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.