#### Answer

Differentiate $F(t)=\frac{At}{Bt^2+Ct^3}$
$F'(t)=\frac{-ABt^2-2ACt^3}{(Bt^2+Ct^3)^2}$

#### Work Step by Step

We are given $F(t)=\frac{At}{Bt^2+Ct^3}$
Assuming $A, B,$ and $C$ are constants, use the quotient rule to differentiate.
$F'(t)=\frac{A(Bt^2+Ct^3)-At(2Bt + 3Ct^2)}{(Bt^2+Ct^3)^2}$
$= \frac{ABt^2+ACt^3-2ABt^2-3ACt^3}{(Bt^2+Ct^3)^2}$
$= \frac{-ABt^2-2ACt^3}{(Bt^2+Ct^3)^2}$