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

Hence, by repeated differentiation of the above differential equation, find the Maclaurin series for y as far as the term in ${x^3}$, showing that two of the terms are zero.

METHOD 1

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

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

now $\frac{{1 + {{\text{e}}^{ - x}}}}{2} = {{\text{e}}^y}$     M1

$\Rightarrow 1 + {{\text{e}}^{ - x}} = 2{{\text{e}}^y}$

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

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

Note: Only one of the two above A1 marks may be implied.

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} = - 1$     AG

Note: Candidates may find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ as a function of x and then work backwards from the given answer. Award full marks if done correctly.

METHOD 2

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

$\Rightarrow {{\text{e}}^y} = \frac{{1 + {{\text{e}}^{ - x}}}}{2}$     M1

$\Rightarrow {{\text{e}}^{ - x}} = 2{{\text{e}}^y} - 1$

$\Rightarrow x = - \ln (2{{\text{e}}^y} - 1)$     A1

$\Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}y}} = - \frac{1}{{2{{\text{e}}^y} - 1}} \times 2{{\text{e}}^y}$     M1A1

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

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} - 1$     AG

[5 marks]

METHOD 1

when $x = 0,{\text{ }}y = \ln 1 = 0$     A1

when $x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} - 1 = - \frac{1}{2}$     A1

$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = - \frac{{{{\text{e}}^{ - y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1A1

when $x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$     A1

$\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{e}}^{ - y}}}}{2}{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2} - \frac{{{{\text{e}}^{ - y}}}}{2}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}$     M1A1A1

when $x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{1}{2} \times \frac{1}{4} - \frac{1}{2} \times \frac{1}{4} = 0$     A1

$y = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \frac{{f'''(0)}}{{3!}}{x^3}$

$\Rightarrow y = 0 - \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots$     (M1)A1

two of the above terms are zero     AG

METHOD 2

when $x = 0,{\text{ }}y = \ln 1 = 0$     A1

when $x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} - 1 = - \frac{1}{2}$     A1

$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{ - {{\text{e}}^{ - y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - {{\text{e}}^{ - y}}}}{2}\left( {\frac{{{{\text{e}}^{ - y}}}}{2} - 1} \right) = \frac{{ - {{\text{e}}^{2y}}}}{4} + \frac{{{{\text{e}}^{ - y}}}}{2}$     M1A1

when $x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = - \frac{1}{4} + \frac{1}{2} = \frac{1}{4}$     A1

$\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \left( {\frac{{{{\text{e}}^{ - 2y}}}}{2} - \frac{{{{\text{e}}^{ - y}}}}{2}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1A1A1

when $x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - \frac{1}{2} \times \left( {\frac{1}{2} - \frac{1}{2}} \right) = 0$     A1

$y = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \frac{{f'''(0)}}{{3!}}{x^3}$

$\Rightarrow y = 0 - \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots$     (M1)A1

two of the above terms are zero     AG

[11 marks]

Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating ${{{\text{e}}^{ - y}}}$ with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.

Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating ${{{\text{e}}^{ - y}}}$ with respect to x. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.