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Given a differentiable real function $f$, the derivative $f'(x)$ is the slope of the tangent to the graph of $f$ at $(x,f(x))$.

Suppose that, instead of the tangent, we look at the secant to the graph of $f$, between the point $(x,f(x))$ and the point $(x^+,f(x^+))$, where $x^+$ is chosen such that $x^+>x$ and the distance between $(x,f(x))$ and $(x^+,f(x^+))$, along the curve of $f$, is some fixed constant $d$:

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We define the "$d$-secant-derivative" of $f$ at $x$ as the slope of that secant. Note that, as $d\to 0$, the $d$-secant-derivative at $x$ approaches $f'(x)$ (if $f$ is differentiable), but here $d$ is constant.

  • Is anything known about this "secant-derivative" operator?
  • Is there a simple way to compute or approximate it, like there are simple rules for computing derivatives?

NOTE: As I am interested in approximations, it does not matter very much what distance measure is used: it can also be Euclidean distance, $\ell_1$ distance or $\ell_\infty$ distance; whatever makes the problem solvable. The reason I chose the distance along the curve is that (I think) it defines $x^+$ uniquely. But, I will also be happy for a solution using e.g. the $\ell_\infty$ metric, where $x+ = \inf\{y\geq x | \ell_\infty(y,x) = d\}$.

Erel Segal-Halevi
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    The point $x^+$ need not be unique. – Jochen Dec 31 '23 at 08:16
  • @Jochen good point. What if we define $x^+$ based on distance along the curve of $f$ (not Euclidean distance)? – Erel Segal-Halevi Dec 31 '23 at 08:22
  • If you define it to the distance along the curve then it would be unique. To compute the secant derivative you need to find the point which you can do by the formula for arc length. – Vivaan Daga Dec 31 '23 at 08:33
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    Once you start on this path, it's tempting to define the right-d-secant-derivative (which is the one you defined), the left-d-secant-derivative (using x- instead of x+), and the symmetric-d-secant-derivative (using both x- and x+, perhaps with d/2). If f is differentiable at a point, then all three should exist (for small d) and have the same limit as d goes to 0. – Stef Dec 31 '23 at 08:40
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    Also note that your choice to use arclength between (x, f(x)) and (x+, f(x+)) instead of horizontal length between x and x+ makes things much more complicated. Instead of having a very simple x+ = x + d, you have made x+ very hard to find. In fact, a function f could be differentiable at x but have infinite arc length in the neighbourhood of x. For instance, take function $f : [0, +\infty) \to \mathbb R$ defined by $f(0) = 0$ and $f(x) = x^2 \sin(1/x^3)$ for $x > 0$. This function is differentiable everywhere, with $f'(0) = 0$, but it has infinite arc length in the neighbourhood of 0. – Stef Dec 31 '23 at 08:52
  • @Stef the problem with $x^+=x+d$ is that, for some functions, the vertical distance may be much larger than the horizontal distance. Using the $\ell_1$ or the $\ell_\infty$ metric would solve this problem, and it is even better for my use-case. But then the point $x^+$ would not be unique. – Erel Segal-Halevi Dec 31 '23 at 08:54
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    @ErelSegal-Halevi Uniqueness is easy to deal with; just take the minimum x+. For instance, $x^+ = \inf { y \geq x ,\mid, (x-y)^2 + (f(x)-f(y))^2 \geq d^2 }$. – Stef Dec 31 '23 at 09:11
  • Have you calculated any examples? – badjohn Dec 31 '23 at 09:25
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    I've deleted my answer (began with: "This is variously called the strong derivative, the strict derivative, the unstraddled derivative, the sharp derivative.") because -- as pointed out by the OP and by @Stef -- the OP's notion is not a limit notion, and also the endpoints of the secant are required to be on opposite sides of $x$. For what it's worth, the OP's notion appears to be a kind of discrete version of the derivative notion mentioned in my comment to Alternate limit characterisations of the derivative. – Dave L. Renfro Dec 31 '23 at 10:10
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    Since $x^+$ need not be unique, I noticed also that the "secant-derivative function" need not be continuous. But I wonder if, for every $x,$ we can choose a corresponding $x^+$ so that the secant-derivative function is continuous. – Adam Rubinson Dec 31 '23 at 12:47

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