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I know that the first derivative of sine of $x$ is cosine of $x$, but I'm really facing a problem trying to understand the other derivatives of the sine function.

The following $2$ statements are both true: $$\frac{d}{dx}\sin(x) = \cos(x)$$ $$\frac{d}{dx}\cos(x) = -\sin(x)$$

And then, according to the constant multiple rule, we can see that the following is also true:

$$\frac{d}{dx}-\sin(x) = -1 \cdot \frac{d}{dx}\sin(x) = -1 \cdot \cos(x) = -\cos(x)$$

Hence:

$$\frac{d}{dx}-\cos(x) = -1\cdot\frac{d}{dx}\cos(x) = -1\cdot(-\sin(x)) = \sin(x) \ \dots$$

Can we keep repeating this process to infinity? And what does it really mean that the fourth derivative of $\sin(x)$ is $\sin(x)$? I've seen a visualisation that explains why the derivative of $\sin(x)$ is $\cos(x)$, so I know why the first derivative of $\sin(x)$ is $\cos(x)$, at least intuitively, but I can't really get the idea of the rest of the derivatives of $\sin(x)$, or why they are repeating, and can we really keep repeating this with no problems?

ACB
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Aueriga
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    Maybe this is helpful: https://math.stackexchange.com/questions/392/intuitive-understanding-of-the-derivatives-of-sin-x-and-cos-x?rq=1 – A. J. Pan-Collantes Mar 04 '23 at 11:30
  • This is like asking "what happens if I keep doubling $2$?" The pedantic answer is "you keep getting $4$", but you meant that we'd double $4$ to get $8$, double $8$ etc. In this case, $\frac{d}{dx}\sin x$ is analogous to $2\times2$. – J.G. Mar 04 '23 at 12:21
  • @J.G. are you sure that comparing differentiation to multiplication is a good idea? I don't think they are similar processes, to differentiate a certain function is to measure certain behaviors in the function, differentiating it again is the measurement of certain behaviors in the measurement of the first function, to keep multiplying a number by a number is just doing the same process no matter what happened in the last time you did this process, or how many times you've done it :D – Aueriga Mar 04 '23 at 15:52
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    @Ryu My point is you made a question about $D^n\sin x$ (where $D:=d/dx$) sound like a question about repeatedly computing $D\sin x$, just as someone might make a question about $2^n2$ sound like a question about repeatedly computing $2\times2$. (True, $D^nf$ "breaks" for large $n$, if $f$ isn't infinitely differentiable, in a way $2^n2$ doesn't.) – J.G. Mar 04 '23 at 16:12
  • @J.G. I understand your point but that's not what I was asking for in my question, I was asking about the meaning behind this process, and I know it repeats when n = 4 but I wanted to know the essence of that rule, not to compare it with something else – Aueriga Mar 04 '23 at 18:28

2 Answers2

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if your question is that whether $\sin(x)$ is differentiable infinite number of times, then the answer is yes! On the other hand, $y = \sin(x)$ is one of the solutions of the differential equation $ \frac{d^2(y)}{dx^2}+y=0 $ which is basically saying that it is a function whose second derivative is the negative of itself.

Similarly, $y = \cos(x)$ and any linear combination of those two functions behave in the same manner.

Can you find any other solution to this differential equation? The solutions will all have the same repeating behaviour and hence must always be differentiable functions. Don't you think..

Solutions of different differential equations highlight all kinds of patterns that some functions inherit.

Note that $\sin(x)$ is also a solution to the differential equation that goes like $ \frac{d^4(y)}{dx^4}-y=0 $ which is saying that its fourth derivative is itself. We say things like, since $(D^2+1)$ is a multiple of $(D^4-1) $ where $D=\frac{d}{dx}$ means that every solution of the first differential equation ($(D^2+1)y=0$) will also be a solution of the second ($(D^4-1)y=0$).

Giorgos Giapitzakis
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Benjamin Kurian
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Yes, you get the derivative as $$\dfrac{d^n \sin(x)}{dx^n} = \sin \left(\frac{\pi n}{2}+x\right)$$ for $n \ge 0$ and $n$ integer.

This just produces a repeating pattern of $$\sin (x),\cos (x),-\sin (x),-\cos (x) \ldots$$

Moo
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    Use induction, please see https://math.stackexchange.com/questions/3103202/what-is-the-nth-derivative-of-sinx – Moo Mar 04 '23 at 11:42