## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Review - Exercises: 1

#### Answer

$y'=4(x^{2}+x^{3})^{3}(2x+3x^{2})$

#### Work Step by Step

Start with function: $y=(x^{2}+x^{3})^{4}$. Let $u = x^{2}+x^{3}$. Substitute u into function: $y=u^{4}$. Differentiate using power rule: $y' = 4u^{3}u'$. $u'=2x+3x^{2}$. Substitute values for $u$ and $u'$ back into equation to arrive at the answer: $y'=4(x^{2}+x^{3})^{3}(2x+3x^{2})$.

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