## Calculus, 10th Edition (Anton)

$\dfrac{da}{dt} = \dfrac{4t^3-6a^2}{4a^3-12at}$
In order to derivate this function you have to apply implicit differentation method. Then derivate the whole equation. Rember the put a' every time you derivate a $4a^3a'-4t^3-12aa't+6a^2=0$ Solve for a' and you have the answer $a'(4a^3-12at)=4t^3-6a^2$ $a' = \dfrac{4t^3-6a^2}{4a^3-12at}$