Is there a difference in domain between the two ways to notate the reciprocal function?

Question

Whilst exploring basic derivation, I noticed something peculiar.

We know that the derivative of any constant is 0. However, does this derivative exist for the whole domain?

Consider the function $f(x)=x^2$. We know $f'(x)=2x$, and $f''(x)=0$.

If you write the functions literally, it would come out to this:

$f(x)=x^2$

$f'(x)=2*x^1=2x$

$f''(x)=1*2x^0=2$

$f'''(x)=0*2x^{-1}=0$

The issue sprouts from this last derivative; since there is an $x^{-1}$, then wouldn't the derivative not exist if $x=0$? If it does exist, this brings up the question if this is mathematical syntax that allows it to exist. Another example is the following two functions, which are equivalent mathematically, but not by function.

$f(x)=\sqrt{\frac{x^3}{x}}$ Domain: ($-\infty, 0) U (0, \infty$)

$g(x)=\frac{\sqrt{x^3}}{\sqrt{x}}$ Domain: ($0,\infty$)

These two functions are taught to be mathematically equivalent, but do not yield the same domain. This is why the question of mathematical syntax has come up. Do these two functions yield the same domain?

$f(x)=1/x$

$g(x)=x^{-1}$

Answer 1

Note that $\frac{d}{dx} x^a = ax^{a-1}$ is not something that is true by fiat or by definition, but the conclusion of a theorem that has some additional assumptions. In fact there is an entire family of theorems, with different assumptions here, including

• If $a$ is a positive integer, then for all real $x$ we have $\frac{d}{dx} x^a = ax^{a-1}$.
• If $a$ is an integer, then for all nonzero real $x$ we have $\frac{d}{dx} x^a = ax^{a-1}$.
• If $x$ is a positive real, then for all real $a$ we have $\frac{d}{dx}x^a = ax^{a-1}$.

and more. But there's no theorem of this form that promises anything in the case where $x$ and $a$ are both zero. (Such a theorem cannot possibly hold, because in that case $ax^{a-1}$ involves a division by zero).

Crucially, this lack of applicable theorems doesn't mean that $\frac{d}{dx} x^0$ fails to exist at $x=0$ -- simply that we can't expect to use the $ax^{a-1}$ formula to find what it is.

So for your $f$, the theorems imply that $f'''(x)=0x^{-1}$ whenever $x$ is nonzero, which is completely true. They don't say anything either way about $f'''(0)$, but that is not the fault of $f'''$, just of trying to use the theorems in a way they don't actually allow.

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For the question at the end: There are various similar but distinct definitions of what the notation $a^b$ means, which assign meaning to different combinations of $a$ and $b$. If we take the practical meaning to be the union of all these definitions (which is happily possible because they all happen to agree whenever more than one of them gives a meaning to some $a^b$), then the outcome is that $x^{-1}$ ends up with exactly the same meaning as $1/x$ -- down to the fact that neither $1/0$ nor $0^{-1}$ is defined.