# Homework Help: Derivatives of ABS values

1. Jan 27, 2006

### 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?

2. Jan 27, 2006

### Hurkyl

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

3. Jan 27, 2006

### 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} \\ &= \lim_{-h \rightarrow 0+} \frac{|h|-|0|}{h} \\ &= \lim_{k \rightarrow 0+} \frac{|-k|-|0|}{-k} \\ &= -\lim_{k \rightarrow 0+} \frac{|k|-|0|}{k}=-1 \end{align*}

Last edited: Jan 27, 2006
4. Jan 28, 2006

### 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...

Last edited: Jan 28, 2006
5. Jan 28, 2006

### dglee

can somebody show me how to solve this?

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

6. Jan 28, 2006

### VietDao29

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

Last edited: Jan 28, 2006
7. Jan 28, 2006

### 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 im an applied mathematics student... this is sad lol. and im a 2nd yr as well.

8. Jan 28, 2006

### VietDao29

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

Last edited: Jan 28, 2006
9. Jan 29, 2006

### 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!