## Calculus 10th Edition

$f'(x)=(2x-5)^2(8x-5)$
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: $u=(2x-5)$; $\dfrac{du}{dx}=2$ $\dfrac{d}{du}v(u)=3u^2$ $\dfrac{d}{dx}v(x)=\dfrac{d}{du}v(u)\times\dfrac{du}{dx}=2(3(2x-5)^2)=6(2x-5)^2.$ $f'(x)=(1)(2x-5)^3+(6(2x-5)^2)(x)=(2x-5)^2(8x-5).$