Originally Posted by
harish21 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?