Suppose that we define a "derivative" in the following way:

$$\mathcal{D^{*^\alpha}}=\lim_{x\to x_0}\frac{f(x)^\alpha-f(x_0)^\alpha}{x-x_0}, $$ where $\alpha$ is a real number.

What would be the rules of derivation of a function (product, sum, composition,...)? What could we say about a "Taylor Polynomial" using this kind of derivative? Is there any advantage in using this object?

  • 1
    $\begingroup$ This is not what you have descriped here, but it might be of interest. en.wikipedia.org/wiki/Fractional_calculus $\endgroup$ – Alice Ryhl Dec 31 '14 at 9:31
  • $\begingroup$ @KristofferRyhl Very interesting. What book would you recommend on this topic? $\endgroup$ – user62029 Dec 31 '14 at 9:54
  • $\begingroup$ @user62029 I have a copy of this one. Take a look at it. $\endgroup$ – Billy Rubina Dec 31 '14 at 10:52
  • $\begingroup$ @Vÿska What background is required to work through that textbook? $\endgroup$ – Dal Dec 31 '14 at 12:46
  • $\begingroup$ @Dal I guess a course in calculus and linear algebra suffices for it. $\endgroup$ – Billy Rubina Jan 1 '15 at 5:32


This just the composition of $f$ with $g$ where $g\colon x\mapsto x^\alpha$. So just apply the chain rule along with the power rule :

$$ \begin{align} \mathcal D^{*^\alpha}&=\frac{\rm d}{{\rm d}x}g(f(x))\left|\right._{x=x_0}\\ &=g'(f(x))\cdot f'(x)\left|\right._{x=x_0}\\ &=\alpha f(x)^{\alpha-1}\cdot f'(x)\left|\right._{x=x_0}\\ \end{align} $$

Use this result and see where it leads.

Remark: $\mathcal D^{*^\alpha}$ can be understood to be the set made by raising the derivative of a specific function to all powers, provided that $\alpha$ is considered to be non-constant.

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    $\begingroup$ I think it should be $\alpha f(x)^{\alpha-1}\cdot f'(x)\left|\right._{x=x_0}$ $\endgroup$ – miracle173 Dec 31 '14 at 9:28
  • $\begingroup$ @miracle173 Thanks for spotting the typo. $\endgroup$ – Workaholic Dec 31 '14 at 9:37


Let $x - x_0 = \Delta x$

Then it follows

$$ \lim_{x \rightarrow x_0} \frac{f(x)^a - f(x_0 )^a }{x - x_0} $$


$$ \lim_{\Delta x\rightarrow 0} \frac{f(x)^a - f(x - \Delta x )^a }{\Delta x} $$

Note that

$$ \lim_{\Delta x \rightarrow 0} \frac{f(x) - f(x - \Delta x ) }{\Delta x} = \frac{df}{dx}$$

Thus we conclude

$$ D^{*a} = \frac{d f(x)^a}{dx} = a \frac{df}{dx} f(x)^{a-1}$$

We now establish the chain rule for $D^{*a}$ (I will use the notation $\frac{D^{*a}}{Dx}$) to denote the argument variable.

$$ \frac{D^{*a}f(g(x))}{Dx} = a \frac{df(g(x))}{dx}f(g(x))^{a-1} $$


$$a \frac{df(g(x))}{dx}f(g(x))^{a-1} = a \frac{dg}{dx} \frac{df(g(x))}{dg(x)} f(g(x))^{a-1} $$

Now using this chain rule its very straightforward to derive the product rules, inverse rules ,etc...

Fun Fact! We can construct some fancy Taylor Series. Suppose we fix a value a.


$$D^{*a}(1) = 0$$

$$f(x) | D^{*a}f = 1 \rightarrow a \frac{df}{dx} f(x)^{a-1} = 1$$

If $f(x) = Cx^b$ then

$$ a C bx^{b-1} C x^{ab-b} = 1 \rightarrow abC^2 = 1, ab-1 = 0 $$

Let $C = 1$, $b = \frac{1}{a}$

Now we find the next f(x) whose $D^a$ is $x^{\frac{1}{a}}$ $$f(x) | D^{*a}f = 1 \rightarrow a \frac{df}{dx} f(x)^{a-1} = x^{\frac{1}{a}}$$ $$ aCbx^{b-1}Ca^{ab-b} = x^{\frac{1}{a}} \rightarrow ab - b = \frac{1}{a} \rightarrow b = \frac{1}{a(a-1)}, abC^2 = 1 \rightarrow C = \sqrt{a-1}$$

Thus we have taylor polynomial-like terms

$$ 1, x^{\frac{1}{a}}, \sqrt{a-1} x^{\frac{1}{a(a-1)}} ...$$

But unless $a = 1$ you can't just add them together, (the formula doesn't distribute over addition) i'm working on how to determine the operator over which this expression distributes.

That is I'm looking for a function

$$G_a(z_1(x), z_2(x))$$ such that

$$D^{*a}G_a(z_1(x),z_2(x)) = G_a(D^{*a}(z_1(x)), D^{*a}(z_2(x)))$$

One of the key derivations here is that

$$ D^{*a} f = f \rightarrow f = \sqrt[a]{\left(1 - \frac{1}{a}\right)x + C}$$

And if we find the the general $G_a$ then we will have an identity of the form

$$ \sqrt[a]{\left(1 - \frac{1}{a}\right)x + C} = G_a(1, G_a(x^{\frac{1}{a}}), G_a(\sqrt{a-1}x^{\frac{1}{a(a-1)}}, G_a(... )))) $$

Which has base case:

$$ e^x = 1 + x + \frac{1}{2}x^2 ... $$

For $a = 1$


First of all, one has to check when forward and backward derivatives are equal.

Let $ \mathcal{D^{*\alpha -}} f (x)= \lim\limits_{\epsilon \rightarrow +0} \frac{f(x)^a - f(x - \epsilon )^a }{\epsilon} $ and

$ \mathcal{D^{*\alpha +}} f (x)= \lim\limits_{\epsilon \rightarrow +0} \frac{f(x +\epsilon)^a - f(x )^a }{\epsilon} $. Then by applying l'Hospital's rule we get

$ \mathcal{D^{*\alpha +}} f (x)= \alpha\, \lim\limits_{\epsilon \rightarrow +0} { f \left( x+\epsilon\right) }^{\alpha-1}\, f^{\prime} \left( x+\epsilon\right) $ and

$ \mathcal{D^{*\alpha -}} f (x) = \alpha\, \lim\limits_{\epsilon \rightarrow +0} { f \left( x-\epsilon\right) }^{\alpha-1}\, f^{\prime} \left( x-\epsilon\right) $.

Therefore, only if $ f^{\prime} \in C^1$ in a closed interval $[x- \epsilon, x+\epsilon]$ one would have $ \mathcal{D^{*\alpha +}} f (x)= \mathcal{D^{*\alpha -}} f (x) = \alpha\, { f \left( x \right) }^{\alpha-1}\, f^{\prime} \left( x \right) $.

Composition rule (under the same hypothesis):

$ \mathcal{D^{*\alpha}} f (g) (x)= \alpha\, { f \left( g \right) }^{\alpha-1}\, {{d}\over{d\,g}}f\left(g\right)\ g^{\prime} \left(x \right) $.

Sum rule:

$ \mathcal{D^{*\alpha +}} (f (x) +g(x) ) = \alpha\, { \left( f \left( x \right) + g \left( x \right) \right) }^{\alpha-1}\, \left( f^{\prime} \left( x \right) + g^{\prime} \left( x \right) \right) $

Product rule:

$\mathcal{D^{*\alpha +}} (f (x) \ g(x) ) = \alpha\, { f \left( x\right) }^{\alpha}\,{ g \left( x\right) }^{\alpha-1}\, g^\prime \left( x\right) +\alpha\,{\mathrm{f}\left( x\right) }^{\alpha-1}\,{ g \left( x\right) }^{\alpha}\, f^\prime \left( x\right) $.

Application to Taylor series:

Since we suppose $ f(x+ \epsilon)=f(x)+ f^\prime(x) \epsilon + R_1(x,\epsilon) $ it follows that $ f(x+ \epsilon)=f(x)+ \dfrac{\mathcal{D^{*\alpha }} f(x) } {\alpha \; f \left( x \right) ^{\alpha-1}} \epsilon + R_1(x,\epsilon) $.


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