Use the quotient rule to show that \(g'(x) = \frac{{1 - 2\ln x}}{{{x^3}}}\) .

The graph of g has a maximum point at A. Find the x-coordinate of A.

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\ln x = \frac{1}{x}\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{x^2} = 2x\) (seen anywhere)     A1A1

attempt to substitute into the quotient rule (do not accept product rule)     M1

e.g. \(\frac{{{x^2}\left( {\frac{1}{x}} \right) - 2x\ln x}}{{{x^4}}}\)

correct manipulation that clearly leads to result     A1

e.g. \(\frac{{x - 2x\ln x}}{{{x^4}}}\) , \(\frac{{x(1 - 2\ln x)}}{{{x^4}}}\) , \(\frac{x}{{{x^4}}}\) , \(\frac{{2x\ln x}}{{{x^4}}}\)

\(g'(x) = \frac{{1 - 2\ln x}}{{{x^3}}}\)     AG     N0

[4 marks]

evidence of setting the derivative equal to zero     (M1)

e.g. \(g'(x) = 0\) , \(1 - 2\ln x = 0\)

\(\ln x = \frac{1}{2}\)     A1

\(x = {{\rm{e}}^{\frac{1}{2}}}\)     A1     N2

[3 marks]

Many candidates clearly knew their quotient rule, although a common error was to simplify \(2x\ln x\) as \(2\ln {x^2}\) and then "cancel" the exponents.

For (b), those who knew to set the derivative to zero typically went on find the correct x-coordinate, which must be in terms of e, as this is the calculator-free paper. Occasionally, students would take \(\frac{{1 - 2\ln x}}{{{x^3}}} = 0\) and attempt to solve from \(1 - 2\ln x = {x^3}\) .