Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises: 48


The derivative is $8x^3-9x^2$.

Work Step by Step

$y=x^2(2x^2-3x)=2x^4-3x^3$ $y'=(2x^4-3x^3)'=(2x^4)'-(3x^3)'=8x^3-9x^2$ 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 $y'=8x^3-9x^2$.
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