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Q.
If $f\left(x\right)=1+x+\frac{x^{2}}{2}+...+\frac{x^{100}}{100}$, then $f'\left(1\right)$ is
Limits and Derivatives
Solution:
We have, $f\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+...+\frac{x^{100}}{100} \quad\ldots\left(i\right)$
Differentiating $\left(i\right)$ w.r.t. $x$, we get
$f'\left(x\right)=0+1+\frac{2x}{2}+\frac{3x^{2}}{3}+...+\frac{100x^{99}}{100}$
$=1+x+x^{2}+x^{3}+...+x^{99}$
$\therefore f'\left(1\right)=1+1+\left(1\right)^{2}+\left(1\right)^{3}+...+\left(1\right)^{99}=100$