Answer
${f^{\left( 7 \right)}}\left( 0 \right) = - 720$
${f^{\left( 8 \right)}}\left( 0 \right) = 0$
Work Step by Step
From Table 2:
${\tan ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}} = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \le 1$.
According to Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
So,
$\dfrac{{{f^{\left( 7 \right)}}\left( 0 \right)}}{{7!}} = - \dfrac{1}{7}$
${f^{\left( 7 \right)}}\left( 0 \right) = - \dfrac{{7!}}{7} = - 6! = - 720$
Since the even terms vanish, ${f^{\left( 8 \right)}}\left( 0 \right) = 0$.
