(i)[4]
Show that the result is $\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}$.
(ii)[4]
Hence show that the curve has two points where the tangent is parallel to the $x$-axis and determine their coordinates.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is defined by the equation $2x^4 + xy^3 + y^4 = 10$.
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This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State or imply $y^3 + 3xy^2 \frac{dy}{dx}$ for the derivative of $xy^3$” …
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