Let $p\_n$ denote the $n$'th prime number. Can anything be said about the existence of a continuous differentiable function $f(x)$ and a positive constant $D > 0$ such that $$\frac{f(p\_{n+1})}{f(p\_n)} < D\cdot \frac{f^{\prime}(p\_n)}{f(p\_n) \log(p\_n)}$$ for all $p\_n$ with $n > N$ where $N$ is an arbitrary fixed positive constant? Further more the derivative must be monotone (increasing or decreasing). One can use eventually the fact that $p\_{n+1} - p\_n < p\_n^\theta$ with $\theta > 0.525$ for $n$ large enough. The motivation for this is the following: 


Supposing that such a function exists and suppose $f'$ is increasing. Also suppose that $p\_{n+1} -p\_n > D$. Then by mean value theorem $\exists c\_n \in (p\_n, p\_{n+1})$ such that $$ f(p\_{n+1}) = (p\_{n+1} - p\_n) f^{\prime}(c\_n) + f(p\_n) \geq D \cdot f'(p\_n) + f(p\_n)$$ $\forall n > N$ hence $$ f(p\_{n+1}) \geq D \sum\_{i=N}^n f'(p\_i) + f(p\_N)$$ Denote $S(p\_n) = \sum\_{i=1}^n f'(p\_i)$ and (if I am not wrong, again) $S(p\_n) \sim \frac{f'(p\_n)}{\log(p\_n)}$ Now this can be used to obtain a contradiction isn't it? 



The existence of a function with application in finding bounded gaps between primes
As you observe, if $f$ and $f'$ are positive and increasing,
$$\frac{f(p\_{n+1})}{f'(p\_n)}>p\_{n+1}-p\_n.$$
The right hand side is not bounded (there exist large gaps between primes), hence the left hand side is not bounded either. This contradicts your assumption, even when you omit $\log(p\_n)$.


