# Homework Help: 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.