Answer
$f'(x)=(x)(x-2)^3(6x-4).$
Work Step by Step
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x^2 ;u’(x)=2x $
$v(x)=(x-2)^4 $
$v'(x)$ is found using the Chain Rule:
$u=(x-2)$; $\dfrac{du}{dx}=1$
$\dfrac{d}{du}v(u)=4u^3$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=4(x-2)^3$
$f'(x)=2x(x-2)^4+4(x-2)^3(x^2)$
$=(x-2)^3(2x(x-2)+4x^2)$
$(x)(x-2)^3(6x-4).$
