If \(y = \ln \left( {\frac{1}{3}(1 + {{\text{e}}^{ - 2x}})} \right)\), show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{2}{3}({{\text{e}}^{ - y}} - 3)\) .

\(y = \ln \left( {\frac{1}{3}(1 + {{\text{e}}^{ - 2x}})} \right)\)

EITHER

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - \frac{2}{3}{{\text{e}}^{ - 2x}}}}{{\frac{1}{3}(1 + {{\text{e}}^{ - 2x}})}}\)     M1A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2{{\text{e}}^{ - 2x}}}}{{1 + {{\text{e}}^{ - 2x}}}}\)     A1

\({{\text{e}}^y} = \frac{1}{3}(1 + {{\text{e}}^{ - 2x}})\)     M1

Now \({{\text{e}}^{ - 2x}} = 3{{\text{e}}^y} - 1\)     A1

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2(3{{\text{e}}^y} - 1)}}{{1 + 3{{\text{e}}^y} - 1}}\)     A1

\( = - \frac{2}{{3{{\text{e}}^y}}}(3{{\text{e}}^y} - 1)\)

\( = - \frac{2}{3}(3 - {{\text{e}}^{ - y}})\)     A1

\( = \frac{2}{3}({{\text{e}}^{ - y}} - 3)\)     AG

OR

\({{\text{e}}^y} = \frac{1}{3}(1 + {{\text{e}}^{ - 2x}})\)     M1A1

\({{\text{e}}^y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{2}{3}{{\text{e}}^{ - 2x}}\)     M1A1

Now \({{\text{e}}^{ - 2x}} = 3{{\text{e}}^y} - 1\)     (A1)

\( \Rightarrow {{\text{e}}^y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{2}{3}(3{{\text{e}}^y} - 1)\)

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{2}{3}{{\text{e}}^{ - y}}(3{{\text{e}}^y} - 1)\)     (A1)

\( = \frac{2}{3}( - 3 + {{\text{e}}^{ - y}})\)     (A1)

\( = \frac{2}{3}({{\text{e}}^{ - y}} - 3)\)     AG

Note: Only two of the three (A1) marks may be implied.

[7 marks]

Solutions were generally disappointing with many candidates being awarded the first 2 or 3 marks, but then going no further.