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

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In summary, if you want to find the derivative of |x|, you need to understand why the derivative of |x| is 1 or -1 at x= 0.
  • #1
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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  • #2
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} $
 
  • #3
Thank you Skeeter, I have managed to find the answer myself, using d(sin(t))/dt = cos(t)
 
  • #4
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).
 
  • #6
hmmm … didn’t notice the absolute value bars. Due for a vision check anyway.
 
  • #7
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 \(\displaystyle \frac{x}{|x|}\) 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]
 
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FAQ: Why is the derivative of |x| not defined at x=0?

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

The derivative of |x| is not defined at x=0 because the function is not differentiable at this point. This means that the slope of the tangent line at x=0 does not exist, and therefore the derivative cannot be computed.

2. Can the derivative of |x| be approximated at x=0?

No, the derivative of |x| cannot be approximated at x=0 because the function is not continuous at this point. This means that the limit of the difference quotient, which is used to approximate the derivative, does not exist.

3. What is the geometric interpretation of the derivative of |x| at x=0?

The geometric interpretation of the derivative of |x| at x=0 is that there is a sharp corner or "cusp" at this point. This is because the function changes direction abruptly at x=0, making it impossible to draw a unique tangent line.

4. Is the derivative of |x| defined at other points?

Yes, the derivative of |x| is defined at all points except x=0. This is because the function is differentiable everywhere except at points where it is not continuous or has a sharp corner.

5. How is the derivative of |x| related to the derivative of x^2?

The derivative of |x| is related to the derivative of x^2 in that they have the same derivative at all points except x=0. This is because the function |x| can be rewritten as x^2 for all values of x except x=0. At x=0, the derivative of x^2 is 0, while the derivative of |x| does not exist.

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