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 \(k\).