Does the Reciprocal of a Derivative Always Hold True?

[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?

 

[Gear300]

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

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.

 

[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?

 

[torquil]

http://en.wikipedia.org/wiki/Inverse_function_theorem

 

[Prologue]

Holy crap, thanks for the link.