This is a Calculus exam task:
Let $f : \Bbb R \to \Bbb R$ be a function defined as $f(x) = \cos(x^3)$. Calculate $f^{(25)}(0)$.
Searching for an answer online only yielded answers to questions such as $\cos^3(x)$ or $\cos(3x)$, but not my example. I've tried deriving the function twice thus getting the following differential equation:
$$y'' = -6x\sin(x^3)-9x^4y$$
While I do get only a single recursion by this method, I get stuck with a $(n-3)^{\text{rd}}$ derivative of $\sin(x^3)$ after applying the Leibniz formula:
$$y^{(n)}=-6\sum_{k=0}^{n-2}\binom{n-2}{k}x^{(k)}\left(\sin(x^3)\right)^{(n-2-k)}-9\sum_{k=0}^{n-2}\binom{n-2}{k}(x^4)^{(k)}y^{(n-2-k)}$$
$$\Rightarrow y^{(n)}(0)=-6(n-2)\left(\sin(x^3)\right)^{(n-3)} -9(n-2)(n-3)(n-4)(n-5)y^{(n-6)}(0)$$
I'm confused over the derivative operator's precedence. My question is whether it is legal math if I immediately plug $x=0$ as the sine's argument and then derive, thus only getting the following:
$$y^{(n)}(0)\stackrel{?}{=}-9(n-2)(n-3)(n-4)(n-5)y^{(n-6)}(0)$$
If the above is true, then the given problem is fairly easy to finish:
$$y^{(25)}(0) = (-9)^4\space\frac{23!}{19\cdot18\cdot13\cdot12\cdot7\cdot6}\space y'(0) = 0$$
However, something tells me that's not the way to go; otherwise all derivatives of anything may as well be 0, as they would be treated as constants when deriving.
I did try to derive the second derivative a few more times, but the Leibniz formula then becomes a hot mess express. When I apply the formula over $y^{(4)}$, which I found was the first one to have the sine function obscured under a $y$ variable, I get a multiple recursion by the means of $y^{(n)}=\ldots y^{(n-4)}+\ldots y^{(n-6)}+\ldots y^{(n-7)}$
Are there any other means of solving this task, perhaps by not using the Leibniz formula at all?
Best Answer
By Taylor's theorem$^1$ a function $f(x)$ infinitely differentiable at $x=0$ can be written as
$$ f(x) = \sum_{k= 0}^{\infty} \frac{f^k(0)x^k}{k!}$$
We know that $$ \cos(x) = \sum_{k = 0 } \frac{(-1)^k x^{2k}}{(2k)!}$$ so it can be shown that $$ f(x) = \cos(x^3) = \sum_{k = 0 } \frac{(-1)^k x^{6k}}{(2k)!}$$.
Since the coefficient of $x^{25}$ is $0$ in the above series, it follows that $f^{25}(0) = 0$
Footnotes: