A curve is defined by the equation $8y\ln x - 2{x^2} + 4{y^2} = 7$. Find the equation of the tangent to the curve at the point where x = 1 and $y > 0$.

$8y \times \frac{1}{x} + 8\frac{{{\text{d}}y}}{{{\text{d}}x}}\ln x - 4x + 8y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     M1A1A1

Note: M1 for attempt at implicit differentiation. A1 for differentiating $8y\ln x$, A1 for differentiating the rest.

when $x = 1,{\text{ }}8y \times 0 - 2 \times 1 + 4{y^2} = 7$     (M1)

${y^2} = \frac{9}{4} \Rightarrow y = \frac{3}{2}{\text{ (as }}y > 0)$     A1

at $\left( {1,\frac{3}{2}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{2}{3}$     A1

$y - \frac{3}{2} = - \frac{2}{3}(x - 1)$ or $y = - \frac{2}{3}x + \frac{{13}}{6}$     A1

[7 marks]

The implicit differentiation was generally well done. Some candidates did not realise that they needed to substitute into the original equation to find $y$. Others wasted a lot of time rearranging the derivative to make $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ the subject, rather than simply putting in the particular values for $x$ and $y$.