Answer
${f^{\left( 8 \right)}}\left( 0 \right) = - 1575$
Work Step by Step
Write $f\left( x \right) = \sqrt {1 - {x^2}} = {\left( {1 + \left( { - {x^2}} \right)} \right)^{1/2}}$.
By Theorem 3:
${\left( {1 + x} \right)^a} = 1 + \dfrac{a}{{1!}}x + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} + \cdot\cdot\cdot$
converges for $\left| x \right| \lt 1$.
Setting $a = \dfrac{1}{2}$ and substituting $ - {x^2}$ for $x$ in the series above gives
$f\left( x \right) = \sqrt {1 - {x^2}} = {\left( {1 + \left( { - {x^2}} \right)} \right)^{1/2}}$
$ = 1 - \dfrac{{\dfrac{1}{2}}}{{1!}}{x^2} + \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)}}{{2!}}{x^4} - \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)}}{{3!}}{x^6} + \cdot\cdot\cdot + {\left( { - 1} \right)^n}\left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^{2n}} + \cdot\cdot\cdot$
converges for $\left| { - {x^2}} \right| \lt 1$.
Thus,
$f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^{2n}}$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$.
According to Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
We set $n=4$ in the series $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\left( {\begin{array}{*{20}{c}}
{1/2}\\
n
\end{array}} \right){x^{2n}}$, such that the coefficient of ${x^8}$ be ${\left( { - 1} \right)^4}\left( {\begin{array}{*{20}{c}}
{1/2}\\
4
\end{array}} \right)$.
Thus,
$\dfrac{{{f^{\left( 8 \right)}}\left( 0 \right)}}{{8!}} = {\left( { - 1} \right)^4}\left( {\begin{array}{*{20}{c}}
{1/2}\\
4
\end{array}} \right)$
${f^{\left( 8 \right)}}\left( 0 \right) = {\left( { - 1} \right)^4}\left( {\begin{array}{*{20}{c}}
{1/2}\\
4
\end{array}} \right)8! = \dfrac{{\dfrac{1}{2}\left( { - \dfrac{1}{2}} \right)\left( { - \dfrac{3}{2}} \right)\left( { - \dfrac{5}{2}} \right)}}{{4!}}8! = - \dfrac{{15}}{{16}}\left( {5\cdot 6\cdot 7\cdot 8} \right)$
So, ${f^{\left( 8 \right)}}\left( 0 \right) = - 1575$.
