Answer
$0.584{\frac{mg}{ml}}$
Work Step by Step
We are given: $C(t)=1+\frac{4t}{1+t^3}-e^{(-0.06t)}(\frac{mg}{ml})$
differentiate with respect to t:
$\frac{dC}{dt}=\frac{4(1+t^3)\frac{dt}{dt}-4t\frac{d(1+t^3)}{dt}}{(1+t^3)^2}+0.06e^{(-0.06t)}$
$\frac{dC}{dt}=\frac{4(1+t^3)-4t\times3t^2}{(1+t^3)^2}+0.06e^{(-0.06t)}$
$\frac{dC}{dt}=\frac{4-8t^3}{(1+t^3)^2}+0.06e^{(-0.06t)}$
change in time is from 20 min to 30 min :
change in time$=30-20=10min=\frac{10}{60}hr=\frac{1}{6}hr$
$dC=C(a')$ (change in time)
$dC=(\frac{4-8{(\frac{1}{6})^3}}{(1+(\frac{1}{6})^2)})+0.06e^{(-0.06\frac{1}{6})}\approx0.584{\frac{mg}{ml}}$
