Wait, wah...? Partial Fractions and Derivative Proof, I think..?

great... just when i was starting to understand math.

Here's an English translation:

- Step 1: Clear the denominator by (1) multiplying each numerator (A1, A2, ..., Aj, ..., An) with every (x - aj) except for the one directly below it and (2) summing them together.
- Step 2: Notice that each term of this pattern: Aj * (x - a[everything else except j]), thus the only term that is left when plugging in x = aj is the one that does not have (x - aj) in it. Incidentally, this is the Aj term.
- Step 3: So P(aj) = Aj * (aj - a1) * (aj - a2) * ... * (aj - aj-1) * (aj - aj+1) * ... * (aj - an)
- Step 4: Take the derivative of Q(x), using product rule after re-ordering Q(x) into (x - aj) * everything else. Q'(x) = (x - aj)' * everything else + (x - aj) * (everything else)'
- Step 5: The second part is 0 when plugging in x = aj, so Q'(aj) = everything else = (aj - a1) * (aj - a2) * ... * (aj - aj-1) * (aj - aj+1) * ... * (aj - an)
- Step 6: P(aj)/Q'(aj) = Aj