Navigation: Wiki Home > Skills > Derivative of the Natural Logarithm
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.
Test yourself: Find $\frac{d}{dx}[\cos(3x)]$.
Answer: $-3\sin(3x)$
If you got this wrong, review the chain rule before continuing.
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.
| Property | Value |
|---|---|
| Concept | Logarithmic Derivatives |
| Chapter | 6.4 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
$$\boxed{\frac{d}{dx}(\ln x) = \frac{1}{x}}$$
This holds for $x > 0$, since $\ln x$ is only defined for positive inputs.
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.
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.
For $x \neq 0$:
$$\frac{d}{dx}(\ln\vert x\vert ) = \frac{1}{x}$$
This works because:
This is important for integration: $\int \frac{1}{x}\,dx = \ln\vert x\vert + C$
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., $(0, \pi)$) |
| $\ln(1 - x^2)$ | $-1 < x < 1$ |
| 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 $n$ |
| $\ln(\sin x)$ | $\cot x$ | $\sin x > 0$ |
| $\ln(\cos x)$ | $-\tan x$ | $\cos x > 0$ |
| $[\ln x]^n$ | $\frac{n[\ln x]^{n-1}}{x}$ | $x > 0$ |
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 $\frac{d}{dx}\left[\ln\left(x + \sqrt{x^2 + 1}\right)\right] = \frac{1}{\sqrt{x^2 + 1}}$.
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 $\ln x$:
The rate depends on where you are ($x$), not on the output value ($\ln x$).
The derivative is $\frac{1}{x}$ in both cases, but the domains differ:
Use this template: When differentiating $\ln(\text{stuff})$:
$$\frac{d}{dx}[\ln(\text{stuff})] = \frac{\text{derivative of stuff}}{\text{stuff}}$$
Example: $\frac{d}{dx}[\ln(x^2 + 1)]$
Always write out "stuff" and "derivative of stuff" separately until it becomes automatic.
The Fraction Pattern: When differentiating $\ln(\text{stuff})$, the answer is always a fraction:
This "derivative over original" pattern is unique to logarithmic functions and explains why logarithms are so useful for simplifying derivatives of products and quotients.
Looking back:
Looking ahead:
Real-world connections:
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|---|---|---|
| Chain Rule | Skills Index | Logarithmic Differentiation |
Last updated: 2026-01-22