if f(x)=1+x+ x^2 2 + + x^ 100 100 ,then f'(1) is equal to:

MCQ

if $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100},$then $f'(1)$ is equal to:

A
$\frac{1}{100}$
✓
$100$
C
$50$
D
$0$

✓

Answer

Correct option: B.

$100$

$\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}$
Differentiate both the sides with respect to $x$, we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big(1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2}\Big)+\dots+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\dots+\frac{1}{100}\frac{\text{d}}{\text{dx}}(\text{x}^{100})$
$=0+1+\frac{1}{2}\times\text{2x}+\dots+\frac{1}{100}\times100\text{x}^{99}$
$=1+\text{x}+\text{x}^2+\dots+\text{x}^{99}$
Putting $x = 1,$ we get
$\text{f}'(\text{x})=1+1+1+\dots+1 (100$ terms$)$
$=100$

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