## Calculus 8th Edition

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