Not quite, it's that as a and b go to infinity,
(\int_{a,b}f(x)f'(x)dx)/(b-a))
goes to zero. \int_{a,b}f(x)f'(x)dx = [ f(x)^2/2 ]^b_a, which is bounded, while b-a is unbounded, QED.
LaTeX to Wiki might work, but LaTeX to LW comment doesn't.
The standard form for correlation coefficient is
cov(x,y)=N(-)
where N is normalisation; it seems that you suppose that =0 and <f'> finite, then. =0 follows from boundedness, but for the derivative it's not clear. If <f'> on (a,b) grows more rapidly than (b-a), anything can happen.
It is a commonplace that correlation does not imply causality, however eyebrow-wagglingly suggestive it may be of causal hypotheses. It is less commonly noted that causality does not imply correlation either. It is quite possible for two variables to have zero correlation, and yet for one of them to be completely determined by the other.