Use implicit differentiation to show that $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4y - 3{x^2}}}{{3{y^2} - 4x}}$.

Find the value of $k$.

$3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4\left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)$    M1A1

$(3{y^2} - 4x)\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4y - 3{x^2}$    A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4y - 3{x^2}}}{{3{y^2} - 4x}}$    AG

[3 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow 4y - 3{x^2} = 0$    (M1)

substituting $x = k$ and $y = \frac{3}{4}{k^2}$ into ${x^3} + {y^3} = 4xy$     M1

${k^3} + \frac{{27}}{{64}}{k^6} = 3{k^3}$    A1

attempting to solve ${k^3} + \frac{{27}}{{64}}{k^6} = 3{k^3}$ for $k$     (M1)

$k = 1.68{\text{ }}\left( { = \frac{4}{3}\sqrt[3]{2}} \right)$    A1

Note:     Condone substituting $y = \frac{3}{4}{x^2}$ into ${x^3} + {y^3} = 4xy$ and solving for $x$.

[5 marks]

Part (a) was generally well done. Some use of partial differentiation accompanied by rudimentary partial derivative notation was observed in a few candidate’s solutions.

In part (b), a large number of candidates knew to use $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ and seemingly understood the required solution plan but were unable to correctly substitute $x = k$ and $y = \frac{{3{k^2}}}{4}$ into the relation and solve for $k$.