How to Take the Derivative of the Logarithm Function

Hongtao Hao / 2022-11-28

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