(i)[4]
Hence, Show that $\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}$.
(ii)[4]
Hence show that there are two points on the curve where the tangent is parallel to the $x$-axis, and determine the coordinates of these points.
Mathematics 9709 · AS & A Level · Differentiation
Hence, Show that $\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}$.
Hence show that there are two points on the curve where the tangent is parallel to the $x$-axis, and determine the coordinates of these points.
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State or indicate $y^3 + 3xy^2 \frac{dy}{dx}$ as the derivative of $xy^3$” …