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.