No, this is not true. You can construct such a function like follows: Take any increasing continuously differentiable function $g$ with the following properties:
(i) $g(x)=0$ for $x\leq 0$ and $g(x)=1$ for $x\geq1$
(ii) $g'(\frac{1}{2})=1$.
You can use some scaled cosine-function as $g$ and cut the ends off. Then define $f$ by
$f(x):=\frac{g(2^n(x-n))}{2^n}+\sum_{k=0}^{n-1}\frac{1}{2^k}$ for $x\in[n,n+1],n\in\mathbb N_0$ and $f(x):=0$ for $x<0$.
Since $g$ is increasing, $f$ is increasing, if it is continuous. But since $f(n)=\frac{1}{2^n}g(0)+\sum_{k=0}^{n-1}\frac{1}{2^k}=\sum_{k=0}^{n-1}\frac{1}{2^k}$ and $f(n+1)=\frac{1}{2^n}+\sum_{k=0}^{n-1}\frac{1}{2^k}=\sum_{k=0}^n\frac{1}{2^k}$ this is the case.
Since $g$ is differentiable everywhere, $f$ is differentiable in $\mathbb R\backslash\mathbb N$. Since $g$ is constant outside of $[0,1]$ we must have $g'(x)=0$ for $x\leq0$ and $x\geq1$ and this implies $f'(n)=0$ for $n\in\mathbb N$. In particular, $f$ is (continuously) differentiable even in $\mathbb N$.
Since $g\leq1$ we have $f\leq\sum_{k=0}^\infty\frac{1}{2^k}=2$, so $f$ is bounded. Since $g'(1/2)=1$ we have $f'(n+\frac{1}{2^{n+1}})=1$. In particular, $f'$ cannot go to $0$ as $x\to\infty$.
Example: Take $g(x)=-\frac{\cos(\pi x)}{2} + \frac{1}{2}$ for $x\in[0,1]$. Then $f$ looks like
