Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 24

Answer

$F'(t)=-\dfrac{AB+2ACt}{(Bt+Ct^{2})^{2}}$

Work Step by Step

$F(t)=\dfrac{At}{Bt^{2}+Ct^{3}}$ (We assume $A$,$B$ and $C$ are constants) Take out common factor $t$: $F(t)=\dfrac{At}{t(Bt+Ct^{2})}=\dfrac{A}{Bt+Ct^{2}}$ Differentiate using the quotient rule: $F'(t)=\dfrac{(Bt+Ct^{2})(A)'-(A)(Bt+Ct^{2})'}{(Bt+Ct^{2})^{2}}=...$ $...=\dfrac{(Bt+Ct^{2})(0)-A(B+2Ct)}{(Bt+Ct^{2})^{2}}=\dfrac{-A(B+2Ct)}{(Bt+Ct^{2})^{2}}=-\dfrac{AB+2ACt}{(Bt+Ct^{2})^{2}}$
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