Is the derivative of infinity zero?

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    Derivative Infinity
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Discussion Overview

The discussion revolves around the concept of taking the derivative of infinity, exploring whether it can be considered a constant or if it is inherently undefined. Participants engage in a debate regarding the mathematical implications and definitions surrounding infinity in the context of calculus.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant argues that the derivative of infinity cannot be taken because infinity is not a number and thus does not have a defined slope.
  • Another participant suggests that infinity could be treated as a constant, leading to the conclusion that its derivative would be zero.
  • A different viewpoint emphasizes that infinity is undefined and that discussing its derivative lacks practical or theoretical purpose.
  • One participant introduces the idea that differentiability requires continuity, implying that if a function approaches infinity, it would be discontinuous and therefore not differentiable.
  • Another participant acknowledges the lack of practicality in the question but expresses appreciation for the interest in the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus; multiple competing views remain regarding the status of infinity in calculus and its implications for differentiation.

Contextual Notes

The discussion highlights the ambiguity surrounding the definition of infinity, its treatment in mathematical contexts, and the conditions required for differentiability, which remain unresolved.

jimmypoopins
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i got into a minor argument with a buddy of mine, he said the derivative of infinity is zero, and i argued that you can't take the derivative of infinity.

my argument was that by definition of derivative there isn't a function that can equal infinity, so you can't take the derivative of it. also, even though infinity isn't a number, theoretically infinity + 1 = infinity so it's increasing, but infinity - 1 = infinity, i.e. you can't find a slope for it at any point.

his argument was that infinity is a constant, so then it is differentiable.

i believe I'm correct but I'm not formally aware as to why, and i was wondering if you guys could give me some insight.

thanks
 
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Infinity isn't even a number. It's like saying "the derivative of chair is 0". You're right in saying that it is undefined.
 
According to my Ti-89 infinity is treated as a sort of constant, and thus the derivative is zero.
 
Infinity is undefined

To be honest the 'derivative' of infinity is very likely to be undefined and it should be of no concern as it has no practical nor theoretical purpose. If you just want a "what if it happened to be practical" answer, anything differentiable must be continuous. If you think of a function y(x) with a vertical asymptote at x = 0 (for example) where lim(x -> 0+) y(x) = (infinity) and lim(x -> 0-) y(x) = (infinity), the function is considered discontinuous at x = 0 because of the infinite limits and is therefore not differentiable (as differentiability requires continuity). This is very informal reasoning but you can see that if at any point in a function (even one defined to be 'infinity', which I'm pretty sure you cannot do) infinity is reached, then there is a discontinuity and therefore no differentiability.
 
thanks for the replies, i think that's a sufficient enough answer for both of us.

yeah, i know it's not practical at all but the question was bothering both of us, even if it wasn't practical at all.
 
Okay, sorry for being blunt, I think it's good that you were interested.
 

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