Let $f$ and $g$ be real nonzero polynomials with nonnegative coefficients such that
- the degree of $g$ equals the degree of $f$ plus one, that is $\deg(f+g) = \deg(f) + 1$,
- $(f+g)(0) \neq 0$.
Let $$ h = \frac{f}{f+g}. $$ In particular, it follows that $h$ is well defined on the interval $[0,+\infty)$, and $$ 0 \le h(x) \le 1, \qquad x \ge 0. $$ Even more, using the partial fractions decomposition, we have that there is a constant $C > 0$ such that the $n$-th derivative of $h$ satisfies $$ |h^{(n)}(x)| \le C \frac{n!}{x^{n+1}}, \qquad x > 0,\ n \in \mathbb{N}. $$
Is it true that we can choose $C$ as above that does not depend on the coefficients of polynomials $f$ and $g$?
I don't think we can. As a counterexample, take $n=0$ and set $f_k(x)=k$ and $g_k(x)=x$, so that we have a sequence of these kinds of functions: $$ h_k(x)=\frac{k}{k+x}. $$ If there were such a $c$, independent of the coefficients of $f$ and $g$, we would have $$ h_k(x)\leq \frac{c}{x}, $$ independently of $k$, that is to say, $x\cdot h_k(x)$ is bounded indepently of $k$. However, we have that $$ \lim_{x\rightarrow \infty}x\cdot h_k(x)=k, $$ so this cannot be true (just take $k$ as for example $\lceil c\rceil+1$ and $x$ sufficiently big).