Using the power series of $\sin x^3$, the value of $f^{(15)}(0)$ is equal to $k\cdot11!$. Find the value of $k$.

Question

I have the following question:

Using the power series of $\sin x^3$, the value of $f^{(15)}(0)$ is equal to $k\cdot11!$. Find the value of $k$.

I tried to write the power series using the one from $\sin(x)$: $$\sin(x^{3})=\sum_{n=1}^{+\infty}(-1)^{n}\frac{x^{6n+3}}{(2n+1)!}$$

So, since $f^{(15)}(0)$ is related to $x^{15}$, so I got $n=2$. But I really don't know how can I use it.

Answer 1

Since $\dfrac{f^{(15)}(0)}{15!}=\dfrac{(-1)^2}{5!}=\dfrac1{5!}$, you have $f^{(15)}(0)=\dfrac{15!}{5!}$. You can get the value of $k$ from this.