Understanding Derivatives of Absolute Values

[dglee]

Can somebody help me review derivatives of abs values?

for example derivative of |sinx| and sin|x| and |x| and what is the formal definition of discontinuity is it that since |x| is discontinous at 0 is because you can not find the slope of |x| at 0?

 

[Hurkyl]

A function is discontinuous at a point P if and only if it is not continuous at P.

 

[StatusX]

None of those functions are discontinuous, and in fact, since |x| is continuous and the composition of two continuous functions is continuous, f(|x|) and |f(x)| are continuous for any continuous function f(x).

The reason the derivative of |x| is not defined at x=0 is because the limit which defines the derivative of a function:

$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

Does not exist for |x| at x=0. This is because it depends on how h approaches 0. If it approaches it from the right, you get:

$$|x|' |_{x=0} = \lim_{h \rightarrow 0+} \frac{|h|-|0|}{h} =1$$

but from the left:

$$\begin{align*}|x|' |_{x=0} &= \lim_{h \rightarrow 0-} \frac{|h|-|0|}{h} \\<br /> &= \lim_{-h \rightarrow 0+} \frac{|h|-|0|}{h} \\<br /> &= \lim_{k \rightarrow 0+} \frac{|-k|-|0|}{-k} \\<br /> &= -\lim_{k \rightarrow 0+} \frac{|k|-|0|}{k}=-1<br /> \end{align*}$$

 

[dglee]

this is embearassing... i understand that

|h|/h = h/h or -h/h at x=0 so that would be undefined. but can we say at a certain point |x| is discontinous at x=0 when we take the derivative at x=0 because we can not evaluate the slope at x=0? is that correct to say?

and that the derivative of sin|x| is discontinous at pi and 2 pi because we can't find the slope at those points because its jagged. meaning pointy...

 

[dglee]

can somebody show me how to solve this?

lim h-->0 (|1+h|-1)/h = 1?

 

[VietDao29]

this is embearassing... i understand that

|h|/h = h/h or -h/h at x=0 so that would be undefined. but can we say at a certain point |x| is discontinous at x=0 when we take the derivative at x=0 because we can not evaluate the slope at x=0? is that correct to say?

It's not very clear what you are asking. Can you make it a bit clearer?

and that the derivative of sin|x| is discontinous at pi and 2 pi because we can't find the slope at those points because its jagged. meaning pointy...

Noooo, the derivative of sin|x| is indeed continuous at $x = \pi$, and $x = 2 \pi$. For x > 0, sin|x| = sin(x), whose derivative is cos(x) (a continuous function). For x < 0, sin|x| = sin(-x) = -sin(x), whose derivative is -cos(x) (a continuous function), ie the derivative of sin|x| is continuous on $] - \infty ; 0[ \ \bigcup \ ]0; + \infty[$. Do you get it?

The derivative of sin|x| is only not continuous (i.e not defined) at x = 0. Do you know why? Can you show it?
But the function sin|x| is continuous for all x, ie, when you graph it, it will be a continuous line.
Do you get it?
If you cannot imagine, try graphing it.
---------------
You should note that:
If the function is not continuous at x₀, then it's not differentiable at x₀.
But if the function is continuous at x₀, then we still cannot be sure that it's differentiable there. (f(x) = |x| is an example, it's everywhere continuous, but not differentiable at x = 0).

can somebody show me how to solve this?

lim h-->0 (|1+h|-1)/h = 1?

What have you done?
I'll give you a hint:
When h approaches 0 (either from left side or right side), what does |1 + h| tend to? Is it positive or negative, or it depends on which side h tends to 0?

 

[dglee]

well opz i mean |sin(x)| is not differentiable at pi 2pi and 2npi. ok i understand now that |x| is continuous everywhere but can not be diff at 0 because we can not estimate a slope at x=0.

(|1+h|-1)/h thus always = to 1 for all h but however you can not plug 0 in or you will get an undefined function?? you know is it possible to do l'hospital's rule? on absolute value limits? i hate absolute values... now its becoming clearer this is sad I am an applied mathematics student... this is sad lol. and I am a 2nd yr as well.

 

[VietDao29]

well opz i mean |sin(x)| is not differentiable at pi 2pi and 2npi. ok i understand now that |x| is continuous everywhere but can not be diff at 0 because we can not estimate a slope at x=0.

Yes |sin(x)| is again everywhere continuous, but is not differentiable at $x = k \pi, \ k \in \mathbb{Z}$.
---------------------
A function is differentiable at x₀ iff:
$$\lim_{h \rightarrow 0 ^ +} \frac{f(x_0 + h) - f(x_0)}{h} = \lim_{h \rightarrow 0 ^ -} \frac{f(x_0 + h) - f(x_0)}{h}$$, ie the limit is the same no matter how h tends to 0 (in other words, no matter how x tends to x₀) (from the left or from the right).

(|1+h|-1)/h thus always = to 1 for all h

This is again wrong. What if h = -2?

but however you can not plug 0 in or you will get an undefined function?? you know is it possible to do l'hospital's rule? on absolute value limits? i hate absolute values... now its becoming clearer this is sad I am an applied mathematics student... this is sad lol. and I am a 2nd yr as well.

You must first understand the definition of limit, look it up in your book. It's important because differentiation is based heavily on limit.
If you say $$\lim_{x \rightarrow \alpha} \mbox{some expression} = L$$, it does not mean that you take $x = \alpha$ and plug it in the expression, it will return you L. It means that as x tends to $\alpha$ (close to $\alpha$, but not $x = \alpha$), the expression: $$\mbox{some expression}$$ will tends to L.
In other word, we are allow to freely choose an $\varepsilon > 0$ (no matter how small it is), there must be a neighbourhood of $\alpha$, that if x is in that neighbourhood, then $$\left| \mbox{some expression} - L \right| < \varepsilon$$. Or, equivalently:
$$\forall \varepsilon > 0, \exists \delta > 0 : \mbox{ if } 0 < |x_0 - \alpha| < \delta \mbox{ then } \left| \mbox{some expression} - L \right| < \varepsilon$$.
Do you understand this?
So if you plan to say that:
$$\lim_{h \rightarrow 0} \frac{|1 + h| - 1}{h} = \frac{|1 + 0| - 1}{0}$$, then you are wrong!
As h tends to 0, both numerator, and denominator tend to 0. That yields the indeterminate form $$\frac{0}{0}$$.
(If it's not an indeterminate form, then you can take $x = \alpha$, and plug in the expression, but in this case, since it's one of the indeterminate forms, you can't).
But before you take the limit, let's answer some questions:
1. As h tends to 0 what does |1 + h| tend to?
2. Is it positive, negative, or depends on the way h tends to 0?
3. So as h tends to 0, |1 + h| = ? (Is it |1 + h| = 1 + h, or |1 + h| = -1 - h)?
Number 3 can be answered after you have answered number 1, and 2.
From there, I think you can solve the problem. Can you?

 

[dglee]

thanks a lot vietdao so when h-->0 it tends to go near 1 and both ways approaching from neg or pos will approach near 1 so it equals 1. yah giving me a lot of mathematical analysis limit definitions. thanks i understand it thanks!