Why is the derivative of |x| not defined at x=0?

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Discussion Overview

The discussion revolves around the differentiation of the absolute value function, specifically |sin(t)| and |x|, and the implications of differentiating these functions at points where they change behavior, such as x=0.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant inquires about differentiating |sin(t)| and whether to use known derivatives or first principles.
  • Another participant suggests using first principles to find the derivative of sin(t) and mentions necessary limits for the process.
  • A participant explains that |sin(t)| can be treated as two separate cases based on the sign of sin(t), leading to different derivatives in each case.
  • A later reply discusses the derivative of |x|, noting that it can be expressed as x/|x| for x non-zero and highlights that the derivative does not exist at x=0 due to differing one-sided limits.
  • Participants emphasize the importance of understanding the reasoning behind derivative formulas rather than merely memorizing them.

Areas of Agreement / Disagreement

Participants generally agree on the approach to differentiate |sin(t)| and the reasoning behind the non-existence of the derivative of |x| at x=0. However, there is no explicit consensus on the best method to explain or derive these concepts.

Contextual Notes

The discussion includes various approaches to differentiation and highlights the need for understanding limits and one-sided behavior, particularly at critical points like x=0.

Purplepixie
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I would like to know how to differentiate |sin(t)| to obtain d(|sin(t)|)/d(t). Thank you!
 
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do you want the fact that $\dfrac{d (\sin{t})}{dt} = \cos{t}$, or do you want to find the derivative using first principles, i.e. from the limit below?

$\displaystyle \dfrac{d (\sin{t})}{dt} = \lim_{h \to 0} \dfrac{\sin(t+h)-\sin{t}}{h}$

Finding the derivative using first principles requires prior knowledge of values for the following limits ...

$\displaystyle \lim_{x \to 0} \dfrac{1-\cos{x}}{x} \text{ and } \lim_{x \to 0} \dfrac{\sin{x}}{x} $
 
Thank you Skeeter, I have managed to find the answer myself, using d(sin(t))/dt = cos(t)
 
To differentiate |sin(t)| separate it into two problems:
1) For sin(t)> 0 (so for t from $2n\pi$ to $(2n+ 1)\pi$ for any integer n) |sin(t)|= sin(t) and the derivative is cos(t).

2) For sin(t)< 0 (so for t from $(2n+ 1)\pi$ to $(2n+ 2)\pi$ for any integer n) |sin(t)|= -sin(t) and the derivative is -cos(t).
 
hmmm … didn’t notice the absolute value bars. Due for a vision check anyway.
 
Purplepixie said:
Thank you Country Boy, this is what I used: https://proofwiki.org/wiki/Derivative_of_Absolute_Value_Function
Yes, the derivative of |x| can be written as [math]\frac{x}{|x|}[/math] for x non-zero and does not exist at x= 0. Do you see why? It is not sufficient to memorize formulas. You need to understand why they are true!

If x> 0 |x|= x so the derivative of x is 1 and [math]\frac{x}{|x|}= \frac{x}{x}= 1[/math]. If x< 0, |x|= -x so the derivative of -x is -1 and [math]\frac{x}{|x|}= \frac{x}{-x}= -1[/math]. For x= 0, the derivative of |x| is given by [math]\lim_{h\to 0}\frac{|h|- 0}{h}= \lim_{h\to 0}\frac{|h|}{h}[/math]. But if h> 0 that fraction is 1 while if h< 0 that fraction is -1. The limit "from the right" (h> 0) is 1 while the limit "from the left" (h< 0) is -1. Since the two "one-sided limits" are different, the limit itself does not exist. |x| has no derivative at x= 0.[/math]
 
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