Quick derivatives

Calculus Level 1

$\begin{aligned}f(x)&=&\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1 \\ f'(0) &=& \ ? \end{aligned}$

100

f'(0)

1 100 -1 2

2 solutions

Shivam Jadhav
May 3, 2015

By differentiating $f(x)$ we get , $f'(x)=x^{99}+x^{98}+.....x+1$ . $f'(0)=1$ .

$\frac{a}{b}=log(\pi)$

Rohan Dahiya - 2 years, 7 months ago

H Brown
Jun 21, 2015

Zero to any (positive integer) power is zero. Zero divided by anything (non-zero) is zero. When x=0, we are left with 0+0+...+0+0+1=1