Derivative

The derivative $f'(x)$, of a differentiable function $f(x)$, is the slope of the tangent line of $f$ at $x$. [TODO]

Note that this definition does not talk about taking a limit as epsilon goes to 0.

For some functions (indeed, many functions), there is a way to write $f'(x)$ in terms of other known functions. That is, there is a relationship between the concept $f'(x)$, and other concepts. This need not be the case in general.

Old stuff [TODO delete]

Let $f:\mathbb {Q} \rightarrow \mathbb {Q}$ be a function, and let $\epsilon :\mathbb {Q} _{>0}$. I define $\Delta _{\epsilon }f:\mathbb {Q} \rightarrow \mathbb {Q}$ by

$$(\Delta _{\epsilon }f)(x):={\frac {f(x+\epsilon )-f(x)}{\epsilon }},\quad x:\mathbb {Q} .$$

I say that $f:\mathbb {Q} \rightarrow \mathbb {Q}$ is differentiable at $x:\mathbb {Q}$, if $(\Delta _{\epsilon }f)(x)-(\Delta _{\epsilon '}f)(x)$ is nill whenever $\epsilon ,\epsilon ':\mathbb {Q} _{>0}$ are both nill.

If $f:\mathbb {Q} \rightarrow \mathbb {Q}$ is differentiable at $x:\mathbb {Q}$, then I define the derivative of $f$ at $x$ as $f'(x):=(\Delta _{\epsilon }f)(x)$, for some nill $\epsilon :\mathbb {Q} _{>0}$.

Criticism

This definition makes things way more complicated. I will demonstrate this with the following example.

Let's suppose that $f(x)=x^{3}$ so $(\Delta _{\epsilon }f)(x)=3x^{2}+3\epsilon x+\epsilon ^{2}$ and $(\Delta _{\epsilon }f-\Delta _{\epsilon '}f)(x)=3(\epsilon -\epsilon ')x+(\epsilon ^{2}-\epsilon '^{2})$. And let's say that we are in a context where anything with absolute value below 0.1 is nill. Let $\epsilon =0.091$ and $\epsilon '=0.001$. Then $(\Delta _{\epsilon }f-\Delta _{\epsilon '}f)(x)=0.27x+0.00828$. This quantity is greater than 0.1, and thus non-nill, when $x>0.339{\overline {703}}$, so we reach the seemingly absurd conclusion that $f(x)=x^{3}$ is not differentiable where $x>0.339{\overline {703}}$.

More generally, if $\nu :\mathbb {Q} _{>0}$ is the nill cutoff (where we also assume $\nu \leq 1$), then $f(x)=x^{3}$ is only "differentiable" in the region

$$-{\frac {1}{3}}(1+\nu )<x<{\frac {1}{3}}(1-\nu )$$

This seems like an absurd conclusion.

Polynomial derivatives

Somewhat interesting, but also somewhat obvious vid https://www.youtube.com/watch?v=oW4jM0smS_E

You can get the tangent line to a polynomial without actually invoking limits. My guess is that you can also get the tangent line to a rational function somehow.