Find the gradient of the curve ${{\text{e}}^{xy}} + \ln \left( {{y^2}} \right) + {{\text{e}}^y} = 1 + {\text{e}}$ at the point (0, 1) .

${{\text{e}}^{xy}} + \ln \left( {{y^2}} \right) + {{\text{e}}^y} = 1 + {\text{e}}$

${{\text{e}}^{xy}}\left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) + \frac{2}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {{\text{e}}^y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ , at (0, 1)      A1A1A1A1A1

$1\left( {1 + 0} \right) + 2\frac{{{\text{d}}y}}{{{\text{d}}x}} + {\text{e}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$

$1 + 2\frac{{{\text{d}}y}}{{{\text{d}}x}} + {\text{e}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{1}{{2 + {\text{e}}}}$   ($= -0.212$)     M1A1     N2

[7 marks]

Implicit differentiation is usually found to be difficult, but on this occasion there were many correct solutions. There were also a number of errors in the differentiation of ${e^{xy}}$ , and although these often led to the correct final answer, marks could not be awarded.