Answer
$0.584$ (mg/ml)
Work Step by Step
Step 1. Given $C(t)=\frac{4t}{1+t^3}+0.06t$, we have $C'(t)=\frac{4+4t^3-4t(3t^2)}{(1+t^3)^2}+0.06=\frac{4-8t^3}{(1+t^3)^2}+0.06$
Step 2. We can find for $t=20min=1/3hr$, $C'(1/3)=\frac{4-8(1/3)^3}{(1+(1/3)^3)^2}+0.06\approx3.504$
Step 3. The change of concentration from $20min$ to $30min(1/2hr)$ would be $\Delta C=C'(1/3)\Delta t=3.504(1/2-1/3)=0.584$ (mg/ml)
