Derivative of logarithm and exponential

[math]\displaystyle{ f(x) = \ln(x) \iff f'(x) = \frac{1}{x} }[/math]

I'm going to explain some property that is pretty simple and yet often overlooked. Let's write down a short sequence of logarithms in base 2:

[math]\displaystyle{ \log_2{1} = 0 }[/math]
[math]\displaystyle{ \log_2{2} = 1 }[/math]
[math]\displaystyle{ \log_2{4} = 2 }[/math]
[math]\displaystyle{ \log_2{8} = 3 }[/math]
[math]\displaystyle{ \log_2{16} = 4 }[/math]

Now the derivative can be defined in terms of a tangent, a ratio rise / run. We define rise as [math]\displaystyle{ \log(x_2) - \log(x_1) }[/math]. While run is [math]\displaystyle{ x_2 - x_1 }[/math]. Notice that we are increasing rise in steps of one unit. While the input is increasing following a powers of 2 rule. If we write the sequence following the formula [math]\displaystyle{ \frac{\log(x_2) - \log(x_1)}{x_2 - x_1} }[/math] we get:

[math]\displaystyle{ \frac{1}{2^0} }[/math], [math]\displaystyle{ \frac{1}{2^1} }[/math], [math]\displaystyle{ \frac{1}{2^2} }[/math], [math]\displaystyle{ \frac{1}{2^3} }[/math], [math]\displaystyle{ \frac{1}{2^4} }[/math]

Notice that the sequence is decrescent, which means that the derivative of a logarithmic function is a decrescent function. Each term of the sequence is the inverse of the corresponding power of 2. The base doesn't matter, all bases should display the same behaviour. I didn't do calculations with decimal numbers but if we consider mean values in between each step the same behaviour should be expected.

This intuitive reasoning should explain why deriving the log yields a function that takes the inverse of a number at each point.