The Reciprocal of a Derivative

In summary, the conversation discusses the relationship between the derivative and inverse of a function. It is noted that the inverse function may not always work in certain situations, such as when the function is not bijective. The use of the square root in the inversion process is also questioned and a link to the inverse function theorem is provided for further understanding.
  • #1
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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
prologue said:
[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
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
Holy crap, thanks for the link.
 

What is the definition of the reciprocal of a derivative?

The reciprocal of a derivative is a mathematical operation that involves taking the inverse of the derivative of a function. In other words, it is the inverse of the slope of a curve at a specific point.

What is the significance of the reciprocal of a derivative?

The reciprocal of a derivative has several applications in mathematics and science. It is commonly used in optimization problems, where finding the maximum or minimum value of a function requires the use of the derivative and its reciprocal.

How is the reciprocal of a derivative calculated?

The reciprocal of a derivative is calculated by taking the inverse of the derivative of a function. This can be done using the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the function.

What is the relationship between the reciprocal of a derivative and the original function?

The reciprocal of a derivative is closely related to the original function as it represents the inverse of the slope of the function at a specific point. It can also be used to determine the behavior of the original function, such as its concavity and points of inflection.

How is the reciprocal of a derivative used in real-world applications?

The reciprocal of a derivative is used in various real-world applications, such as in physics, engineering, and economics. It is used to analyze the rate of change of a physical quantity, determine the optimal conditions for a system, and predict the behavior of a market or economy.

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