If f(x)=(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) , then the value of ( textf)' (5) is equal to

Question

If f(x)=(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) , then the value of ( textf)' (5) is equal to (A) 0 (B) 120 (C) 24 (D) 5

Tardigrade Admin · Accepted Answer

f prime(x)=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5) + ldots+(x-1)(x-2)(x-3)(x-4) Putting x=5, we get, f prime(5)=0+0+0+0+(5-1)(5-2)(5-3)(5-4) f prime(5)=24

Q. If $f (x) = (x - 1) (x - 2) (x - 3) (x - 4) (x - 5)$ , then the value of $(f)^{^{'}} (5)$ is equal to

$0$

$120$

$24$

$5$

$f^{'} (x) = (x - 2) (x - 3) (x - 4) (x - 5) + (x - 1) (x - 3) (x - 4) (x - 5)$ $+ \dots + (x - 1) (x - 2) (x - 3) (x - 4)$
Putting $x = 5$ , we get,
$f^{'} (5) = 0 + 0 + 0 + 0 + (5 - 1) (5 - 2) (5 - 3) (5 - 4)$
$f^{'} (5) = 24$

Q. If f(x)=(x−1)(x−2)(x−3)(x−4)(x−5) , then the value of (f)′(5) is equal to

0

120

24

5

f′(x)=(x−2)(x−3)(x−4)(x−5)+(x−1)(x−3)(x−4)(x−5) +…+(x−1)(x−2)(x−3)(x−4) Putting x=5, we get, f′(5)=0+0+0+0+(5−1)(5−2)(5−3)(5−4) f′(5)=24

Q. If $f (x) = (x - 1) (x - 2) (x - 3) (x - 4) (x - 5)$ , then the value of $(f)^{^{'}} (5)$ is equal to

$0$

$120$

$24$

$5$

$f^{'} (x) = (x - 2) (x - 3) (x - 4) (x - 5) + (x - 1) (x - 3) (x - 4) (x - 5)$ $+ \dots + (x - 1) (x - 2) (x - 3) (x - 4)$
Putting $x = 5$ , we get,
$f^{'} (5) = 0 + 0 + 0 + 0 + (5 - 1) (5 - 2) (5 - 3) (5 - 4)$
$f^{'} (5) = 24$