Show that the points (0, 0) and (\(\sqrt {2\pi } \) , \( - \sqrt {2\pi } \)) on the curve \({{\text{e}}^{\left( {x + y} \right)}} = \cos \left( {xy} \right)\) have a common tangent.

attempt at implicit differentiation     M1

\({{\text{e}}^{\left( {x + y} \right)}}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = - \sin \left( {xy} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right)\)     A1A1

let \(x = 0\), \(y = 0\)     M1

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

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1\)     A1

let \(x = \sqrt {2\pi } \) , \(y = - \sqrt {2\pi } \)

\({{\text{e}}^0}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = - \sin \left( {- 2\pi } \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right) = 0\)

so \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 1\)     A1

since both points lie on the line \(y = - x\) this is a common tangent     R1

Note: \(y = - x\) must be seen for the final R1. It is not sufficient to note that the gradients are equal.

[7 marks]

Implicit differentiation was attempted by many candidates, some of whom obtained the correct value for the gradient of the tangent. However, very few noticed the need to go further and prove that both points were on the same line.