Answer
${f^{\left( 3 \right)}}\left( 0 \right) = - \dfrac{7}{4}$
Work Step by Step
From Table 2 of Section 11.7, we get
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
And also
${\left( {1 + x} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} = 1 + ax + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot$,
converges for $\left| x \right| \lt 1$.
Setting $a = \dfrac{1}{2}$, we get
$\sqrt {1 + x} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}\\
n
\end{array}} \right){x^n} = 1 + \dfrac{x}{2} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^2} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot$
Thus,
$f\left( x \right) = \left( {\sin x} \right)\sqrt {1 + x} $
$ = \left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \cdot\cdot\cdot} \right)\left( {1 + \dfrac{x}{2} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^2} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot} \right)$
Since we are to find ${f^{\left( 3 \right)}}\left( 0 \right)$, we need only the coefficient of ${x^3}$. Inspecting the multiplication of series above, we get the terms in ${x^3}$:
$f\left( x \right) = \cdot\cdot\cdot + \left( { - \dfrac{{{x^3}}}{{3!}} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^2}} \right) + \cdot\cdot\cdot$
$f\left( x \right) = \cdot\cdot\cdot + \left( { - \dfrac{1}{6} - \dfrac{1}{8}} \right){x^3} + \cdot\cdot\cdot$
$f\left( x \right) = \cdot\cdot\cdot + \left( { - \dfrac{7}{{24}}} \right){x^3} + \cdot\cdot\cdot$
According to the Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
So, $\dfrac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}} = - \dfrac{7}{{24}}$.
Therefore, ${f^{\left( 3 \right)}}\left( 0 \right) = - \dfrac{7}{4}$.
