Question

f(x)=x(x-1)(x-2)...(x-19)(x-20)
Найти f'(0)

Answer 1

Обозначим $h(x)=(x-1)(x-2)...(x-19)(x-20)$
$f(x)=x*h(x)$

$f'(x)=(x*h(x))=(x)'*h(x)+x*h'(x)=\\1*h(x)+x*h'(x)=h(x)+x*h'(x)$
$f'(0)=h(0)+0*h'(x)=h(0)$
$f'(0)=h(0)=(0-1)(0-2)*...*(0-19)*(0-20)=\\(-1)*(-2)*(-3)*...*(-19)*(-20)=\\1*2*3*..*19*20=20!=2432902008176640000$