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Derivative using first principle definition

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the derivative of f(x) = x^3 - 3x^2


    2. Relevant equations
    (f(x+h) - f(x))/h


    3. The attempt at a solution
    Just wondering if someone can check my solution.
    (f(x+h)- f(x))/h= ((x+h)^3- 3(x+h)^2- (x^3-3x^2))/h
    = (x^3+ 3x^2 h+3xh^2+ h^3-3(x^2+ 2xh+ h^2 )- x^3+3x^2)/h
    = (x^3+3x^2 h+3xh^2+h^3-3x^2-6xh-3h^2-x^2+3x^2)/h
    = (x^3+3x^2 h+3xh^2+h^3-6xh-3h^2-x^2)/h
    = x3 + 3x2 + 3xh + h2 – 6x – 3h –x2
    = x3 + 2x2 + 3xh + h2 – 6x – 3h

    lim┬(h→0)⁡〖x^3+ 2x^2- 6x〗

    f’(x) = x^3+ 2x^2- 6x
     
  2. jcsd
  3. May 6, 2013 #2

    Dick

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    You are doing the right general thing but the algebra starts going badly after the third line. How come the x^3 and the 3x^2 didn't just cancel? Then I completely start losing track.
     
  4. May 6, 2013 #3
    I thought they could only cancel out if they were of the same degree?
     
  5. May 6, 2013 #4
    It's 3x^2h so I think I can't cancel it out without a negative 3x^2h?
     
  6. May 6, 2013 #5
    Sorry, I copy and pasted it from word and didn't edit it, I see now what you mean
     
  7. May 6, 2013 #6
    The third line should read:
    (x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^2 + 3x^2)/h
     
  8. May 6, 2013 #7

    SammyS

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    There's still at least one error/typo there .

    Should be :

    (x^3 + 3x^2h + 3xh^2 + h^3 - 3x^2 - 6xh - 3h^2 - x^3 + 3x^2)/h

    Right ?
     
  9. May 7, 2013 #8
    Oh! Typo. So f(x+h) - f(x) should be 3x^2 + 3xh + h^2 - 6x - 3h
    And the limit would be
    3x^2 - 6x
    ?
     
  10. May 7, 2013 #9

    Dick

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    That's really (f(x+h)-f(x))/h. But yes, that's the right limit.
     
  11. May 7, 2013 #10
    Great, thank you both :)
     
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