(i)[4]
Hence show that $\displaystyle \frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 - x^2}$.
(ii)[4]
Hence determine the exact coordinates of the two points on the curve where the gradient of the normal is $1$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is defined by the equation $x^2(x + 3y) - y^3 = 3$.
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This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Give or indicate $3y^2\frac{dy}{dx}$ as the derivative of $y^3$” …
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