how-is-the-solution-of-this-qustion-f-x-x-x-1-x-2-x-3-x-4-x-100-f-x-f-1-

Question Number 178250 by zaheen last updated on 14/Oct/22

h o w i s t h e s o l u t i o n o f t h i s q u s t i o n

f (x) = x (x - 1) (x - 2) (x - 3) (x - 4) \cdot \dots \dots . \cdot (x - 100)

f^{^{'}} (x) = ? f^{'} (1) = ?

Answered by CElcedricjunior last updated on 14/Oct/22

f (x) = x (x - 1) (x - 2) (x - 3) \dots . (x - n)

f (x) = \underset{k = 1}{\prod^{100}} x (x - k) = x (x - 100)!

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

\frac{x (x - 1) (x - 2) \dots (x - n)}{x - 1} + \frac{x (x - 1) \dots (x - n)}{x - 2}

+ \dots . . + \frac{x (x - 1) (x - 2) \dots (x - 100)}{x - 100}

f^{'} (x) = \underset{k = 0}{\sum^{100}} \frac{x (x - 100)!}{x - k}

f^{'} (1) = 1 \times (- 1) \times (- 2) \times \dots . . (1 - 100)

f^{'} (1) = - 99!

\dots \dots . . \dots . l e c e l e b r e c e d r i c j u n i o r \dots \dots .

Answered by Sheshdevsahu last updated on 14/Oct/22

f^{'} (x) = {1 . (x - 1) (x - 2) \dots . . (x - 100)} + {x . 1 . (x - 2) \dots . (x - 100)} + \dots \dots . {x (x - 1) (x - 2) \dots . (x - 99) . 1}

f^{'} (1) = (0) + {1.1 (1 - 2) (1 - 3) \dots . (1 - 100)} + \dots . . (0) \dots . + (0)

f^{'} (1) = 1 (- 1) (- 2) (- 3) \dots . . (- 98) (- 99)

f^{'} (1) = - 99! (A n s)

f^{'} (x) = \frac{x!}{x} + \frac{x!}{x - 1} + \dots \dots . \frac{x!}{x - 100}