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.