# Find the nth derivative of $\cos(x^3)$

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?

• Why derive a differential equation? why not simply perform the derivative? $$3^n \cos \left(\frac{\pi n}{2}+3 x\right)$$ Aug 16, 2021 at 16:16
• I believe the derivative of $\cos(x^3)$ is $-3x^2\sin(x^3)$. The power of 3 prevents me from using the standard formula $(\cos x)^{(n)}=\cos(x+\frac{n\pi}{2})$. Aug 16, 2021 at 16:24
• This question is similar to yours. Aug 16, 2021 at 16:27

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:

1. For precise statement look here
• Sure, or in this special case, we can see it is 0 simply due to evenness of the function Aug 16, 2021 at 16:36
• @peek-a-boo is correct, of course, but this method can also be used to find $f^{(24)}(0)$, which is not zero. Aug 16, 2021 at 17:54
• It is because the odd order derivative of an even function vanishes at zero. Aug 16, 2021 at 18:00