High Order Derivative

Question

Find the ninth derivative of the following function at $x=0$: $$f(x) =\frac{\cos\left(4x^4\right)-1}{x^7}$$

So I did all the manipulations and I got the following Maclaurin Series:

$$\sum _{n=0}^{\infty }\,(-1)^n\frac{16x^{8n-7}}{(2n)!}-1$$

So to have $8n-7=9$, I got that $n=2$. Therefore, I thought that:

$$(-1)^2\frac{16x^9}{24}-1=\frac{16x^9}{24}-1$$

And, therefore, the ninth derivative would be:

$$\frac{16\cdot 9!}{24}-1=241919$$

But this was incorrect. Any help?

Answer 1

$$ \cos\left(4x^4\right)-1=\sum_{n=1}^\infty(-1)^n\frac { (4x^4)^{2n}}{(2n)!}$$

So

$$f(x) = \sum_{n=1}^\infty(-1)^n\frac { 16^n x^{8n-7}}{(2n)!} $$

Answer 2

When you substitute $(4x^4)$ for $x$ you get $(4x^4)^{2n}$ =$16^nx^{8n}$ You forgot the power for $16$