Find \(f'(x)\) .

Let \(g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\) , for \(x > 0\) .

Show that \(g'(x) = \frac{1}{{x(x + 1)}}\) .

Let \(h(x) = \frac{1}{{x(x + 1)}}\) . The area enclosed by the graph of h , the x-axis and the lines \(x = \frac{1}{5}\)  and \(x = k\) is \(\ln 4\) . Given that \(k > \frac{1}{5}\) , find the value of k .

METHOD 1

evidence of choosing quotient rule     (M1)

e.g. \(\frac{{u'v - uv'}}{{{v^2}}}\)

evidence of correct differentiation (must be seen in quotient rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1\)

correct substitution into quotient rule     A1

e.g. \(\frac{{(x + 1)6 - 6x}}{{{{(x + 1)}^2}}}\) , \(\frac{{6x + 6 - 6x}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)    A1     N4

[5 marks]

METHOD 2

evidence of choosing product rule     (M1)

e.g. \(6x{(x + 1)^{ - 1}}\) , \(uv' + vu'\)

evidence of correct differentiation (must be seen in product rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ - 1}} = - 1{(x + 1)^{ - 2}} \times 1\)

correct working     A1

e.g. \(6x \times - {(x + 1)^{ - 2}} + {(x + 1)^{ - 1}} \times 6\) , \(\frac{{ - 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)   A1     N4

[5 marks]

METHOD 1

evidence of choosing chain rule     (M1)

e.g. formula, \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct reciprocal of \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}\) is \(\frac{{x + 1}}{{6x}}\) (seen anywhere)     A1

correct substitution into chain rule     A1

e.g. \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}\) , \(\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)\)

working that clearly leads to the answer     A1

e.g. \(\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)\) , \(\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)\) , \(\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

METHOD 2

attempt to subtract logs     (M1)

e.g. \(\ln a - \ln b\) , \(\ln 6x - \ln (x + 1)\)

correct derivatives (must be seen in correct expression)     A1A1

e.g. \(\frac{6}{{6x}} - \frac{1}{{x + 1}}\) , \(\frac{1}{x} - \frac{1}{{x + 1}}\)

working that clearly leads to the answer     A1

e.g. \(\frac{{x + 1 - x}}{{x(x + 1)}}\) , \(\frac{{6x + 6 - 6x}}{{6x(x + 1)}}\) , \(\frac{{6(x + 1 - x)}}{{6x(x + 1)}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

valid method using integral of  h(x) (accept missing/incorrect limits or missing \({\text{d}}x\) )     (M1)

e.g. \({\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x} \) , \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} \) 

recognizing that integral of derivative will give original function     (R1)

e.g. \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct substitution and subtraction     A1

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln (1)\)

setting their expression equal to \(\ln 4\)     (M1) 

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln (1) = \ln 4\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4\) , \(\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4} \)

correct equation without logs     A1

e.g.\(\frac{{6k}}{{k + 1}} = 4\) , \(6k = 4(k + 1)\) 

correct working     (A1)

e.g. \(6k = 4k + 4\) , \(2k = 4\)

\(k = 2\)    A1     N4

[7 marks]

In part (a), most candidates recognized the need to apply the quotient rule to find the derivative, and many were successful in earning full marks here.

In part (b), many candidates struggled with the chain rule, or did not realize the chain rule was necessary to find the derivative. Again, some candidates attempted to work backward from the given answer, which is not allowed in a "show that" question. A few clever candidates simplified the situation by applying properties of logarithms before finding their derivative.

For part (c), many candidates recognized the need to integrate the function, and that their integral would equal \(\ln 4\) . However, many did not recognize that the integral of h is g . Those candidates who made this link between the parts (b) and (c) often carried on correctly to find the value of k , with a few candidates having errors in working with logarithms.