Answer
$f'(x)=(2x-5)^2(8x-5)$
Work Step by Step
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x ;u’(x)=1 $
$v(x)=(2x-5)^3 $
$v'(x)$ is found using the Chain Rule:
$u=(2x-5)$; $\dfrac{du}{dx}=2$
$\dfrac{d}{du}v(u)=3u^2$
$\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=2(3(2x-5)^2)=6(2x-5)^2.$
$f'(x)=(1)(2x-5)^3+(6(2x-5)^2)(x)=(2x-5)^2(8x-5).$
