Find \(f'(1)\).

Write down \(g'(1)\).

Find \(g(1)\).

Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).

choosing chain rule     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u' = 4\)

correct derivative of \(f\)     A2

eg\(\,\,\,\,\,\)\(\frac{1}{2}{(4x + 5)^{ - \frac{1}{2}}} \times 4,{\text{ }}f'(x) = \frac{2}{{\sqrt {4x + 5} }}\)

\(f'(1) = \frac{2}{3}\)    A1     N2

[4 marks]

recognize that \(g'(x)\) is the gradient of the tangent     (M1)

eg\(\,\,\,\,\,\)\(g'(x) = m\)

\(g'(1) = 3\)    A1     N2

[2 marks]

recognize that R is on the tangent     (M1)

eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch

\(g(1) = 9\)    A1     N2

[2 marks]

\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (seen anywhere)     A1

\(h(1) = 3 \times 9{\text{ }}( = 27)\) (seen anywhere)     A1

choosing product rule to find \(h'(x)\)     (M1)

eg\(\,\,\,\,\,\)\(uv' + u'v\)

correct substitution to find \(h'(1)\)     (A1)

eg\(\,\,\,\,\,\)\(f(1) \times g'(1) + f'(1) \times g(1)\)

\(h'(1) = 3 \times 3 + \frac{2}{3} \times 9{\text{ }}( = 15)\)     A1

EITHER

attempt to substitute coordinates (in any order) into the equation of a straight line     (M1)

eg\(\,\,\,\,\,\)\(y - 27 = h'(1)(x - 1),{\text{ }}y - 1 = 15(x - 27)\)

\(y - 27 = 15(x - 1)\)     A1     N2

OR

attempt to substitute coordinates (in any order) to find the \(y\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(27 = 15 \times 1 + b,{\text{ }}1 = 15 \times 27 + b\)

\(y = 15x + 12\)     A1     N2

[7 marks]

Part a) was relatively well answered – the obvious errors seen were not using the chain rule correctly and simple fraction calculations being wrong.

In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this question. There was lack of understanding of tangents, gradients and their relationship to the original function, \(g\). A working sketch may have been beneficial but few were seen and many did a lot more work than required.

In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this question. There was lack of understanding of tangents, gradients and their relationship to the original function, \(g\). A working sketch may have been beneficial but few were seen and many did a lot more work than required.

In part d) although candidates recognized \(h(x)\) as a product and may have correctly found \(h(1)\), they did not necessarily use the product rule to find \(h'(x)\), instead incorrectly using \(h'(x) = f'(x) \times g'(x)\). It was rare for a candidate to get as far as finding the equation of a straight line but those who did usually gained full marks.