Given-f-x-x-x-1-x-2-x-n-find-the-value-of-f-0-

Question Number 108043 by ZiYangLee last updated on 14/Aug/20

Given f (x) = x (x + 1) (x + 2) \dots (x + n)

find the value of f^{'} (0) .

Answered by hgrocks last updated on 14/Aug/20

\ln (f (x)) = \underset{k = 0}{\sum^{n}} \ln (x + k)

\frac{f^{'} (x)}{f (x)} = \frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x + 2} + \dots \dots . \frac{1}{x + n}

\frac{f^{'} (0)}{f (0)} = \frac{1}{x} + 1 + \frac{1}{2} + \frac{1}{3} + \dots \dots \dots \frac{1}{n}

f^{'} (0) = \underset{k = 1}{\prod^{n}} {(x + k)}_{x = 0} + f (0) (H_{n})

= 1.2.3 \dots . . n + 0 \times H_{n}

= n!

Commented by ZiYangLee last updated on 14/Aug/20

wow \dots

Answered by Her_Majesty last updated on 14/Aug/20

f (x) = x (x + 1) (x + 2) \dots (x + n) =

= x^{n + 1} + (\dots .) x^{2} + n! x

f^{'} (x) = (n + 1) x^{n} + 2 (\dots) x + n!

f^{'} (0) = n!