# Thread: partial differentiation

1. ## partial differentiation

The partial fraction decomposition of can be written in the form of where

2. Originally Posted by sobadin
The partial fraction decomposition of can be written in the form of where

first, "partial differentiation" is something completely different. you are not differentiating here at all. in fact, this technique is usually used to aid in integration.

here's a start:

$\displaystyle \frac {x^2 + 75}{x^3 + x^2} = \frac {x^2 + 75}{x^2(x + 1)}$

$\displaystyle \Rightarrow \frac {x^2 + 75}{x^3 + x^2} = \frac Ax + \frac B{x^2} + \frac C{x + 1}$ ...........(look up the partial fraction decomposition to see how we got this)

multiply both sides by $\displaystyle x^2(x + 1)$

$\displaystyle \Rightarrow x^2 + 75 = Ax(x + 1) + B(x + 1) + Cx^2$

there are several ways to proceed here. one is to pick x-values that cause all but one variable to be wiped out and solve for each variable one by one. for example, if we plug in x = 0, we get

$\displaystyle 75 = B$ .........and we've found B. do a similar trick to find A and C

another way is to expand the right side, group the like powers of x and equate coefficients. lets find B again using this method

$\displaystyle x^2 + 75 = Ax^2 + Ax + Bx + B + Cx^2$

$\displaystyle \Rightarrow x^2 + 75 = (A + C)x^2 + (A + B)x + B$

B is the constant term on the right, so it must be the same as the constant term on the left to have equality, thus, again, B = 75

now finish up with the method of your choice