Find the first two derivatives of $f\left(x\right)$ and hence find the Maclaurin series for $f\left(x\right)$ up to and including the $x^2$ term.

Show that the coefficient of $x^3$ in the Maclaurin series for $f\left(x\right)$ is zero.

Using the Maclaurin series for $\text{arctan}x$ and ${\text{e}}^{3x}-1$, find the Maclaurin series for $\text{arctan}\left({\text{e}}^{3x}-1\right)$ up to and including the $x^3$ term.

Hence, or otherwise, find $\underset{x\to 0}{\text{lim}}\frac{f\left(x\right)-1}{\text{arctan}\left({\text{e}}^{3x}-1\right)}$.

attempting to use the chain rule to find the first derivative     M1

$f^{\prime }\left(x\right)=\left(\text{cos}x\right){\text{e}}^{\text{sin}x}$       A1

attempting to use the product rule to find the second derivative      M1

$f^{″}\left(x\right)={\text{e}}^{\text{sin}x}\left(\text{co}{\text{s}}^{2}x-\text{sin}x\right)$ (or equivalent)        A1

attempting to find $f\left(0\right)$, $f^{\prime }\left(0\right)$ and $f^{″}\left(0\right)$       M1

$f\left(0\right)=1$; $f^{\prime }\left(0\right)=\left(\text{cos}0\right){\text{e}}^{\text{sin}0}=1$; $f^{″}\left(0\right)={\text{e}}^{\text{sin}0}\left(\text{co}{\text{s}}^{2}0-\text{sin}0\right)=1$        A1

substitution into the Maclaurin formula $f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2\text{!}}{f}^{″}\left(0\right)+\dots$       M1

so the Maclaurin series for $f\left(x\right)$ up to and including the $x^2$ term is $1+x+\frac{x^2}{2}$      A1

[8 marks]

METHOD 1

attempting to differentiate $f^{″}\left(x\right)$       M1

$f^{‴}\left(x\right)=\left(\text{cos}x\right){\text{e}}^{\text{sin}x}\left(\text{co}{\text{s}}^{2}x-\text{sin}x\right)-\left(\text{cos}x\right){\text{e}}^{\text{sin}x}\left(2\text{sin}x+1\right)$ (or equivalent)        A2

substituting $x=0$ into their $f^{‴}\left(x\right)$       M1

$f^{‴}\left(0\right)=1\left(1-0\right)-1\left(0+1\right)=0$

so the coefficient of $x^3$ in the Maclaurin series for $f\left(x\right)$ is zero    AG

METHOD 2

substituting $\text{sin}x$ into the Maclaurin series for ${\text{e}}^x$       (M1)

${\text{e}}^{\text{sin}x}=1+\text{sin}x+\frac{\text{si}{\text{n}}^{2}x}{2\text{!}}+\frac{\text{si}{\text{n}}^{3}x}{3\text{!}}+\dots$

substituting Maclaurin series for $\text{sin}x$       M1

${\text{e}}^{\text{sin}x}=1+\left(x-\frac{x^3}{3\text{!}}+\dots \right)+\frac{{\left(x-\frac{x^3}{3\text{!}}+\dots \right)}^2}{2\text{!}}+\frac{{\left(x-\frac{x^3}{3\text{!}}+\dots \right)}^3}{3\text{!}}+\dots$     A1

coefficient of $x^3$ is $-\frac{1}{3\text{!}}+\frac{1}{3\text{!}}=0$     A1

so the coefficient of $x^3$ in the Maclaurin series for $f\left(x\right)$ is zero    AG

[4 marks]

substituting $3x$ into the Maclaurin series for ${\text{e}}^x$       M1

${\text{e}}^{3x}=1+3x+\frac{{\left(3x\right)}^2}{2\text{!}}+\frac{{\left(3x\right)}^3}{3\text{!}}+\dots$      A1

substituting $\left({\text{e}}^{3x}-1\right)$ into the Maclaurin series for $\text{arctan}x$    M1

$\text{arctan}\left({\text{e}}^{3x}-1\right)=\left({\text{e}}^{3x}-1\right)-\frac{{\left({\text{e}}^{3x}-1\right)}^3}{3}+\frac{{\left({\text{e}}^{3x}-1\right)}^5}{5}-\dots$

$=\left(3x+\frac{{\left(3x\right)}^2}{2\text{!}}+\frac{{\left(3x\right)}^3}{3\text{!}}+\dots \right)-\frac{{\left(3x+\frac{{\left(3x\right)}^2}{2\text{!}}+\frac{{\left(3x\right)}^3}{3\text{!}}+\dots \right)}^3}{3}+\dots$    A1

selecting correct terms from above      M1

$=\left(3x+\frac{{\left(3x\right)}^2}{2\text{!}}+\frac{{\left(3x\right)}^3}{3\text{!}}\right)-\frac{{\left(3x\right)}^3}{3}$

$=3x+\frac{9x^2}{2}-\frac{9x^3}{2}$     A1

[6 marks]

METHOD 1

substitution of their series       M1

$\underset{x\to 0}{\text{lim}}\frac{x+\frac{x^2}{2}+\dots }{3x+\frac{9x^2}{2}+\dots }$       A1

$=\underset{x\to 0}{\text{lim}}\frac{1+\frac{x}{2}+\dots }{3+\frac{9x}{2}+\dots }$

$=\frac{1}{3}$     A1

METHOD 2

use of l’Hôpital’s rule      M1

$\underset{x\to 0}{\text{lim}}\frac{\left(\text{cos}x\right){\text{e}}^{\text{sin}x}}{\frac{3{\text{e}}^{3x}}{1+{\left({\text{e}}^{3x}-1\right)}^2}}$  (or equivalent)    A1

$=\frac{1}{3}$     A1

[3 marks]