I didn't really know what to call this question since I didn't know what this is called nor what to look up.
What I'm talking about is this ratio: $$f^{\Delta}(x)=\lim \limits_{h \to 0}{\frac{\sqrt{(f(x+h)-f(x))^2+h^2}}{h}}$$
Also, sorry about the $f^{\Delta}(x)$ notation but that's all I could think of to notate this.
I'm not so sure if this idea already exists. It's looking at taking a small step in $x$ for some $f(x)$ and seeing the distance between $(x, f(x))$ and $(x+\delta, f(x+\delta))$ and then dividing it all by $\delta$. In the above definition I decided to use $h$ above because that's what the limit definition of derivative uses.
This formula is key for my question.
When I worked out this limit for some powers of $x$, let's say $x^{n}$, I noticed a pattern.
$$f(x) = x^n\quad \iff \quad f^{\Delta}(x) = \sqrt{n^2x^{2(n-1)}+1} \quad where \quad n\space\epsilon\space\mathbb Z$$
So here's the question itself: Is there a proof for this? Does this pattern hold?
I also think it works for all $n$ where $n$ is a real number, but I'm not sure since I haven't worked some examples of that out, I only graphed them.
Please tell me if my notation is off somewhere, I'm new to mathematical logic.
Trivial sidenote:
I used the distance formula to derive this, it's supposed to formulate the idea of how long or effecient a path is. Say, if I had point A and point B, if I were to walk in a sine curve along that path instead of a straight line, at any given point, how much time / energy / some metric, am I wasting?
The ratio you are considering equals the secant of the angle $\angle BAC$ in the triangle formed with points $A=(x,f(x))$, $B=(x+h,f(x+h))$ and $C=(x+h,f(x))$. For small $h$ we can linearly approximate $f(x+h)$ by using $$f(x+h)\approx f(x)+hf'(x)$$ So our triangle can be approximated by another with the point $B=(x+h,f(x)+hf'(x))$ instead. As $h$ tends to zero this approximation becomes equal to the original triangle as any curve is effectively a straight line for small enough changes in $x$. The secant of this new triangle is given by $$\sec{(\angle BAC)}=\frac{\sqrt{(hf'(x))^2+h^2}}{h}=\frac{h\sqrt{(f'(x))^2+1}}{h}=\sqrt{(f'(x))^2+1}$$ and as we take $h\to0$ this secant equals the original secant so $$\lim_{h\to0}\frac{\sqrt{(f(x+h)-f(x))^2+h^2}}{h}=\sqrt{(f'(x))^2+1}$$