Question

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

• A. $\frac{1}{100}$
  
• B. 100
  
• C. 50
  

Answer 1

The correct option is B 100


$f\left(x\right)=1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100}$
Differentiating both sides with respect to x, we get
$f^{\prime }\left(x\right)=\frac{d}{dx}(1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100})$
$=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2}\right)+...+\frac{d}{dx}\left(\frac{x^{100}}{100}\right)$
$=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{1}{2}\frac{d}{dx}\left(x^2\right)+...+\frac{1}{100}\frac{d}{dx}\left(x^{100}\right)$
$=0+1+\frac{1}{2}\times 2x+...+\frac{1}{100}\times 100x^{99}$
$=1+x+x^2+...+x^{99}$
Putting x=1, we get
$f^{\prime }\left(1\right)=1+1+1+...+1$ (100 terms)
=100
Hence, the correct answer is option (b).