Adding changes or derivatives between two points I've playing with the limit definition of derivative and I've to somewhat confusing conclusions. 
To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?! 
So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.
So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$
$\lim _{h\to \:0}\left(\frac{\left(x+h\right)^2-x^2}{x}\right) = 2x$
This is the expected answer
However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer
$\lim _{h\to \:0}\left(\frac{\left(x+6h\right)^2-\left(x+5h\right)^2}{x}\right) = 2x$
It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!
$\lim _{h\to \:0}\left(\frac{\left(x\right)^2-\left(x-h\right)^2}{x}\right) = 2x$
I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments? 
I hope I made myself clear!
Thanks in advance!
 A: Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$ 
We can draw a straight line between these two points, and it's gradient is given by:
$$m=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$
for any $x_1, x_2 \in \mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)
We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)
Then we have:
$$m=\frac{f(x_1+h)-f(x_1)}{x_1+h-x_1}=\frac{f(x_1+h)-f(x_1)}{h}$$
If $f(x)=x^2$, we have $m=\frac{(x_1+h)^2-x_1^2}{h}=\frac{h^2+2hx_1}{h}=2x_1+h$
What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.
We can say that $$\lim_{h\to 0}{(2x_1+h)}=2x_1$$
In fact:
$$\lim_{h\to0} (kh)=0$$
is true for any constant $k$
and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.
A: This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $h\rightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.
