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F(x) = c => f'(x) = 0 ?

  1. Nov 2, 2004 #1
    f(x) = c => f'(x) = 0 ??

    Hi again! Now I can't understand [itex]f(x) = c\Rightarrow f'(x) = 0[/itex], where c is a constant. I think I should be undefined.
    y = c
    [itex]\Delta x[/itex]= c
    [tex]\frac{\Delta y}{\Delta x} = \frac{c}{\Delta x}[/tex]
    [tex]st(\frac{c}{\Delta x}) = Undefined [/tex]
    What am I doing wrong this time?
     
    Last edited: Nov 2, 2004
  2. jcsd
  3. Nov 2, 2004 #2

    arildno

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    y=c, but [tex]\bigtriangleup{y}=c-c=0[/tex]
     
  4. Nov 2, 2004 #3

    Fredrik

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    If f(x)=c for all x, then

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(h)}{h}=\lim_{h\rightarrow 0}\frac{c-c}{h}=\lim_{h\rightarrow 0}0=0[/tex]
     
  5. Nov 2, 2004 #4
    Arghhh! :mad: That demostrates the importance of written every stage out.

    y = c
    y + [itex]\Delta y[/itex] = c + c
    [itex]\Delta y[/itex] = c - c = 0
    [tex]\frac{\Delta y}{\Delta x} = \frac{0}{\Delta x} = 0[/tex]
    Thus f'(x) = 0

    Sorry, Fredrik. Haven't learned limits yet. Thanks for trying to explain.

    Thanks!
     
  6. Nov 3, 2004 #5

    HallsofIvy

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    Haven't learned limits yet???

    Then what are you doing with the derivative? That's a lot like working with fractions before you have learned to multiply!

    (Do you know the "slope" of a straight line? What's the slope of a horizontal straight line?)
     
  7. Nov 3, 2004 #6
    I think he said in another thread that he was reading about non-standard analysis.
     
  8. Nov 3, 2004 #7
    Sure I know that a constant function has zero slope, but I want to solve it the algebra-way.

    The book I'm reading is "Elementary Calculus: An Approach Using Infinitesimals", which, as the titel states, uses infinitesimals to introduce both derivatives and integrals - before limits.
     
  9. Nov 3, 2004 #8
    Hmm. When I think about it: [itex]\Delta y + y[/itex] can't be c + c. Then
    [tex]\Delta y = c + c - c = c[/tex]
    Argh!
     
  10. Nov 3, 2004 #9
    again, [itex]y+\Delta y=c[/itex], [itex]\Delta y =0[/itex]

    You have to be more carefull with algebra if you dont want to go insane (trust me)!
     
  11. Nov 3, 2004 #10

    HallsofIvy

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    And how do they actually DEFINE "infinitesmal"?
     
  12. Nov 3, 2004 #11
    It's a lot easier to understand derivatives if you know limits. I suggest you hit that before trying to understand the derivative of a constant.
     
  13. Nov 4, 2004 #12
    As an infinite small number (short: infinitesimal).
     
  14. Nov 4, 2004 #13

    Hurkyl

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    Why should [itex]\Delta x = c[/itex]?!?!

    I think you've totally forgotten what [itex]\Delta[/itex] usually means:

    [tex](\Delta p)(q) = p(q + v) - p(q)[/tex]

    For some nonzero v. (For your purposes, v should be a nonzero infinitessimal)
    (And yes, I know I used unusual variable names. :tongue:)
     
  15. Nov 4, 2004 #14

    Hurkyl

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    I believe I've seen the text danne is using, it's not bad. It doesn't rigorously define infinitessimals, but normal calc textbooks don't rigorously define the real numbers either. :tongue:

    In nonstandard analysis, the derivative of a standard function f is defined by:

    [tex]
    f'(x) = \mathrm{Std} \frac{f^*(x^*+h) - f^*(x^*)}{h}
    [/tex]

    whenever the right hand side is independent of your choice of nonzero infinitessimal h. "Std" means "round to the nearest real number", and the * denotes the nonstandard version of that symbol.

    Note that the notion of limit, here, has been replaced with the notion of nearest standard number.
     
    Last edited: Nov 4, 2004
  16. Nov 4, 2004 #15
    Hmm. Isn't it [itex] (\Delta p)(q) = p(q) + p(v) - p(q)[/itex].
    [tex]y = f(x)[/tex]
    [tex]y + \Delta y = f(x) + f(\Delta x)[/tex]
    [tex]\Delta y = f(x) + f(\Delta x) - f(x) = f(\Delta x)[/tex]
    And then, because f(x) = c
    [tex]\Delta y = c + c - c = c[/tex]

    I think you denote your std() with st() and it's called the "Standard Part" of the hyperreal (reals + infinitesimals) number.
     
  17. Nov 4, 2004 #16
    hmmm ...isn't it (Delta p)(q) = q(p+v) - q(p) .
     
  18. Nov 4, 2004 #17
    Oops. My mistake. I should have checked that up before bother you. :(
     
  19. Nov 4, 2004 #18

    Hurkyl

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    I think you're thinking of the more explicit notation:

    [tex]
    (\Delta_v p)(q) = p(q + v) - p(q)
    [/tex]
     
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