Confused about partial derivative to function

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1. Oct 1, 2016

toforfiltum

1. The problem statement, all variables and given/known data
Let $f(x,y) = \|x \| - \|y\| - |x| - |y|$ and consider the surface defined by the graph of $z=f(x,y)$. The partial derivative of $f$ at the origin is:

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

$f_{y}(0,0) = lim_{h \rightarrow 0} \frac{ f(0,0 + h) - f(0,0)}{h} = lim_{ h \rightarrow 0} \frac{ |-| h\| -|h|}{h} = lim_{h \rightarrow 0} 0 = 0$

2. Relevant equations

3. The attempt at a solution
I do not understand where $lim_{h \rightarrow 0} \frac{\|h\| -|h|}{h}$ come from? Can anyone tell me how this comes about? I'm confused.

And also, where does the negative sign come from in the partial derivative of $y$?

Why is there $\|h\|$ and $|h|$ in the limit equation? What is the difference?

Thanks.

2. Oct 1, 2016

Staff: Mentor

What are $f(0+h,y)\, , \,f(0,y)$ and then $f(0+h,0)\, , \,f(0,0)\,$?
The same questions for $f(x,0+h)\, , \,f(x,0)$ and then $f(0,0+h)\, , \,f(0,0)\,$?
It is simply the definition of $f$ plus $\|0\|=|0|=0\,$.
I don't know. You haven't defined either.

3. Oct 1, 2016

haruspex

Why are there both in the definition of f?
As far as I am aware, there are two reasons for mixing these notations.
1. || x || represents an arbitrary norm, which may be different from the Euclidean norm.
2. || x || is used for vector x and | x | for scalar x.

By the way, are you sure you have quoted all the signs correctly in the definition of f? It looks strange.

4. Oct 1, 2016

toforfiltum

Ah, sorry, I looked it over and realised I made a typo.

The function is $f(x,y)= | |x| - |y| | - |x| -|y|$

5. Oct 1, 2016

LCKurtz

So those aren't norms at all, it is just$$|~( |x| - |y|) ~| -|x|-|y|$$Are you still confused about where the
$$\frac {f(0+h,0)-f(0,0)}{h}=\frac {|h| - |h|}{h}$$comes from?

6. Oct 1, 2016

toforfiltum

I think I'm actually confused as to what $h$ really is. Why does $f(0+h,0)$ and $f(0,0)$ both yield $|h|$? It doesn't make sense to me. It's like saying $f(0+h,0) = f(0,0)$, which seems bizarre.

7. Oct 1, 2016

Staff: Mentor

No, $f(0,0)=0$. Both $h$ come from $f(0+h,0)$. Just substitute $x=0+h$ and $y=0$ in the formula for $f$.
$h$ is simply a small positive real number that gets smaller and smaller in the limit, i.e. it tends to $0$.

8. Oct 1, 2016

LCKurtz

$h$ is just a variable. What do you get when you calculate $f(0+h,0)$ from the formula$$f(x,y)=|~( |x| - |y|) ~| -|x|-|y|$$

9. Oct 1, 2016

toforfiltum

Ah, I got it now. @fresh_42 and you have helped to clear the confusion. Thanks.

10. Oct 1, 2016

Staff: Mentor

To be exact:
$h$ doesn't have to be positive. It could as well tend from the negative side towards $0$.
I only made this restriction because the word small I used would have become ambiguous for negative $h\,$. $\,-5$ is smaller than $-4$ and getting smaller would have meant $-\infty$. So I found it easier to restrict myself to positive $h$.

It is important to keep this in mind for future examples and other cases. In principal it could even be that $\lim_{h \rightarrow +0}$ is different from $\lim_{h \rightarrow -0}$ although it isn't the case here.

11. Oct 2, 2016

Ray Vickson

If $f(x,y) = |~( |x| - |y|) ~| -|x|-|y|$, what makes you think that $f(0,0) = |h|$?

12. Oct 2, 2016

toforfiltum

Just a silly mistake on my part, sorry.

13. Oct 2, 2016

toforfiltum

Yes, thank you for the reminder.