Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 36


$$ y'=(x+1)^2(x+4)^3(7x+16) .$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(x^n)'=nx^{n-1}$ Since $ y=(x+1)^3(x+4)^4$, then by the product and chain rules, the derivative $y'$ is given by $$ y'=3(x+1)^2(1)(x+4)^4+4(x+1)^3(x+4)^3(1)=3(x+1)^2(x+4)^4+4(x+1)^3(x+4)^3 =(x+1)^2(x+4)^3(3(x+4)+4(x+1))=(x+1)^2(x+4)^3(7x+16) .$$
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