Differentiate the equation with respect to the variable and express in terms of . (Use D for ).
$\displaystyle x^4 + 3xy^2 = 8$
$\displaystyle \frac{d}{dt}(x^4 + 3xy^2) = \frac{d}{dt}(8)$
$\displaystyle \frac{d}{dt}(x^4) + \frac{d}{dt}(3xy^2) = 0$
$\displaystyle \frac{d}{dx}(x^4)\,\frac{dx}{dt} + 3x\,\frac{d}{dt}(y^2) + y^2\,\frac{d}{dt}(3x) = 0$
$\displaystyle 4x^3\,\frac{dx}{dt} + 3x\,\frac{d}{dy}(y^2)\,\frac{dy}{dt} + y^2\,\frac{d}{dx}(3x)\,\frac{dx}{dt} = 0$
$\displaystyle 4x^3\,\frac{dx}{dt} + 6xy\,\frac{dy}{dt} + 3y^2\,\frac{dx}{dt} = 0$
$\displaystyle (4x^3 + 3y^2)\,\frac{dx}{dt} + 6xy\,\frac{dy}{dt} = 0$
$\displaystyle 6xy\,\frac{dy}{dt} = -(4x^3 + 3y^2)\,\frac{dx}{dt}$
$\displaystyle \frac{dy}{dt} = -\left(\frac{4x^3 + 3y^2}{6xy}\right)\frac{dx}{dt}$.