The Continuous Differential Operator I was playing around a bit this morning until I realized that one could write the arbitrary derivative of $\cos x$ as
$$(\cos x)^{(n)} = \cos\big(\frac{\pi}{2}(n-1) \big)\sin x + \cos\big(\frac{\pi}{2}n\big)\cos x$$ 
Similarly, writing $$\sin x=\cos \left(x-\frac{\pi}{2}\right)$$
we see that $$\frac{d^n}{dx^n}\sin x = \frac{d^n}{dx^n} \cos \left(x-\frac{\pi}{2}\right)$$
which evaluates to (I think I'm using the chain rule properly in this case):
$$\cos\big(\frac{\pi}{2}(n-1) \big)\sin \left(x-\frac{\pi}{2}\right) + \cos\big(\frac{\pi}{2}n\big)\cos\left(x-\frac{\pi}{2}\right)$$
simplifying
$$(\sin x)^{(n)} = \cos\big(\frac{\pi}{2}(n-1) \big)\cos x + \cos\big(\frac{\pi}{2}n\big)\sin x$$
where we have that $n \in \mathbb{R}$
These two results can be expressed as a complex exponential too. 
Because these functions happens to be continuous, in some ways the differential operator is now continuous for cosine and sine. 
I'm wondering if anyone can interpret this for me a I haven't studying real analysis and only have a nonrigorous calculus background. 
Question
After doing some reading, I realized that because every function can be written as a Fourier transform, one ought to be able to simply use the above definitions to take the nth derivative of the fourier transform of an arbitrary function to an arbitrary degree of precision (by choosing how many terms to include).  
Is the hypothesis true? If so, can you summarize it in a way I would be able to understand and use. Assume I know the definition of the fourier transform. All I am looking for now is a general equation for the continuous differential operator of a function. 
 A: You have a set of functions $\{ \sin x, \cos x \}$ that is closed under differentiation because $\sin ' = \cos$ and $\cos' = -\sin$. So you are restricting differentiation to the two-dimensional subspace $M$ generated by these functions, which consists of all $f=A\cos x + B \sin x$ with constants $A$ and $B$. In this context $\frac{d}{dx}$ is a linear operator on a two-dimensional vector space. No matter what norm you put on this space, the topology will be the same because of it being finite dimensional. So $\frac{d}{dx}$ will be continuous no matter what norm you choose for this subspace of functions. You can do the same with $\{ \cos(nx),\sin(nx) \}$ for $n=1$ or $n=2$, or $n=3$, and you can combine any number of these as well. You'll end up with a space of functions consisting of linear combinations of such functions, and differentiation will map that space back to itself.
The matrix representation of $\frac{d}{dx}$ acting on a basis $\{\cos(nx),\sin(nx)\}$ has the form
$$
          \left[\begin{array}{cc}0 & -n \\ n & 0\end{array}\right]=n\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]
$$
which is $n$ times a counter-clockwise rotation by $90^o$. The $k$th power of a such a matrix can be written as $n^k$ times a rotation by $k \times 90^o$, which leads to a natural extension to a real power $r$ which is $n^r$ times a rotation about $r \times 90^o$:
$$
    n^{r}R(r) = n^{r}\left[\begin{array}{cc}
                   \sin(\frac{\pi}{2}r) & -\cos(\frac{\pi}{2}r) \\
                   \cos(\frac{\pi}{2}r) & \sin(\frac{\pi}{2}r)
                 \end{array}\right]
$$
Multiplication by $i$ also serves to rotate counterclockwise in the complex plane by $90^o$, which has a natural extension to multiplication by $e^{ir\theta}$, which is why the exponential power of the trigonometric functions is a bit easier to deal with.
In terms of the exponential Fourier series
$$
      \frac{d^r}{d\theta^r}e^{in\theta}=(in)^re^{in\theta} = (ne^{i\pi/2})^re^{in\theta} = n^r e^{ir\pi/2}e^{in\theta}
$$
So,
$$
     \frac{d^r}{d\theta^r}\sum_{n=-\infty}^{\infty}c_n e^{in\theta} = \sum_{n=-\infty}^{\infty}n^r e^{i\pi r/2} e^{in\theta}.
$$
This is a reasonable definition of fractional derivatives for $r > 0$ that agrees with the usual definition for $r=1,2,3,\cdots$.
The same sort of thing works for the Fourier transform, too:
\begin{align}
       \frac{d^r}{dx^r}f & = \frac{d^r}{dx^r} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixs}\hat{f}(s)ds \\
     & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(is)^r e^{isx}\hat{f}(s)ds.
\end{align}
These ideas fall under the general heading of Functional Calculus for operators.
