Find \(f'(x)\) .

Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .

evidence of choosing product rule     (M1)

eg   \(uv' + vu'\)

correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\)     (A1)(A1)

\(f'(x) = {x^2}\cos x + 2x\sin x\)     A1 N4

[4 marks]

substituting \(\frac{\pi }{2}\) into their \(f'(x)\)     (M1)

eg   \(f'\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)

correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\)     (A1)

eg   \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\) 

\(f'\left( {\frac{\pi }{2}} \right) = \pi \)     A1 N2

[3 marks]

Many candidates correctly applied the product rule for the derivative, although a common error was to answer \(f'(x) = 2x\cos x\) .

Candidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was made in the original function. Some candidates did not know the values of sine and cosine at \({\frac{\pi }{2}}\) .