A function's derivative being not defined for some X but having a limit

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Discussion Overview

The discussion revolves around the differentiability of a function at a point where its derivative is not defined, specifically at X=0. Participants explore the implications of limits and definitions in the context of calculus, particularly focusing on whether a function can be considered differentiable if its derivative approaches a limit but is not defined at that point.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the function can be concluded as not differentiable at X=0 or if the derivative can be considered as 0 based on the limit approaching 0.
  • Another participant asserts that differentiability at X=0 requires the existence and equality of limits of the difference quotient, regardless of the lack of a closed form for the derivative.
  • A participant emphasizes that a function cannot be differentiable at points where it is not defined, suggesting a specific definition for the function that would allow for differentiability at X=0.
  • Further clarification is provided that if a function is defined at X=0 and the limit of the derivative exists, then it can be considered differentiable, regardless of the absence of a closed expression.

Areas of Agreement / Disagreement

Participants express disagreement regarding the implications of a derivative not being defined at a point. While some argue that this indicates the function is also not defined at that point, others suggest that a defined function can still have an undefined derivative.

Contextual Notes

Participants discuss the necessity of defining the function appropriately at X=0 for differentiability to be meaningful, highlighting the dependence on definitions and the limits involved.

Mcp
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Let's say I have a function whose derivative is (tan(x)-sin(x))/x. It is not defined for X=0 but as X approaches 0 the derivative approaches 0, so should I conclude that my function is not differentiable at X=0 or should I conclude that the derivative at X=0 is 0.
 
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The function is differentiable at ##x=0## if the limits ##\lim_{h \to \pm 0} \dfrac{f(x+h)-f(x)}{h}## exist and are equal. That you cannot find a closed form which reflects this existence is irrelevant.
 
A function cannot be differentiable at points where it is not defined.

However if you define f(x) = \begin{cases}<br /> \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, &amp; x \neq 0, \\<br /> 0, &amp; x = 0\end{cases}<br /> then f will be differentiable and <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.<br />
 
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pasmith said:
A function cannot be differentiable at points where it is not defined.

However if you define f(x) = \begin{cases}<br /> \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, &amp; x \neq 0, \\<br /> 0, &amp; x = 0\end{cases}<br /> then f will be differentiable and <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.<br />
Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?
 
Mcp said:
Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?
No. The other way around. No function defined, no derivative. If the function is defined and the derivative (as limit) exists, then it is differentiable, regardless of there is a closed expression or not. In your case we have
$$
f'(x) = \begin{cases} \dfrac{\tan(x)-\sin(x)}{x} & \text{ if }x\neq 0\\ 0 & \text{ if } x=0 \end{cases}
$$
The function ##f(x)## whose derivative is ##f'(x)## has to be defined at ##x=0## in an appropriate, i.e. differentiable way. If ##f(x)## isn't defined at ##x=0##, then the question whether ##\left. \dfrac{d}{dx}\right|_{x=0}f(x)## exists or not doesn't make sense.
 
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