Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 82

$f^{58}(0)=0$

Maclaurin series of $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(0)x^{n}}{n!}$ Coefficient of $x^{58}=\frac{f^{58}(0)}{n!}$ Note that the powers of $x$ in the expansion of $(1+x^{3})^{30}$ must be multiples of $58$. But $58$ is not a multiple of $3$. Thus, $0=\frac{f^{58}(0)}{n!}$ Hence, $f^{58}(0)=0$