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Derivatives and Integrals

  1. Jun 28, 2008 #1
    I just want to see if I got this correct. From what all I've read it seems that I have most of it understood, but eh, I don't trust my judgement...

    Lets say we have [tex]f(x) = {{3x^3 + 8x^2 + 7x + 12} \over {4x^2 - 12x - 15}}[/tex]

    And the derivative...

    {d \over dx} f(x) = \lim _{h \rightarrow 0} {{f(x+h) - f(x)} \over h} =
    {d \over dx} (3x^3 + 8x^2 + 7x + 12)
    {d \over dx} (4x^2 - 12x - 15)
    }} = f'(x)

    Thus the integral would be...

    \int {f'(x)} \textbf{ }dx = f(x)

    And if the constants are unknown, thus letting a and b represent the constants...

    \int {f'(x)} \text{ } dx = {{3x^3 + 8x^2 + 7x + a} \over {4x^2 - 12x + b}}
  2. jcsd
  3. Jun 28, 2008 #2


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    I'm afraid that what you have written isn't correct. You might want to have a read of these tutorials for Differentiation (work in progress) & Integration.
  4. Jun 28, 2008 #3


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    Staff Emeritus
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    No. the derivative of f(x)/g(x) is (f'(x)g(x)- f(x)g'(x))/g2(x), not f'/g'.

    Nor can you integrate f/g by integrating numerator and denominator separately.
  5. Sep 14, 2009 #4
    Hi Mol_Bolom
    when you find the derivative by definition
    (d/dx)(f(x)=lim(f(x+h)-f(x))/h you can find the limit of d/dh(f(x+h)-f(x))
    h__ 0 ---------------
    for example if F(x)=x^2 you have

    d/dx(F(x))=lim ((x+h)^2-x^2)/h =
    =lim [2(x+h)-0]/1 by diff. both num. and denum. w.r.t h and take the limit as h
    approaches 0 you get
    Best wishes
    Riad Zaidan
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