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Confused about partial derivative to function

  1. Oct 1, 2016 #1
    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. jcsd
  3. Oct 1, 2016 #2

    fresh_42

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    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.
     
  4. Oct 1, 2016 #3

    haruspex

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    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.
     
  5. Oct 1, 2016 #4
    Ah, sorry, I looked it over and realised I made a typo.

    The function is ##f(x,y)= | |x| - |y| | - |x| -|y|##
     
  6. Oct 1, 2016 #5

    LCKurtz

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    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?
     
  7. Oct 1, 2016 #6
    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.
     
  8. Oct 1, 2016 #7

    fresh_42

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    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##.
     
  9. Oct 1, 2016 #8

    LCKurtz

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    ##h## is just a variable. What do you get when you calculate ##f(0+h,0)## from the formula$$
    f(x,y)=|~( |x| - |y|) ~| -|x|-|y|$$
     
  10. Oct 1, 2016 #9
    Ah, I got it now. @fresh_42 and you have helped to clear the confusion. Thanks.
     
  11. Oct 1, 2016 #10

    fresh_42

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    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.
     
  12. Oct 2, 2016 #11

    Ray Vickson

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    If ##f(x,y) = |~( |x| - |y|) ~| -|x|-|y|##, what makes you think that ##f(0,0) = |h|##?
     
  13. Oct 2, 2016 #12
    Just a silly mistake on my part, sorry.:wink:
     
  14. Oct 2, 2016 #13
    Yes, thank you for the reminder.
     
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