Найти f´(3) и f´(1), если f(x)=(x-1)^2(x-3)

Решите задачу:

$f(x)=(x-1)^2(x-3)$

$f'(x)=[(x-1)^2(x-3)]'=[(x-1)^2]'*(x-3)+(x-1)^2*[x-3]'=$

$=2*(x-1)^{2-1}*[x-1]'*(x-3)+(x-1)^2*[(x)'-(3)']=$

$=2*(x-1)^{1}*[(x)'-(1)']*(x-3)+(x-1)^2*[x^{1-1}-0]=$

$=2*(x-1)*[1-0]*(x-3)+(x-1)^2*[1-0]=$

$=2(x-1)(x-3)+(x-1)^2$

$f'(1)=2*(1-1)*(1-3)+(1-1)^2=0-0=0$

$f'(3)=2*(3-1)*(3-3)+(3-1)^2=0+2^2=4$