I am in a little over my head. This all began with my reading how each level of pascals triangle adds to $2^n$, where n=row# starting with n=0. I then though, "wouldn't it be clever if the rows added to something else--like say $3^n$ instead?" Or even better generalize it for any constant, $a^n$.
All that was needed was to Multiply any given number in pascal's triangle by $a^n/2^n$
$$ \begin{array}{rcccccccccc} & & & & & & 1\\\ & & & & & \frac{a}{2} & & \frac{a}{2}\\\ & & & & \frac{a^2}{4} & & \frac{a^2}{2} & & \frac{a^2}{4}\\\ & & & \frac{a^3}{8} & & \frac{3*a^3}{8} & & \frac{3*a^3}{8} & & \frac{a^3}{8}\\\ & & \frac{a^4}{16} & & \frac{a^4}{4} & & \frac{3*a^4}{8} & & \frac{a^4}{4} & & \frac{a^4}{16}\\\ & \frac{a^5}{32} & & \frac{5*a^5}{32} & & \frac{5*a^5}{16} & & \frac{5*a^5}{16}& & \frac{5*a^5}{32} & & \frac{a^5}{32}\\\ & \frac{a^6}{64}& &\frac{3*a^6}{32}& &\frac{15*a^6}{64}& &\frac{5*a^6}{16}& &\frac{15*a^6}{64}& &\frac{3*a^6}{32}& &\frac{a^6}{64}\\\ & & ... & & & &... & & & & ... & \end{array} $$
Adding any row should give $a^n$. Placing said coefficients in front of a binomial expansion and solving for the binomial expression yields $$(a^n/2^n)*(x+y)^n$$
Letting a=2 makes pascals triangle, but every other value of "a" distorts every value and relation (except n=0 row for obvious reasons).
This triangle creates some interesting relations that are shared with Pascal's triangle and are immediately obvious: every term in the middle column may be divided by "a" to yield the term above and to the right or left of it--just like Pascals triangle (a=2).
Next is a really fascinating fluke: $$ \frac{\partial \frac {a^4}{4}}{\partial a}=a^3 $$, which is the sum of the line above it. $$ \frac{\partial \frac {a^2}{4}}{\partial a}=\frac {a}{2} $$, which is found in the line above it.
I realize that "a" must have a definite value as a coefficient in order to have meaning, and it is not itself a function, but it seems curious that derivative relationship would show up in the relations between the coefficients of this modified triangle. This serendipitous relation fascinates me to the point of asking "whats up?" here. Is this a derivative? Have derivative relationships popped up organically elsewhere in function theory?
Pascal's triangle is fascinating and encodes many profound mathematical relationships. However, I think all you've observed is that if $f(a)=a^n$ then $f'(a)=na^{n-1}$. This explains why if you differentiate a term $\frac{a^n}{2^n}{n\choose k}$ in one row with respect to $a$ you'll get something resembling something in the row right above.
In particular you get $a^{n-1} \frac{n!\cdot n}{2^n(n-k)!k!}$. For small n it might seem that there are patterns, particularly in the rows that are powers of 2 since the $2^n$ might go away.
In general, however, this is not a meaningful manipulation of Pascal's triangle, especially since you've stolen away one of its most beautiful features: the fact that ${{n-1}\choose k-1}+{{n-1}\choose k}={n \choose k}$, equivalently that each entry is the sum of the two entries directly above.
As to your derivative question, I'm not sure what you're expecting... I think what you've observed unfortunately happens to be purely incidental. But it's great that you're noticing patterns - that's how most great ideas in math come about! Derivatives have wide-ranging and incredibly useful applications. If by "derivative relationships" you mean differential equations, then heck yeah, they come up all the time in analysis and applied mathematics.