1. Find (dy)/(dx) given: (a) xy^3-2x^2y^2+x^4=1

To find $\frac{dy}{dx}$, we need to differentiate the given equation with respect to $x$. Let's start by differentiating both sides of the equation:<br /><br />$\frac{d}{dx}(xy^3 - 2x^2y^2 + x^4) = \frac{d}{dx}(1)$<br /><br />Using the product rule for differentiation, we have:<br /><br />$y^3 + 3xy^2\frac{dy}{dx} - 4xy^2 - 4x^2y\frac{dy}{dx} + 4x^3 = 0$<br /><br />Now, we can solve for $\frac{dy}{dx}$ by isolating it on one side of the equation:<br /><br />$3xy^2\frac{dy}{dx} - 4x^2y\frac{dy}{dx} = -y^3 + 4xy^2 - 4x^3$<br /><br />$\frac{dy}{dx}(3xy^2 - 4x^2y) = -y^3 + 4xy^2 - 4x^3$<br /><br />$\frac{dy}{dx} = \frac{-y^3 + 4xy^2 - 4x^3}{3xy^2 - 4x^2y}$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is given by:<br /><br />$\frac{dy}{dx} = \frac{-y^3 + 4xy^2 - 4x^3}{3xy^2 - 4x^2y}$