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_2(x_2) - \log_2(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_2(x_2) - \log_2(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 crescent exponential function yields a crescent function which is also an exponential function. The opposite, deriving a decrescent exponential yields a decrescent function.

Now if we apply brute force and calculate the derivatives of [math]\displaystyle{ 2^x }[/math], [math]\displaystyle{ (2.1)^x }[/math], ..., [math]\displaystyle{ (2.9)^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 an exponential 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. The formal proof of the derivative of exp requires us to know how to compute a limit of an infinite sum. In the case of the log the derivative requires the application of many algebraic properties and substitutions to reach the result. Using implicit differentiation and the fact that exp is the inverse of ln and vice-versa makes the proof trivial. But it works by assuming that we already know the derivatives, which is a circular logic.

Arbitrary bases for log and exp

When we have [math]\displaystyle{ f(x) = a^x }[/math] with [math]\displaystyle{ a \gt 1 }[/math], differentiating it yields [math]\displaystyle{ f'(x) = a^x \ln(a) }[/math]. All we have to do is to remember that a number can be expressed in the form of a log in any base, in particular the [math]\displaystyle{ e }[/math] base:

[math]\displaystyle{ a = e^{\ln(a)} \iff \log_e(a) = \log_e(a) }[/math]

Multiply both sides by [math]\displaystyle{ 1^x }[/math]:

[math]\displaystyle{ a^x = e^{x\ln(a)} }[/math]

Now the function is [math]\displaystyle{ f(x) = e^{x\ln(a)} }[/math], therefore:

[math]\displaystyle{ f'(x) = e^{x\ln(a)} }[/math] (in case you are confused, [math]\displaystyle{ \ln(a) }[/math] is a constant because the variable is [math]\displaystyle{ x }[/math])

[math]\displaystyle{ f'(x) = \ln(a) e^{x\ln(a)} }[/math] (apply the chain rule to derive [math]\displaystyle{ e^{g(x)} }[/math])

[math]\displaystyle{ f'(x) = \ln(a) \cdot a^x }[/math]

Now for the case of [math]\displaystyle{ f(x) = \log_a(x) }[/math] we rely on the change of base:

[math]\displaystyle{ \log_a(x) = \frac{\ln(x)}{\ln(a)} }[/math]

As seen before [math]\displaystyle{ \ln(a) }[/math] is a constant:

[math]\displaystyle{ \frac{\ln(x)}{\ln(a)} = \ln(x)\frac{1}{\ln(a)} }[/math]

With this the derivative should be immediate:

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

Can a log have functions on both the base and the variable? Yes, it's the case of [math]\displaystyle{ f(x)^{g(x)} }[/math]. To differentiate this we have to rely on the [math]\displaystyle{ \ln(x) }[/math] once again for the same reasons as before:

[math]\displaystyle{ f(x)^{g(x)} = h(x) }[/math]

[math]\displaystyle{ \ln{\left(f(x)^{g(x)}\right)} = \ln{(h(x))} }[/math]

[math]\displaystyle{ g(x) \ln(f(x)) = \ln(h(x)) }[/math]

[math]\displaystyle{ e^{g(x) \ln(f(x))} = h(x) }[/math]

Now to compute [math]\displaystyle{ h'(x) }[/math] we proceed as we'd do with [math]\displaystyle{ e^{a(x)} }[/math] and apply the chain rule:

[math]\displaystyle{ h'(x) = [g(x) \ln(f(x))]' \cdot e^{g(x) \ln(f(x))} }[/math] (notice there is also a product rule because of the product of functions)

In case you got lost, recall how we read a log and write exp from it and from exp write the log

Note: we have a single variable case here. Don't confuse it with a two variable function as in [math]\displaystyle{ f(x,y) = x^y }[/math]. I used [math]\displaystyle{ h(x) }[/math] just to avoid using [math]\displaystyle{ y }[/math] and maybe creating a confusion between single and two variable functions.

I missed one case and it's [math]\displaystyle{ \ln(g(x)) }[/math]. With the chain rule and knowing that the derivative of [math]\displaystyle{ \ln(x) }[/math] is [math]\displaystyle{ 1/x }[/math]:

[math]\displaystyle{ [\ln(g(x))]' = g'(x)[\ln(g(x))]' }[/math]

Therefore:

[math]\displaystyle{ [\ln(g(x))]' = \frac{g'(x)}{g(x)} }[/math]

This pattern of having a quotient where the numerator is the denominator's derivative is very common.