The Reciprocal of a Derivative

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Main Question or Discussion Point

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

[tex]\frac{ds}{dx}=\frac{1}{\frac{dx}{ds}}[/tex]

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

[tex]s(x)=x^{2}[/tex]

so that

[tex]\frac{ds}{dx}=2x[/tex]

Now to get the inversion, use [tex]x=\pm\sqrt{s}[/tex]

[tex]\frac{dx}{ds}=\pm\frac{d}{ds}\sqrt{s}=\pm\frac{1}{2\sqrt{s}}[/tex]

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,

[tex]\frac{dx}{ds}=\frac{1}{2x}[/tex]

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

[tex]\frac{dx}{ds}=\pm\frac{1}{2\sqrt{s}}[/tex]

to get

[tex]\frac{dx}{ds}=\pm\frac{1}{2\sqrt{x^{2}}}[/tex]

which results in

[tex]\frac{dx}{ds}=\pm\frac{1}{2x}[/tex]

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

[tex]\frac{1}{\frac{dx}{ds}}=\pm 2x[/tex]

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?
 
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  • #2
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[tex]\frac{dx}{ds}=\pm\frac{1}{2\sqrt{x^{2}}}[/tex]
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.
 
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  • #3
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So it should be something more like:

[tex]\frac{dx}{ds}=\pm\frac{1}{2\sqrt{x^{2}}}=\pm\frac{1}{2|x|}=\frac{1}{2x}[/tex]

That takes care of the case that I brought up but does it always work?
 
  • #5
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Holy crap, thanks for the link.
 

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