Derivative of the Natural Logarithm
Why Does $\ln x$ Have Such a Simple Derivative?
The natural logarithm is the inverse of $e^x$, the one exponential function whose derivative equals itself. This special relationship gives $\ln x$ the remarkably clean derivative $1/x$. No other logarithm has this property: base-10 or base-2 logs all have an extra constant factor. This is precisely why mathematicians prefer natural logarithms in calculus.
When you differentiate expressions involving products, quotients, or powers, the chain rule combined with $\frac{d}{dx}(\ln x) = \frac{1}{x}$ transforms complicated derivatives into manageable algebra.
Prerequisite Map
Before You Start: Prerequisite Check
Can you apply the chain rule?
Test yourself: Find $\frac{d}{dx}[\cos(3x)]$.
Answer: $-3\sin(3x)$
If you got this wrong, review the chain rule before continuing.
Can you use implicit differentiation?
Test yourself: If $xy = 1$, find $\frac{dy}{dx}$.
Answer: Differentiate both sides: $y + x\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{y}{x}$.
If this is unfamiliar, review implicit differentiation.
Quick Reference
| Property | Value |
|---|---|
| Concept | Logarithmic Derivatives |
| Chapter | 6.4 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
The Fundamental Formula
$$\boxed{\frac{d}{dx}(\ln x) = \frac{1}{x}}$$
This holds for $x > 0$, since $\ln x$ is only defined for positive inputs.
Why It Works
Let $y = \ln x$. Then by the definition of logarithm:
$$e^y = x$$
Differentiating both sides implicitly:
$$e^y \frac{dy}{dx} = 1$$
Solving for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$
The key insight: since $e^y = x$, we can replace $e^y$ with $x$ directly.
The Chain Rule Form
When differentiating $\ln(g(x))$ where $g(x)$ is a differentiable function:
$$\boxed{\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}}$$
This formula appears constantly in applications. Notice the pattern: the derivative of what’s inside goes on top, while the original inside function goes on the bottom.
Handling Absolute Values
For $x \neq 0$:
$$\frac{d}{dx}(\ln\vert x\vert ) = \frac{1}{x}$$
This works because:
- When $x > 0$: $\ln\vert x\vert = \ln x$, so the derivative is $\frac{1}{x}$
- When $x < 0$: $\ln\vert x\vert = \ln(-x)$, and by chain rule: $\frac{1}{-x} \cdot (-1) = \frac{1}{x}$
This is important for integration: $\int \frac{1}{x}\,dx = \ln\vert x\vert + C$
Domain Restrictions: Watch Out!
Critical point: $\ln x$ is only defined for $x > 0$. When differentiating, always check that your input stays positive.
| Situation | Valid Domain |
|---|---|
| $\ln x$ | $x > 0$ |
| $\ln(x^2)$ | $x \neq 0$ (since $x^2 > 0$ when $x \neq 0$) |
| $\ln(\sin x)$ | Where $\sin x > 0$ (e.g., $(0, \pi)$) |
| $\ln(1 - x^2)$ | $-1 < x < 1$ |
Common Patterns
| Expression | Derivative | Valid When |
|---|---|---|
| $\ln x$ | $\frac{1}{x}$ | $x > 0$ |
| $\ln(ax)$, $a > 0$ | $\frac{1}{x}$ | $x > 0$ |
| $\ln(x^n)$ | $\frac{n}{x}$ | depends on $n$ |
| $\ln(\sin x)$ | $\cot x$ | $\sin x > 0$ |
| $\ln(\cos x)$ | $-\tan x$ | $\cos x > 0$ |
| $[\ln x]^n$ | $\frac{n[\ln x]^{n-1}}{x}$ | $x > 0$ |
Practice Problems
Find the derivative of $f(x) = \ln(5x)$.
Find $\frac{d}{dx}\left[\ln(x^2 + 4)\right]$.
Find $\frac{d}{dx}[\ln(\tan x)]$.
Find $\frac{d}{dx}\left[x^2 \ln(3x + 1)\right]$.
Show that $\frac{d}{dx}\left[\ln\left(x + \sqrt{x^2 + 1}\right)\right] = \frac{1}{\sqrt{x^2 + 1}}$.
Still Confused?
Why is the derivative $\frac{1}{x}$ and not $\frac{1}{\ln x}$?
A common confusion! The derivative formula $\frac{d}{dx}(\ln x) = \frac{1}{x}$ might seem strange because $x$ appears, not $\ln x$.
Think of it this way: The derivative measures the rate of change. For $\ln x$:
- Near $x = 1$: $\ln x$ changes at rate $1/1 = 1$
- Near $x = 10$: $\ln x$ changes at rate $1/10 = 0.1$ (slower)
The rate depends on where you are ($x$), not on the output value ($\ln x$).
When do I use $\ln x$ vs $\ln\vert x\vert $?
- Use $\ln x$ when you know $x > 0$ (e.g., $x$ represents time, population, distance)
- Use $\ln\vert x\vert $ when $x$ could be negative (e.g., in integration, or when $x$ is unrestricted)
The derivative is $\frac{1}{x}$ in both cases, but the domains differ:
- $\ln x$ is defined for $x > 0$
- $\ln\vert x\vert $ is defined for $x \neq 0$
I keep making errors with the chain rule. Help!
Use this template: When differentiating $\ln(\text{stuff})$:
$$\frac{d}{dx}[\ln(\text{stuff})] = \frac{\text{derivative of stuff}}{\text{stuff}}$$
Example: $\frac{d}{dx}[\ln(x^2 + 1)]$
- stuff = $x^2 + 1$
- derivative of stuff = $2x$
- Answer: $\frac{2x}{x^2 + 1}$
Always write out “stuff” and “derivative of stuff” separately until it becomes automatic.
Common Misconceptions
$\frac{d}{dx}[\ln(g(x))] = \frac{1}{g(x)}$, forgetting the chain rule factor $g'(x)$.
This is the composition-is-not-chaining error. The chain rule gives $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$; the numerator is the derivative of the inner function, not 1. For $f(x) = \ln(x^2 + 4)$, the derivative is $\frac{2x}{x^2 + 4}$, not $\frac{1}{x^2 + 4}$. Differentiating the antiderivative $\ln(x^2 + 4)$ without the factor $2x$ in the numerator would not recover the original expression.
Mastery Checklist
Mental Model
The Fraction Pattern: When differentiating $\ln(\text{stuff})$, the answer is always a fraction:
- Top: derivative of the stuff
- Bottom: the original stuff
This “derivative over original” pattern is unique to logarithmic functions and explains why logarithms are so useful for simplifying derivatives of products and quotients.
Connections
Looking back:
- The proof uses implicit differentiation on $e^y = x$
- The chain rule is essential for most applications
Looking ahead:
- Logarithmic differentiation uses this formula to simplify complex products and powers
- Integration reverses this: $\int \frac{1}{x}\,dx = \ln\vert x\vert + C$
Real-world connections:
- In physics, logarithmic derivatives measure relative rates of change
- In economics, $\frac{d}{dx}[\ln f(x)]$ gives the elasticity of $f$
| Previous | Up | Next |
|---|---|---|
| Chain Rule | Skills Index | Logarithmic Differentiation |
Last updated: 2026-01-22