## Calculus 10th Edition

The derivative is $8x^3-9x^2$.
$y=x^2(2x^2-3x)=2x^4-3x^3$ $y'=(2x^4-3x^3)'=(2x^4)'-(3x^3)'=8x^3-9x^2$ First distribute the term $x^2$ onto the expression to get $2x^4-3x^3$ then using the power rule $((x^n)'=nx^{n-1})$ to differentiate the individual terms which by adding you get $y'=8x^3-9x^2$.