How to study the derivability of a function?

[seaglespn]

I have f(x) = x / (1 + |x|)
The task is to study f's derivability and calculate f'.

I don't have any experience in this type of derivates, pls help .

 

[Office_Shredder]

I would start by graphing it. Where does the absolute value cause problems? Where does it act like a normal, smooth function?

 

[HallsofIvy]

Look at the "parts". A fraction is differentiable (by the "quotient rule") as long as both numerator and denominator are differentiable and the denominator is not 0. Where, if ever, is 1+ |x| equal to 0?

The numerator is easy: the derivative of x is 1 for all values of x so it is always differentiable.

The denominator is not much harder: a sum of functions is differentiable as long as each addend is. 1, of course, is a constant and so differentiable for all x. What about |x|? Where is that differentiable or not differentiable? (It might help to graph |x| and remember that differentiable functions are often called "smooth".)

 

[seaglespn]

well, if x<0 then |x|=-x, else if x>=0 |x|=x...
So is never negative, always possitive, a function witch draws a line, twice :), so this was the big deal? :D
Now the question is seams easy to me... the denominator is never 0, because |x| cannot be -1, so that function is always differentiable.
And f' = 1 / (1+x)^2 , if x>=0
or if x<0 f' = 1 / (1-x)^2

 

[seaglespn]

Thank you guys for another promt ,very needed , help.

 

[Office_Shredder]

Now the question is seams easy to me... the denominator is never 0, because |x| cannot be -1, so that function is always differentiable.

Did you graph the function?

 

[arildno]

Here's how you could do this using the definition of the derivative:
1. The only possible problem might be at x=0, since at every other point, there will be an open region about it where you may remove the absolute value sign.
Since both x/(1+x) and x/(1-x) are differentiable functions at all points of their domains, x/(1+|x|) is also differentiable there.

2. We are left with the case x=0.
By the definition of the derivative, if it exists, we have:
$$f'(0)=\lim_{x\to{0}}\frac{\frac{x}{1+|x|}-0}{x}=\lim_{x\to{0}}\frac{1}{1+|x|}=1$$
That is, f is differentiable at x=0 as well.

 

[HallsofIvy]

well, if x<0 then |x|=-x, else if x>=0 |x|=x...
So is never negative, always possitive, a function witch draws a line, twice :), so this was the big deal? :D [\quote]
Yes, it is a big deal! The graph of |x| is two straight lines, but the graph of x/(1+|x|) is not.

Now the question is seams easy to me... the denominator is never 0, because |x| cannot be -1, so that function is always differentiable.
And f' = 1 / (1+x)^2 , if x>=0
or if x<0 f' = 1 / (1-x)^2

No, that's wrong. It is true that the denominator is never 0 and that the numerator, x, is always differentiable, but the denominator, 1+ |x|, is not! What do you know about the derivative of |x|?

 

[seaglespn]

No, that's wrong. It is true that the denominator is never 0 and that the numerator, x, is always differentiable, but the denominator, 1+ |x|, is not! What do you know about the derivative of |x|?

Well, I don't know how to derivate |x| , but I thought that first I need to discuss |x| , and thus to create two functions, and after that I need to see if first limit from left (x>0) equals limit in 0 and also equals limit from rigth( in case x<0 ), if this is correct, then I do the derivate for each function...

I might be wrong... but as far as I remember a function is differentiable if is continuos, or "smooth" : like f:A->B must be "smooth" on B to be differentiable.Or that the graph of f doesn't have any interruption.

So I derivate badly, or just I do the right work but (as always in my case ) I just run for the answer leaving something behind?

 

[HallsofIvy]

I might be wrong... but as far as I remember a function is differentiable if is continuos, or "smooth" : like f:A->B must be "smooth" on B to be differentiable.Or that the graph of f doesn't have any interruption.

No, "continous" is NOT the same as "differentiable". In order to be differentiable at x=a, a function MUST be continuous but continuous does not guarantee differentiable (differentiable is a "stronger" condition than continuous). f(x)= |x| is the typical example. If x> 0, then f(x)= x so the derivative is 1. If x< 0, f(x)= -x so the derivative is -1. What about x= 0?
Several people have suggested you graph y= |x|. If you have done so, look at x=0 on the graph and think about the word "smooth". Remember that the derivative is "the slope of the tangent line". What is the tangent line to y= |x| at x=0?

 

[Signifier]

The derivative of |x| is sgn(x).

 

[HallsofIvy]

No, it is not! sgn(x) is defined to be 1 if x> 0, 0 if x= 0, -1 if x< 0.
That is not the derivative of |x|.

 

[seaglespn]

In 0 f=0
f'=1; so the angle made by the tangent is 45.

As far as I can tell, and see, the function is continuous, I can't find or remember a larger definition on differentiability... I don't know what else has a function to be in order to be differentiable... help ?!

 

[HallsofIvy]

Are you serious? You really don't know what "differentiable" means?

A function is differentiable at x= a if it has a derivative there.
Using the basic formula for derivative, the derivative of y= |x| at x= 0 would have to be
$$\lim_{h\rightarrow 0}\frac{|0+h|}{h}= \lim_{h\rightarrow 0}\frac{|h|}{h} if the limit exists. What is that?<br /> <br /> What is the "limit from the left" (i.e. assuming h is negative)? What is the "limit from the right" (i.e. assuming h is positive)?$$

 

[Signifier]

How is the derivative of |x| not sgn(x)? It is, at least, the derivative of |x| for all x but x = 0. Consider when x > 0. Then |x| = x and d(|x|)/dx = dx/dx = 1 = sgn(x) (as x > 0). Consider when x < 0. then |x| = -x and d(|x|)/dx = -(dx/dx) = -1 = sgn(x) (as x < 0). However, I suppose that because |x| is not differentiable at 0, but sgn(x) returns a real at 0 that sgn(x) is not really the derivative of |x|... it's close, nevertheless...

Using d(|x|)/dx = sgn(x), I find the derivative of x / (1 + |x|) to be 1 / (1 + 2|x| + x^2), which matches all the values returned by my calculator at least.

 

[Office_Shredder]

However, I suppose that because |x| is not differentiable at 0, but sgn(x) returns a real at 0 that sgn(x) is not really the derivative of |x|... it's close, nevertheless...

|x| has no close approximation to its derivative at 0, seeing how there clearly is nothing even resembling a tangent line there

 

[seaglespn]

Are you serious? You really don't know what "differentiable" means?

If I was knowing the answer I wasn't asking for help.
And I know what means "differentiable", I said that I don't know what else I've got to prove other than that that function in continuous in order to demonstrate it's differentiability.
So can anyone show me the correct solution to this problem... cus I'm getting confused.

 

[HallsofIvy]

I'm sorry, exactly what course is this for?? It's a calculus problem- you are asked to determine where a function is or is not differentiable but you don't know what "differentiable" means?? Wouldn't the first thing you do in a case like that is look up "differentiable" in your textbook? How can you expect to do a problem if you don't know the meanings of all the words in the problem? Certainly knowing the specific definition of "differentiable" is necessary in calculus!

No, a continuous function is not necessarily differentiable as you've been told several times now. The absolute value function is the simplest example of that.

A function is "differentiable" at x= a if and only if this limit (the derivative itself) exists:
[tex]\lim_{h\rightarrow 0}\frac{f(a+h)- fa)}{h}[/itex]<br /> <br /> For a= 0, f(x)= |x|, that reduces to <br /> $$\lim_{h\rightarrow 0}\frac{|h|}{h}$$<br /> <br /> I've already suggested that you look at "right" and "left" limits:<br /> $$\lim_{h\rightarrow 0^+}\frac{|h|}{h}$$<br /> and<br /> $$\lim_{h\rightarrow 0^-}\frac{|h|}{h}$$<br /> since then you can replace |h| with h and -h respectively.<br /> <br /> If you honestly do not know the definition of "differentiable" and/or cannot do those limits, it would not help you to have some on just do the problem for you.[/tex]

 

[HallsofIvy]

How is the derivative of |x| not sgn(x)? It is, at least, the derivative of |x| for all x but x = 0.

Exactly. Since the two functions (derivative of |x| and sgn(x)) are equal everywhere except at x=0, they are not "equal".

 

[seaglespn]

Well, I don't know how to derivate |x| , but I thought that first I need to discuss |x| , and thus to create two functions, and after that I need to see if first limit from left (x>0) equals limit in 0 and also equals limit from rigth( in case x<0 ), if this is correct, then I do the derivate for each function...

This is what I've told earlier... I just forgot to mention that there must EXIST the limit in f(x), where x is the point we're checking out...

So please don't get me wrong... but there was a single point to correct me, you wasn't supose to totally disagree with me, I wasn't after that sure about anything, because is obvious that you're good at it ...
Cheers!

 

[HallsofIvy]

I don't see that you have actually done anything that has been suggested.

What is $\lim_{h\rightarrow 0^+}\frac{h}{h}$ and \$\lim_{h\rightarrow 0^-}\frac{-h}{h}$?