Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 30

Answer

$h'(t)=\dfrac{2t^3(4-t^3)}{(t^3+2)^3}.$

Work Step by Step

Using the quotient rule: $h'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$. $u(t)=t^4 ;u'(t)=4t^3$. $v(t)=(t^3+2)^2 $ $v'(t)$ is found using the chain rule: $u=t^3+2$; $\dfrac{du}{dt}=3t^2$ $\dfrac{d}{du}v(u)=2u$ $\dfrac{d}{dt}v(t)=\dfrac{d}{du}v(u)\times\dfrac{du}{dt}=6t^2(t^3+2).$ $h'(t)=\dfrac{(4t^3)(t^3+2)^2-(6t^2(t^3+2))(t^4)}{(t^3+2)^4}$ $=\dfrac{2t^3(4-t^3)}{(t^3+2)^3}.$
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