Let us take an example:

Let $$X(t)=t^{3}\to{T}(x)=\sqrt[3]{x}$$

Then, we have:
$$\frac{dT}{dx}=\frac{1}{3}x^{-\frac{2}{3}}$$
whereas:
$$\frac{dX}{dt}=3t^{2},\frac{dX}{dt}\mid_{t=T(x)}=3x^{\frac{2}{3}}$$
and the reciprocal of this agreeing with the former expression.

4. Dec 6, 2009

### Allday

thanks Halls and arildno, that makes it clear.