Method 1: If you have memorised the standard power series for \(\raise 0.3pt{e^x}\) (as above) you can just substitute \(3x\) for \(\raise 0.3pt{x}\) and obtain the answer almost immediately. This method receives full credit in SQA marking schemes.

$$ \begin{eqnarray} f(x) &=& e^{3x}\\[6pt] &=& 1+3x+\small\frac{{3x}^2}{2!}\normalsize+\small\frac{{3x}^3}{3!}\normalsize+\tiny\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2!}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{3!}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \\[6pt] \end{eqnarray} $$

Method 2: If you haven't memorised the power series for \(\raise 0.3pt{e^x}\small,\) you can use Maclaurin expansion from scratch:

$$ \begin{matrix} f(x)=e^{3x} \:&\: f(0)=1 \\ f'(x)=3e^{3x} \:&\: f'(0)=3 \\ f''(x)=9e^{3x} \:&\: f''(0)=9 \\ f'''(x)=27e^{3x} \:&\: f'''(0)=27 \\ \end{matrix} $$

Then we substitute into the general Maclaurin series:

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\tiny\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2!}\normalsize\tiny\,\normalsize x^2+\small\frac{27}{3!}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \\[6pt] &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \end{eqnarray} $$

Method 1: If you have memorised the standard power series for \(\raise 0.3pt{sin\,x}\) you can just substitute \(4x\) for \(\raise 0.3pt{x}\small.\)

$$ \begin{eqnarray} f(x) &=& sin\,4x\\[6pt] &=& 4x-\small\frac{(4x)^3}{3!}\normalsize+\tiny\,\cdots \\[6pt] &=& 4x-\small\frac{64}{3!}\normalsize x^3+\tiny\,\cdots \\[6pt] &=& 4x-\small\frac{32}{3}\normalsize x^3+\tiny\,\cdots \\[6pt] \end{eqnarray} $$

Method 2: If you haven't memorised the power series for \(\raise 0.3pt{sin\,x}\small,\) use Maclaurin expansion from scratch.

$$ \begin{matrix} f(x)=sin\,4x \:&\: f(0)=0 \\ f'(x)=4\,cos\,4x \:&\: f'(0)=4 \\ f''(x)=-\!16\,sin\,4x \:&\: f''(0)=0 \\ f'''(x)=-\!64\,cos\,4x \:&\: f'''(0)=-\!64 \\ \end{matrix} $$

Then substitute into the general Maclaurin series:

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\frac{f'''(0)}{3!}\normalsize x^3\!\small+\normalsize\!\small\,\tiny\cdots \\[6pt] &=& 0+4x+0-\small\frac{64}{3!}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \\[6pt] &=& 4x-\small\frac{32}{3}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \end{eqnarray} $$

In the first two examples, we obtained the following:

$$ \begin{eqnarray} e^{3x} &=& 1+3x+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^2+\small\frac{9}{2}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots\\[6pt] sin\,4x &=& 4x-\small\frac{32}{3}\normalsize\tiny\,\normalsize x^3+\tiny\,\cdots \\[6pt] \end{eqnarray} $$

To obtain the Maclaurin expansion for the product of these two functions, we simply multiply their respective expansions, ignoring anything that will multiply to a higher power of \(\raise 0.3pt{x}\) than \(3\small.\)

$$ \begin{eqnarray} && e^{3x}\tiny\,\normalsize sin\,4x\\[6pt] &=& \small\left(\!\normalsize1\!\small+\normalsize\!3x\!\small+\normalsize\!\small\frac{9}{2}\normalsize x^2\!\small+\normalsize\!\small\frac{9}{2}\normalsize x^3\!\small+\normalsize\!\tiny\,\cdots\small\!\right)\left(\!\normalsize4x\small-\normalsize\!\small\frac{32}{3}\normalsize x^3\!\small+\normalsize\!\tiny\,\cdots\small\!\right)\normalsize \\[6pt] &=& 1.4x+3x.4x+\small\frac{9}{2}\normalsize x^2.4x-1.\small\frac{32}{3}\normalsize x^3+\normalsize\!\tiny\,\cdots \\[6pt] &=& 4x+12x^2+18x^3-\small\frac{32}{3}\normalsize x^3+\normalsize\!\tiny\,\cdots \\[6pt] &=& 4x+12x^2+\small\frac{54}{3}\normalsize x^3-\small\frac{32}{3}\normalsize x^3+\normalsize\!\tiny\,\cdots \\[6pt] &=& 4x+12x^2+\small\frac{22}{3}\normalsize x^3+\normalsize\!\tiny\,\cdots \end{eqnarray} $$

The beauty of this solution is that we don't have to go to the bother of differentiating \(\raise 0.3pt{e^{3x}\tiny\,\normalsize sin\,4x\small,}\) finding Maclaurin series for each expression within the derivative and then combining them.

Instead, we just differentiate the Maclaurin series for \(\raise 0.3pt{e^{3x}\tiny\,\normalsize sin\,4x}\) term-by-term. Simple!

$$ \begin{gather} e^{3x}\tiny\,\normalsize sin\,4x=4x+12x^2+\small\frac{22}{3}\normalsize x^3+\normalsize\!\tiny\,\cdots\\[6pt] \small\frac{d}{dx}\left(\normalsize e^{3x}\tiny\,\normalsize sin\,4x\small\right)\normalsize=4+24x+22x^2+\tiny\,\cdots \\[6pt] \end{gather} $$

\(\large\frac{d}{dx}\normalsize(tan\,x)=sec^{2}\,x\) so we can integrate term-by-term to obtain the Maclaurin expansion for \(\raise 0.2pt{tan\,x}\small.\)

$$ \begin{eqnarray} \tan\,x&=& \!\!\int\small\left(\normalsize 1\small+\normalsize x^2+\small\frac{2}{3}\normalsize x^4\small+\normalsize\tiny\ \cdots\normalsize\small\right)\normalsize\,dx \\[6pt] &=& x+\small\frac{1}{3}\normalsize x^3+\small\frac{2}{15}\normalsize x^5+\tiny\ \cdots\\[6pt] \end{eqnarray} $$

Finally we substitute substitute \(\raise 0.2pt{2x}\) for \(\raise 0.3pt{x}\small,\) and simplify:

$$ \begin{eqnarray} \tan\,2x &=& 2x+\small\frac{1}{3}\normalsize (2x)^3+\small\frac{2}{15}\normalsize (2x)^5+\tiny\ \cdots\\[6pt] &=& 2x+\small\frac{1}{3}\normalsize (8x^3)+\small\frac{2}{15}\normalsize (32x^5)+\tiny\ \cdots\\[6pt] &=& 2x+\small\frac{8}{3}\normalsize x^3+\small\frac{64}{15}\normalsize x^5+\tiny\ \cdots\\[6pt] \end{eqnarray} $$

This question combines Maclaurin series and the quotient rule for differentiation.

$$ \begin{eqnarray} f(x)\!&=&\!f(0)\!\small+\normalsize\!f'(0)x\!\small+\normalsize\!\small\frac{f''(0)}{2!}\normalsize x^2\!\small+\normalsize\!\small\,\tiny\cdots \\[6pt] &=& 1+\small\left(\!\frac{1}{1\!+\!1^4}\!\right)\normalsize x+\frac{f''(0)}{2!}\normalsize x^2\!\normalsize+\tiny\,\cdots \end{eqnarray} $$

Before we go any further, we need to differentiate to obtain \(f''(x)\small.\)

$$ \begin{eqnarray} f''(x)\!&=& \small\frac{\{1\!+\!(x\!+\!1)^4\}.1-(x\!+\!1)\{4(x\!+\!1)^3\}}{\{1\!+\!(x\!+\!1)^4\}^2}\\[6pt] \end{eqnarray} $$

There is no need to simplify the expression above, because its only purpose is to evaluate \(f''(0)\small.\)

$$ \begin{eqnarray} f''(0)\!&=& \small\frac{1+1^4-(1)(4(1^3))}{(1+1^4)^2}\\[6pt] &=& \small\frac{1+1-4}{2^2}\\[6pt] &=& \small\frac{-2}{4}\\[6pt] &=& -\!\small\frac{1}{2}\\[6pt] \end{eqnarray} $$

Putting this together, we obtain:

$$ \begin{eqnarray} f(x)\!&=& 1+\small\left(\!\frac{1}{1\!+\!1^4}\!\right)\normalsize x-\small\frac12\normalsize\left(\!\small\frac{x^2}{2!}\normalsize\!\right)\!\normalsize+\tiny\,\cdots \\[6pt] &=& 1+\small\frac12\,x-\small\frac14\,x^2\normalsize+\tiny\,\cdots \end{eqnarray} $$