Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 41

Answer

$G'(y)=-\dfrac{3ABy^2}{(Ay^3+B)^2}$

Work Step by Step

$G(y)=\dfrac{B}{Ay^3+B}$ $A$ and $B$ are constants while $y$ is the variable. This is important to know because the derivative of a constant is always 0. Use the quotient rule: $\Bigg(\dfrac{f}{g}\Bigg)'=\dfrac{gf'-fg'}{g^2}$ so... $G'(y)=\dfrac{(Ay^3+B)(0)-(B)(3Ay^2+0)}{(Ay^3+B)^2}$ Perform multiplication: $G'(y)=\dfrac{-3ABy^2}{(Ay^3+B)^2}$' which can be rewritten as: $G'(y)=-\dfrac{3ABy^2}{(Ay^3+B)^2}$
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