Can you get an infinite number of derivatives from $\frac{d}{dx}\sin(x)$? 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:
$\frac{d}{dx}-\sin(x) = -1 * \frac{d}{dx}\sin(x) = -1 * \cos(x) = -\cos(x)$
Hence:
$\frac{d}{dx}-\cos(x) = -1*\frac{d}{dx}\cos(x) = -1*-\sin(x) = \sin(x) ....$
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?
 A: 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 ($(D^4-1)y=0$).
A: 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 $$\sin (x),\cos (x),-\sin (x),-\cos (x) \ldots$$
