## Calculus 10th Edition

The derivative is $12x^3-12x^2+15$.
Product Rule: y'=$((u(x)(v(x))’=u'(x)v(x)+u(x)v’(x))$ $u(x)=(3x-4) ;u’(x)=(3)$ $v(x)=(x^3+5) ;v’(x)=(3x^2)$ $y'=(3)(x^3+5)+(3x-4)(3x^2)=$ $(3x^3+15)+(9x^3-12x^2)=$ $12x^3-12x^2+15$