Question

If $f\left(x\right)=1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100}$, then $f^{\prime }\left(1\right)$ is

• A. $\frac{1}{100}$
• B. $100$
• C. Does not exist
• D. $0$

Answer 1

Answer: (B)

We have, $f\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+...+\frac{x^{100}}{100}\dots \left(i\right)$
Differentiating $\left(i\right)$ w.r.t. $x$, we get
$f^{\prime }\left(x\right)=0+1+\frac{2x}{2}+\frac{3x^2}{3}+...+\frac{100x^{99}}{100}$
$=1+x+x^2+x^3+...+x^{99}$
$\therefore f^{\prime }\left(1\right)=1+1+{\left(1\right)}^2+{\left(1\right)}^3+...+{\left(1\right)}^{99}=100$