## Calculus 8th Edition

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