## Calculus: Early Transcendentals 8th Edition

$F'(t)=-\dfrac{AB+2ACt}{(Bt+Ct^{2})^{2}}$
$F(t)=\dfrac{At}{Bt^{2}+Ct^{3}}$ (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: $F'(t)=\dfrac{(Bt+Ct^{2})(A)'-(A)(Bt+Ct^{2})'}{(Bt+Ct^{2})^{2}}=...$ $...=\dfrac{(Bt+Ct^{2})(0)-A(B+2Ct)}{(Bt+Ct^{2})^{2}}=\dfrac{-A(B+2Ct)}{(Bt+Ct^{2})^{2}}=-\dfrac{AB+2ACt}{(Bt+Ct^{2})^{2}}$