What is the derivative of `$f(x) = log(x)$`

? This post tries to prove that it is `$\frac{1}{x}$`

when the log base is `$e$`

.

## Derivative of exponential function and the definition of e #

Before talking about the derivative of log function, let us review the definition of `$e$`

. It is related to the derivative of the exponential function, for example, `$f(x) = 2^x$`

.

According to the definition of derivative, we know that the derivative of `$f(x) = 2^x$`

should be:

`$$\lim_{dx \to 0}\frac{2^{x+dx} - 2^x}{dx} = \lim_{dx \to 0}\frac{(2^{dx} - 1) \cdot 2^x}{dx}$$`

And we know that as `$dx$`

approaches `$0$`

, `$\frac{2^{dx} - 1}{dx}$`

is approaching a constent (`$ln(2)$`

). That is to say, the derivative of `$f(x) = 2^x$`

is `$2^x$`

itself multiplied by a constant.

Then we wonder, is there a case where the drivatie of `$f(x) = t^x$`

is itself? That is to say, when will the constant be `$1$`

? And that is the definition of `$e$`

! We stipulate that if the derivative of `$f(x) = t^x$`

is `$t^x$`

itself, then we denote that special `$t$`

with `$e$`

.

Therefore, we have:

`$$\lim_{dx \to 0} \frac{e^{dx} - 1}{dx} = 1$$`

So, as `$dx$`

approaches `$0$`

, we have

`$$e^{dx} = dx + 1$$`

Therefore, we have

`$$\lim_{dx \to 0}(dx + 1)^{\frac{1}{dx}} = e$$`

If we replace `$dx$`

with `$n$`

, we have:

`$$\lim_{n \to 0}(n + 1)^{\frac{1}{n}} = e$$`

## Direct derivation #

With this background information, we should be able to derive the derivative of the log function. We will take the natural log function as an example, i.e., we have `$f(x) = ln(x)$`

.

`$$\begin{aligned} f^\prime(x) &= \lim_{dx \to 0}\frac{ln(x+dx) - ln(x)}{dx} \\ & = \lim_{dx \to 0}\frac{ln(\frac{x+dx}{dx})}{dx} \\ &= \lim_{dx \to 0}\frac{ln(1 + \frac{dx}{x})}{dx} \\ &= \lim_{dx \to 0}\frac{1}{dx}ln(1 + \frac{dx}{x}) \\ &= \lim_{dx \to 0}{ln(1 + \frac{dx}{x})}^{\frac{1}{dx}}\end{aligned}$$`

Nowt, it’s becoming interesting. It looks similar to the definition of `$e$`

, right? But not quite the same. Let’s say we denote `$\frac{dx}{x}$`

as `$m$`

, then the above equation can be written as

`$$f^\prime(x) = {ln(m + 1)}^{\frac{x}{n}} = {ln(m + 1)}^{\frac{1}{m}\cdot \frac{1}{x}} = \frac{1}{x}{ln(m + 1)}^{\frac{1}{m}} $$`

You might jump and say that we can use the definition of `$e$`

directly, but to do that, we need to prove that `$m$`

here and the `$n$`

in the difinition of `$e$`

is the same.

In the definition of `$e$`

`$$\lim_{n \to 0}(n + 1)^{\frac{1}{n}} = e$$`

We have `$n$`

which is approaching `$0$`

, and `$\frac{1}{n}$`

approaching positive infinity. Since `$m = \frac{dx}{x}$`

, and `$dx$`

is approaching `$0$`

, so `$m$`

is approaching `$0$`

and `$\frac{1}{m} = \frac{x}{dx}$`

is approaching positive infinity. Therefore, we can say with certainty that as `$dx$`

approaches `$0$`

, `${(m + 1)}^{\frac{1}{m}}$`

is approaching `$e$`

. Thus, `${ln(m + 1)}^{\frac{1}{m}}$`

is approaching 1. Therefore, `$f^\prime(x)$`

is approaching `$\frac{1}{x}$`

, which means that **the derivative of the natural log function is $\frac{1}{x}$**.

## A more clever method: inverse function. #

In case you are not familiar with the properties of the derivatives of inversion functions, let’s have an example.

Suppose we have `$y = x^2$`

, so we have `$x = \sqrt{y}$`

. We know that `$y^{\prime} = 2x$`

. And that `$x^{\prime} = \frac{1}{2\sqrt{y}} = \frac{1}{2x}$`

.

We find that `$y^{\prime} \cdot x^{\prime} = 1$`

.

Now, let’s say `$y = e^x$`

, so we have `$x = ln(y)$`

. Since `$y^{\prime} = e^x$`

, we have

`$$x^{\prime} = \frac{1}{e^x} = \frac{1}{e^{ln(y)}} = \frac{1}{y}$$`

Therefore, if `$f(x) = ln(x)$`

, we have

`$$f^{\prime}(x) = \frac{1}{x}$$`

Last modified on 2023-01-18