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Thread: differentiating partial fraction

  1. #16
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    thank you very much i understand it now i wasnt taking the other constant A in to account.
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  2. #17
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by harish21 View Post
    Yes.

    Your question is telling you to decompose the function into partial fractions.

    $\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)}$

    Notice that both numerator and denominators are polynomials of degree 2. SO you can decompose the given function by:

    $\displaystyle y = \frac{4u^2+3u-2} {(u+1)(2u+3)} = {A} + \frac{B}{u+1} + \frac{C}{2u+3}$

    This gives

    $\displaystyle {4u^2+3u-2} = A(u+1)(2u+3) + B(2u+3) + C(u+1)$

    Now to find A, B, and C,

    first let u=1, this gives B = -1

    then let $\displaystyle u = \frac{-3}{2}$, this gives C = -5

    and then let u = 1, this gives A=2.

    So your function is decomposed into partial fractions as:

    $\displaystyle \frac{4u^2+3u-2} {(u+1)(2u+3)} = {2} - \frac{1}{u+1} - \frac{5}{2u+3}$

    Now,you are supposed to find the derivative of the right hand side of the equation, which is the function decomposed into partial fractions.

    You have posted this question under "Calculus" section of this forum. So, I am assuming you have enough knowledge of derivatives.

    However, I did the derivative finding part yesterday in my previous post (using the quotient rule).

    Remember: Derivative of a constant (which is ,2) = 0

    Is it clear now?
    To give a little background on this. The reason you need the A is because the method of Partial Fractions doesn't work unless the numerator is at least one degree lower then the denominator.

    This can also be accomplished by adding and subtracting a variable or value to the numerator such that a portion of the numerator is now equal to the denominator, and we can factor that out as 1. This will result in a numerator of a lesser order and the partial fractions method can be applied.
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  3. #18
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    just to make sure i differentiated the partial fraction and got

    $\displaystyle 1/(u+1)^2 + 10/(2u+3)^2$

    is that correct?
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  4. #19
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by sigma1 View Post
    just to make sure i differentiated the partial fraction and got

    $\displaystyle 1/(u+1)^2 + 10/(2u+3)^2$

    is that correct?
    Yes that is correct.
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  5. #20
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by sigma1 View Post
    just to make sure i differentiated the partial fraction and got

    $\displaystyle 1/(u+1)^2 + 10/(2u+3)^2$

    is that correct?
    Check back on my first post on this thread, and you'll see thats exactly what I got!
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