$\underline{\textbf {Answer: C}}$
$\underline{\mathbf{Solution:}}$
On differentiating $\mathbf{f(x)}$ $\mathbf{w.r.t.}\;$ $\mathbf x$, we get:
$$\mathrm {f'(x) = 0 + 1 +\underbrace{ x^1 + x^2 + x^3+\ldots+x^{2017}}_{\text{ = 2017 terms}}}$$
$$\mathrm{\therefore f'(x) = 1 + x^1 + x^2 + x^3 +\ldots+ x^{2017}}$$
Now substitute $\color{green} {\mathrm{x = 1}}$, we get:
$$\mathrm{f'(x) = 1 + \underbrace{1 + 1 + \ldots+ 1}_\text{ = 2017 times 1 = $2017\times1 = 2017$}}$$
$$\mathrm{\Rightarrow f'(x) = 1 + 2017 = 2018}$$
$\therefore \textbf {C}$ is the right option.