Computing $n$-th derivative $f(x)= \frac{\cos(5x^4)-1}{x^7}$ using MacLaurin Series.

Question

The prompt is to find the $9$-th derivative of the function f(x) defined as,
$$f(x)= \frac{\cos(5x^4)-1}{x^7}$$
at $x = 0$.

We are suggested to use the MacLaurin Series for $f(x)$.

Answer 1

$$f(x) = \frac{\cos(5x^4)-1}{x^7} = \sum_{n=1}^{+\infty} \frac{(-1)^n \cdot 5^{2n} \cdot x^{8n-7}}{(2n)!} $$ Therefore if $k \not\equiv -7 \mod 8$ we have $f^{(k)}(0)=0$.

Otherwise $\displaystyle f^{(8n-7)}(0) = \frac{(-25)^n \cdot (8n-7)!}{(2n)!}$

So $\displaystyle f^{(9)}(0) = \frac{625 \cdot 9!}{24}$.

I used $\displaystyle \cos(x) = \sum\limits_{n=0}^{+\infty} (-1)^n \cdot \frac{x^{2n}}{(2n)!}$ and $\displaystyle f(x) = \sum\limits_{n=0}^{+\infty} f^{(n)}(0) \cdot \frac{x^n}{n!}$