TITLE: Fractional Order Derivative
QUESTION [7 upvotes]: A friend and I were talking about derivatives, and he asked an interesting question. Since both of us have not taken our calculus courses yet, neither of us were sure of the answer. His question has two parts: 1. can you have a fractional order derivative i.e. could you have $\frac{d^ny(x)}{dx^n}$ where, for example, $n=\frac{1}{2}$? And, 2. can you have a variable as the order of a derivative i.e. $\frac{d^ny(x)}{dx^n}$ where n=some variable? I have doubts as to whether either are even possible, but I decided to post the qestion to find out for sure.

REPLY [4 votes]: Let me add this "destructive" answer as an addition. We take a conceptual approach to the "half" derivative $Hf$ of $f$.
What properties should a "half derivative" $Hf$ of $f$ have?

(1) Clearly, we want $HHf = f'$ for all twice half-differentiable functions $f$, so applying the half derivative twice should give the ordinary derivative.
However, there are many ways of defining a suitable half derivative with property (1), so this doesn't give a unique definition of $H$. We need more properties.
The next thing which comes to my mind is
(2) For the ordinary $n$-th derivative, very basic differentiation rules state that
$$(f+g)^{(n)} = f^{(n)} + g^{(n)}\quad\text{and}\quad(\lambda f)^{(n)} = \lambda f^{(n)}.$$
It's only natural to ask for the same for the half derivative. So $H$ should be linear, that is
$$H(f+g) = Hf + Hg\quad\text{and}\quad H(\lambda f) = \lambda Hf$$
for all half-differentiable functions $f,g$ and all $\lambda\in\mathbb{R}$.
(In other words, the set of all half-differentiable functions forms a vector space and $H$ is a linear operator on it.)
So do (1) and (2) give us a unique definition of a half derivative?
Unfortunately, we will see that the answer is "no", as these properties turn out to be contradictory if we additionally demand only the following tiny bit
(3) All constant functions as well as the identity function $\operatorname{id} : x\mapsto x$ are three times half-differentiable.

Let us derive the contradiction from (1), (2) and (3).
For the zero function $\mathbf{0} : x \mapsto 0$ we have by additivity (part of the linearity property)
$$
H\mathbf{0} = H(\mathbf{0} + \mathbf{0}) = H\mathbf{0} + H\mathbf{0},
$$
so subtracting $H\mathbf{0}$ on both sides gives
$$
H\mathbf{0} = \mathbf{0}.
$$
Now for the constant function $\mathbf{1} : x\mapsto 1$ we have
$$
(H\mathbf{1})' = HH(H\mathbf{1}) = H(HH\mathbf{1}) = H(\mathbf{1}') = H\mathbf{0} = \mathbf{0},$$
so $H\mathbf{1}$ is a constant function, say $H\mathbf{1} = (x \mapsto a)$ with $a\in\mathbb{R}$.
Now by linearity
$$ \mathbf{0} = \mathbf{1}' = HH\mathbf{1} = H(x\mapsto a) = H(a\cdot\mathbf{1}) = a\cdot H\mathbf{1} = a\cdot (x\mapsto a) = (x\mapsto a^2),$$
showing that $a = 0$ and therefore
$$H\mathbf{1} = \mathbf{0}.$$
Therefore, by linearity any constant function $x\mapsto b$ with $b\in\mathbb{R}$ has the half derivative
$$
H(x\mapsto b) = H(b\cdot \mathbf{1}) = b\cdot H\mathbf{1} = b\mathbf{0} = \mathbf{0}
$$
Now
$$(H\operatorname{id})' = HH(H\operatorname{id}) = H(HH\operatorname{id}) = H(\operatorname{id}') = H\mathbf{1} = \mathbf{0},$$
so $H\operatorname{id} = (x \mapsto c)$ is constant and therefore its half derivative is $\mathbf{0}$.
This gives the final contradiction
$$
\mathbf{0} = H(H\operatorname{id}) = HH\operatorname{id} = \operatorname{id}' = \mathbf{1}.
$$

Conclusion. For me, there is no convincing concept of a half derivative. Of course, people can define fractional derivatives in this way or another and derive conclusions, but for it being really a "fractional derivative", something is missing (as (1), (2) and (3) cannot all be true), and there is no way out by the above contradiction.