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Definition and Power Rule

  1. Oct 24, 2009 #1
    Its more a simplifying problem...

    I was trying to differentiate this using definition principal

    1-x^1/2

    But I got stuck here :

    (1-(x+h)^1/2 - (1-x^1/2))/h

    I mean how do you explan something to the power of half or infact any fraction? I know I can change it to 1/sqrt(x+h)... but it just makes things more complicated... coz I can't get rid of the h then...

    But using power rule is simple... 1-x^1/2 = -1/2X^-1/2

    On top of that, I want to ask... are the constant numbers ignorable in differentiation? For example 4 - x^2 is -2x, I can just apply power rule on -x^2 and get the same answer... If I were to differentiate this by definition, can I just do -x^2 and ignore 4 too?
     
  2. jcsd
  3. Oct 24, 2009 #2

    lurflurf

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    write sqrt(x+h)-sqrt(x) as
    [(x+h)-x]/[sqrt(x+h)+sqrt(x)]
    by using
    a-b=(a^2-b^2)/(a+b)

    We do not ignor constants, but differentiation is linear that is
    (a f+b g)'=a f'+b g'
    when a and b are constants and f and g are functions
     
    Last edited: Oct 24, 2009
  4. Oct 24, 2009 #3
    Erm... so I am guessing you cancelled the 1 away already before coming to this step? But if I do this its a fraction over h...

    Coz the equation will become {[(x+h)-x]/[sqrt(x+h)+sqrt(x)]}/h and I still don't know how to get rid of h...

    Erm... don't understand. Well I just did a differentiation using definition just now, and I realized you cannot remove the constant beforehand but when using Power Rule, it seems it can be ignored.

    And how do you tell if it linear anyway?
     
  5. Oct 24, 2009 #4
    What does (x+h)-x simplify to? After that, can you cancel the h?
     
  6. Oct 24, 2009 #5

    lurflurf

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    -yes I cancelled the one
    -the h's cancel after the x's cancel in (x+h)-x
    -one can prove the differentiation is linear or take it as an axiom
    have you seen limit theorems like
    lim[a f(x)]=a lim f(x)
    and
    lim [f(x)+g(x)]=lim f(x)+lim g(x)
    if so it will be easy to show
    (a f+b g)'=a f'+b g'
    by using the limit definition of differentiation
    -can you finish the differentiation of sqrt(x)?
     
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