It’s obviously not true in general, but is it true if, for example, there exists a function $f$ such that $x$ appears in $(p(x) - q(x))$ only within an arithmetic argument of $f$?
Edit:
Background:
I was trying to establish that the rule for multiplying complex numbers be obtained from that fact that the derivative of the sine is the cosine. The significance would be that this would mean that all of trigonometry is implied by the fact that sin’ = cos, because if you have the complex number system, you have trigonometry. (The formula for the ADDITION of complex numbers is self-evident; it is only the formula for the MULTIPLICATION of complex numbers that needs justification.) However, I ran into a snag. Here is my attempt:
sin’ = cos
implies (by l’Hopital’s Rule)
lim as x goes to 0 of sin(x)/x = 1
implies (by changing the variable name from ‘x’ to ‘h’)
lim as h goes to 0 of sin(h)/h = 1
implies
lim as h/2 goes to 0 of sin(h/2)/(h/2) = 1
implies
cos(x) = (lim as h/2 goes to 0 of sin(h/2)/(h/2))(lim as h/2 goes to 0 of cos(x + h/2))
implies
cos(x) = lim as h/2 goes to 0 of ((sin(h/2)/(h/2))cos(x + h/2)
implies
cos(x) = lim as h/2 goes to 0 of 2(sin(h/2)/h)cos(x + h/2)
implies
cos(x) = lim as h goes to 0 of 2(sin(h/2)/h)cos(x + h/2)
implies
cos(x) = lim as h goes to 0 of (2sin(h/2)cos(x + h/2))/h
implies (since sin' = cos)
lim as h goes to 0 of (2sin(h/2)cos(x + h/2))/h
= lim as h goes to 0 of (sin(x + h) – sin(x))/h
At this point I want to conclude that for all x and h 2sin(h/2)cos(x + h/2) = sin(x + h) – sin(x), from which one could then deduce the formula for the sine of a difference. One could also then deduce the formula for the cosine of a difference. From these would follow the multiplication rule of complex numbers simply by associativity.
It seems to me that the fact that the two functions in question, namely, 2sin(h/2)cos(x + h/2), and sin(x + h)- sin(x) are as “lean” as possible might allow for the conclusion of their equality, based on their common limiting behavior. Anyway, that’s where I was coming from with my question. I didn’t give the backround at first because I wanted the cast this net as wide as possible first, without anyone being influenced by the particular functions I had in mind.