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Let $f(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}...+\dfrac{x^{2018}}{2018}.$ Then $f’(1)$ is equal to 

  1. $0$
  2. $2017$
  3. $2018$
  4. $2019$
in Calculus by Boss (17.5k points)
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1 Answer

+3 votes
$\underline{\textbf {Answer: C}}$

$\underline{\mathbf{Solution:}}$

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.
by Boss (19.2k points)
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Nicely answered.
+1
Thanks!
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