Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 327: 15


$y' = \frac{3(3x+ 2)}{(3x+2)^3+1}$

Work Step by Step

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $y = \int ^{3x+2}_{1}\frac {t}{1+t^3}dt$ Since the upper bound is a function we need to use the chain rule: u = 3x+2 u' = 3 $y = \int ^{u}_{1}\frac {t}{1+t^3}dt$ Since $F(u) = \int^{u}_{0}f(t)dt$ $F'(u) = f(u)$ Thus, $y' = \frac{u}{1+u^3}u'$ $y' = \frac{3x+ 2}{(3x+2)^3+1}3$ $y' = \frac{3(3x+ 2)}{(3x+2)^3+1}$
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