Show that $f''(x) = - 2{{\rm{e}}^x}\sin x$ .

Determine the Maclaurin series for $f(x)$ up to and including the term in ${x^4}$ .

By differentiating your series, determine the Maclaurin series for ${{\rm{e}}^x}\sin x$ up to the term in ${x^3}$ .

$f'(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x$     A1

$f''(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\cos x$     A1

$= - 2{{\rm{e}}^x}\sin x$     AG

[2 marks]

$f'''(x) = - 2{{\rm{e}}^x}\sin x - 2{{\rm{e}}^x}\cos x$     A1

${f^{IV}}(x) = - 4{{\rm{e}}^x}\cos x$     A1

$f(0) = 1$, $f'(0) = 1$ ,$f''(0) = 0$, $f'''(0) = - 2$, ${f^{IV}}(0) = - 4$     (A1)

the Maclaurin series is

${{\rm{e}}^x}\cos x = 1 + x - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6} +  \ldots$     M1A1

Note: Accept multiplication of series method.

[5 marks]

differentiating,

${{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x = 1 - {x^2} - \frac{{2{x^3}}}{3} +  \ldots$     M1A1

${{\rm{e}}^x}\sin x = 1 + {x^{}} - \frac{{{x^3}}}{3} +  \ldots  - 1 + {x^2} + \frac{{2{x^3}}}{3} +  \ldots$     M1

$= x + {x^2} + \frac{{{x^3}}}{3} +  \ldots$     A1

[4 marks]