Confused about partial derivative to function

[toforfiltum]

Homework Statement

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:

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

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

Homework Equations

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.

 

[fresh_42]

Homework Statement

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:

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

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

Homework Equations

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 $y$?

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 $\|0\|=|0|=0\,$.

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

Thanks.

I don't know. You haven't defined either.

 

[haruspex]

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

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.

 

[toforfiltum]

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

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

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

 

[LCKurtz]

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

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

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?

 

[toforfiltum]

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?

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.

 

[fresh_42]

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.

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 $0$.

 

[LCKurtz]

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.

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

 

[toforfiltum]

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

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

 

[fresh_42]

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

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.

 

[Ray Vickson]

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.

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

 

[toforfiltum]

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

Just a silly mistake on my part, sorry.

 

[toforfiltum]

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.

Yes, thank you for the reminder.