# The Reciprocal of a Derivative

1. Feb 28, 2010

### Prologue

I have seen this sort of thing over the past few years and it is bothering me, something like this

$$\frac{ds}{dx}=\frac{1}{\frac{dx}{ds}}$$

But it seems to me that this sort of thing only works in certain situations. For example, take s(x) to be

$$s(x)=x^{2}$$

so that

$$\frac{ds}{dx}=2x$$

Now to get the inversion, use $$x=\pm\sqrt{s}$$

$$\frac{dx}{ds}=\pm\frac{d}{ds}\sqrt{s}=\pm\frac{1}{2\sqrt{s}}$$

Now, we know that x is +/- sqrt(s), and I suppose that you could look at it like this: when x is positive, its value is given by +sqrt(s) and when x is negative it given by -sqrt(s) and so we can just directly replace +/- sqrt(s) with x.

So,

$$\frac{dx}{ds}=\frac{1}{2x}$$

But, if you back it up a bit and instead substitute $$s=x^{2}$$ into

$$\frac{dx}{ds}=\pm\frac{1}{2\sqrt{s}}$$

to get

$$\frac{dx}{ds}=\pm\frac{1}{2\sqrt{x^{2}}}$$

which results in

$$\frac{dx}{ds}=\pm\frac{1}{2x}$$

because x squared is always positive. Then invert it to get

$$\frac{1}{\frac{dx}{ds}}=\pm 2x$$

Which doesn't agree with the earlier method. So, depending on how you look at it, it seems that you get different answers. However, this isn't a problem if s is bijective. So, does this 'only' work when the function is bijective? Or, possibly, there is something wrong with my way of viewing these relationships and the +/- isn't there in the final result? What do you think about this inversion in general?

Last edited: Feb 28, 2010
2. Feb 28, 2010

### Gear300

This would also equal to 1/(2x). The square root takes the principle root. Thus the range of Sqr(x2) does not cover the original domain of x. The +/- Sqr(x2) covers for that.

Last edited: Feb 28, 2010
3. Feb 28, 2010

### Prologue

So it should be something more like:

$$\frac{dx}{ds}=\pm\frac{1}{2\sqrt{x^{2}}}=\pm\frac{1}{2|x|}=\frac{1}{2x}$$

That takes care of the case that I brought up but does it always work?

4. Feb 28, 2010

### torquil

5. Feb 28, 2010

### Prologue

Holy crap, thanks for the link.