Let $f(x) = {{\rm{e}}^x}\cos x$ . Find the gradient of the normal to the curve of f at $x = \pi$ .

evidence of choosing the product rule     (M1)

$f'(x) = {{\rm{e}}^x} \times ( - \sin x) + \cos x \times {{\rm{e}}^x}$ $( = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x)$     A1A1 

substituting $\pi$     (M1)

e.g.  $f'(\pi ) = {{\rm{e}}^\pi }\cos \pi  - {{\rm{e}}^\pi }\sin \pi$ , ${{\rm{e}}^\pi }( - 1 - 0)$ , $- {{\rm{e}}^\pi }$

taking negative reciprocal      (M1)

e.g. $- \frac{1}{{f'(\pi )}}$

gradient is $\frac{1}{{{{\rm{e}}^\pi }}}$     A1     N3 

[6 marks]

Candidates familiar with the product rule easily found the correct derivative function. Many substituted $\pi$ to find the tangent gradient, but surprisingly few candidates correctly considered that the gradient of the normal is the negative reciprocal of this answer.