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:


$$\Rightarrow y^{(n)}(0)=-6(n-2)\left(\sin(x^3)\right)^{(n-3)} -9(n-2)(n-3)(n-4)(n-5)y^{(n-6)}(0)$$

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:


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?

  • 1
    $\begingroup$ Why derive a differential equation? why not simply perform the derivative? $$3^n \cos \left(\frac{\pi n}{2}+3 x\right)$$ $\endgroup$ Aug 16, 2021 at 16:16
  • 1
    $\begingroup$ 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 $(\cos x)^{(n)}=\cos(x+\frac{n\pi}{2})$. $\endgroup$
    – kyrillox
    Aug 16, 2021 at 16:24
  • $\begingroup$ This question is similar to yours. $\endgroup$
    – Axion004
    Aug 16, 2021 at 16:27

1 Answer 1


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 $f^{25}(0) = 0$


  1. For precise statement look here
  • 1
    $\begingroup$ Sure, or in this special case, we can see it is 0 simply due to evenness of the function $\endgroup$
    – peek-a-boo
    Aug 16, 2021 at 16:36
  • 1
    $\begingroup$ @peek-a-boo is correct, of course, but this method can also be used to find $f^{(24)}(0)$, which is not zero. $\endgroup$ Aug 16, 2021 at 17:54
  • $\begingroup$ It is because the odd order derivative of an even function vanishes at zero. $\endgroup$
    – Axion004
    Aug 16, 2021 at 18:00

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