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

[Purplepixie]

I would like to know how to differentiate |sin(t)| to obtain d(|sin(t)|)/d(t). Thank you!

 

[skeeter]

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} $

 

[Purplepixie]

Thank you Skeeter, I have managed to find the answer myself, using d(sin(t))/dt = cos(t)

 

[HOI]

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).

 

[Purplepixie]

Thank you Country Boy, this is what I used: https://proofwiki.org/wiki/Derivative_of_Absolute_Value_Function

 

[skeeter]

hmmm … didn’t notice the absolute value bars. Due for a vision check anyway.

 

[HOI]

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 \(\displaystyle \frac{x}{|x|}\) 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 \(\displaystyle \frac{x}{|x|}= \frac{x}{x}= 1\). If x< 0, |x|= -x so the derivative of -x is -1 and \(\displaystyle \frac{x}{|x|}= \frac{x}{-x}= -1\). For x= 0, the derivative of |x| is given by \(\displaystyle \lim_{h\to 0}\frac{|h|- 0}{h}= \lim_{h\to 0}\frac{|h|}{h}\). 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]