Answer
$\frac{dx}{dt} = \frac{-9}{2}$
Work Step by Step
Given that $\frac{dy}{dt} = \frac{1}{2}$ and $x^2 y^3=\frac{4}{27}$
on differentiating with respect to time
$x^2\frac{dy^3}{dt} + y^3\frac{d x^2}{dt} = \frac{d4}{27dt}$
$3x^2y^2\frac{dy}{dt} + 2y^3x\frac{dx}{dt} =0 $
when x=2,
$y^3= \frac{4}{27\times 4}$
so $y=\frac{1}{3}$
then $ 3\times 4\times\frac{1}{18} - 2\times \frac{1}{27}(2)\frac{dx}{dt}=0$
$\frac{dx}{dt} = \frac {-9}{2}$
Thus, the answer is $\frac{dx}{dt} = \frac{-9}{2}$
