Derivative of logarithm and exponential

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

I'm going to explain a 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. It shouldn't come as a surprise because the rate of change of a logarithmic function does decrease over time, approaching zero at infinity.

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

[math]\displaystyle{ f(x) = e^x \iff f'(x) = e^x }[/math]

Let's see what happens if we apply [math]\displaystyle{ \frac{2^{x_2} - 2^{x_1}}{x_2 - x_1} }[/math] from 0 to 5:

1, 2, 4, 8, 16

It forms a sequence where each term is a power of 2. Unlike the derivative of a log that yields a decrescent function, deriving a power function yields a crescent function which is also a power function. It's natural because the rate of change is crescent and does follow a power of 2 rule in this case.

Now if we apply brute force and calculate the derivatives of [math]\displaystyle{ 2^x }[/math], [math]\displaystyle{ (2.1)^x }[/math], [math]\displaystyle{ (2.2)^x }[/math], ..., [math]\displaystyle{ (2.9)^x }[/math], [math]\displaystyle{ 3^x }[/math] in Wolfram alpha. We are going to notice that the derivatives are of the form [math]\displaystyle{ \log_e{a} \cdot a^x }[/math], with [math]\displaystyle{ \log_e{a} }[/math] being a number close to 1. When does it happen to be exactly 1? When we have [math]\displaystyle{ \log_e{e} = 1 }[/math]. With this reasoning we conclude that there is a certain base for which the derivative of a power function happens to be equal to itself.

If we apply the same brute force reasoning in Wolfram for the derivatives of [math]\displaystyle{ \log_2{x}, \log_{2.1}{x} ..., \log_{2.9}{x}, \log_3{x} }[/math] we see that they are all of the form [math]\displaystyle{ 1/x\log_e{a} }[/math]. With this reasoning we get that when the base is [math]\displaystyle{ e }[/math], the derivative of the natural logarithm is the inverse of [math]\displaystyle{ x }[/math].

Note: the reasoning above to find the derivatives of exponential and logarithm isn't a formal one. I followed a reasoning similar to what we do with the Bisection method to find roots of equations in numerical methods.