Whats the derivative of the absolute value to the power of p?

In summary, the derivative of |x|^p is differentiable at x=0 if p is even, but not differentiable at x=0 if p is odd.
  • #1
michonamona
122
0

Homework Statement



Let p>0. What is the derivative of [tex] |x|^{p}[/tex]?

Homework Equations


The Attempt at a Solution

I know that if p is even, then the derivative is just [tex] px^{p-1} [/tex]. But what if x is odd? Would it turn out to be some piecewise function, such as

[tex] px^{p-1},\: if x \geq 0[/tex] and [tex]-px^{p-1}, \: if x < 0 [/tex]

Any insights?

Thanks,
M
 
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  • #2
Is the function |x|^n continuous everywhere ? How about differentiable ?
 
  • #3
Ok, |x|^p is definitely continuous everywhere. I graphed it and it looks differentiable at x=0 (seems to be the local min). Is the following a valid proof?

Show that |x|^p, p>1 is differentiable at x=0

[tex] lim_{h->0^{+}} \frac{|0+h|^{p} - |0|^p}{h}= lim_{h->0^{+}} \frac{|h|^{p}}{h}=lim_{h->0^{+}} |h|^{p-1} = 0 [/tex], since h is always positive as it approaches 0 from the right. Similarly

[tex]lim_{h->0^{-}} \frac{|0+h|^{p} - |0|^p}{h}=lim_{h->0^{-}} \frac{|h|^{p}}{h}=lim_{h->0^{-}} |h|^{p}h^{-1} [/tex].

Since h is always negative as it approaches zero from the right, then

[tex]lim_{h->0^{-}} |h|^{p}h^{-1}=lim_{h->0^{-}} (-h)^{p}h^{-1}=lim_{h->0^{-}} (-1)^{p}h^{p}h^{-1} = lim_{h->0^{-}} (-1)^{p}h^{p-1}=0 [/tex]

Since the left and right limits are equal then the limit exist and is finite at x=0. Therefore we can take the derivative of |x|^p as p|x|^(p-1).

Is this ok?

Thank you,
M
 
  • #4
For p=1 the function is not even differentiable at 0.

The slope to the left of zero is negative that to the right of zero. Hence the "left and right" derievatives are unequal.
 
  • #5
What happens when [itex] 0<p\leq 1 [/itex] ?
 
  • #6
michonamona said:
...

[tex] lim_{h->0^{+}} \frac{|0+h|^{p} - |0|^p}{h}= lim_{h->0^{+}} \frac{|h|^{p}}{h}=lim_{h->0^{+}} |h|^{p-1} \,,[/tex] since h is always positive as it approaches 0 from the right.
[tex]=\left\{\begin{array}{cc}0 \,,&\mbox{ if } p>1\\
1, & \mbox{ if } p=1\\
\to+\infty, & \mbox{ if } p<1 \end{array}\right.[/tex]​
[tex]lim_{h->0^{-}}\ \frac{|0+h|^{p} - |0|^p}{h}=lim_{h->0^{-}}\ \frac{|h|^{p}}{h}=lim_{h->0^{-}}\ |h|^{p}h^{-1} .[/tex]

Since h is always negative as it approaches zero from the [STRIKE]right[/STRIKE] left, then

[tex]lim_{h->0^{-}}\ |h|^{p}h^{-1}=lim_{h->0^{-}}\ (-h)^{p}h^{-1}[/tex]

M
Hello M.

[tex]\lim_{h->0^{-}}\ (-h)^{p}h^{-1}=\lim_{h->0^{-}}\ (-h)^{p}(-1)(-h)^{-1}=\lim_{h->0^{-}}\ (-1)(-h)^{p-1} [/tex]
[tex]=\left\{\begin{array}{cc}0 \,,&\mbox{ if } p>1\\
-1, & \mbox{ if } p=1\\
\to-\infty, & \mbox{ if } p<1 \end{array}\right.[/tex]​


You still need to find [tex]\textstyle \frac{d}{dx}\ \left|x\right|^p\,[/tex] for x≠0 .

It shouldn't be a problem for x>0 .

For x<0, remember that |x|= ‒x .
 
Last edited:
  • #7
If [tex]\textstyle p[/tex] is real or rational, and [tex]x<0[/tex] then you should avoid writing [tex]x^{p}.[/tex]

[tex]\text{If }\ x<0\,,\ \text{ then using the chain rule: }\ \frac{d}{dx}\ \left(-x\right)^p=p\left(-x\right)^{p-1}\cdot\frac{d}{dx}\left(-x\right)=-p\left(-x\right)^{p-1}\,.[/tex]
 

1. What is the formula for finding the derivative of the absolute value to the power of p?

The formula for finding the derivative of the absolute value to the power of p is: d/dx |x|^p = px|x|^(p-1)

2. How do you interpret the derivative of the absolute value to the power of p?

The derivative of the absolute value to the power of p represents the rate of change of the function at a given point. It tells us how quickly the function is increasing or decreasing at that point.

3. Can you provide an example of finding the derivative of the absolute value to the power of p?

Sure, let's say we have the function f(x) = |x|^3. Using the formula, we can find the derivative as: f'(x) = 3x|x|^(3-1) = 3x|x|^2 = 3x^3.

4. What is the relationship between the absolute value and the derivative of the absolute value to the power of p?

The absolute value and the derivative of the absolute value to the power of p are closely related because the absolute value function itself is not differentiable at the point where x = 0. However, the derivative of the absolute value to the power of p is defined and differentiable at this point.

5. How can the derivative of the absolute value to the power of p be applied in real-world situations?

The derivative of the absolute value to the power of p can be applied in various real-world situations, such as calculating the velocity of an object at a specific point in time or determining the rate of change of a quantity in a given situation. For example, it can be used in physics to calculate the acceleration of an object or in economics to analyze the demand for a product.

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