Solutions - 0240-09

$$\begin{array}{c}{x}^2\sin \left(5x^3\right)=x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(5x^3\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{5^{2n+1}}{(2n+1)!}{x}^{6n+5}\end{array}$$

The Coefficient-Derivative Identity says that $f^{(k)}(0)={\displaystyle \frac{a_k}{k!}}$ where $a_k$ is the coefficient of the $k^{\text{th}}$ power $x^k$.

Solve:

$$x^{35}=x^{6n+5}\gg \gg 35=6n+5\gg \gg n=5$$

So, for the coefficient $a_{35}$:

$$\begin{array}{c}{(-1)}^5\frac{5^{2\cdot 5+1}}{(2\cdot 5+1)!}\gg \gg -\frac{5^{11}}{11!}\\ \\ \gg \gg f^{(35)}(0)=35!a_{35}\gg \gg -5^{11}\frac{35!}{11!}\end{array}$$