# Derivative of ln(2x): Why It Is 1/x, With Examples

> The derivative of ln(2x) is 1/x, the same as ln x. See two proofs, a graph of why the slopes match, and how ln(2x + 1) is different.

Source: https://practicecalculus.com/derivatives/ln-2x
Updated: 2026-11-06

$$ \frac{d}{dx}\,\ln(2x) = \frac{1}{x} $$

## At a glance

- **Derivative:** $\dfrac{1}{x}$ for $x > 0$
- **General form:** $\dfrac{d}{dx}\ln(kx) = \dfrac{1}{x}$ for any constant $k > 0$
- **Key log rule:** $\ln(2x) = \ln 2 + \ln x$
- **Domain of ln(2x):** $x > 0$
- **Compare:** $\dfrac{d}{dx}\ln(2x + 1) = \dfrac{2}{2x + 1}$

## Key takeaways

- The derivative of $\ln(2x)$ is $\frac{1}{x}$, exactly the same as the derivative of $\ln x$.
- Reason: $\ln(2x) = \ln 2 + \ln x$, and the constant $\ln 2$ has derivative $0$.
- The chain rule gives the same thing: $\frac{2}{2x} = \frac{1}{x}$.
- This only works for a multiple of $x$. For $\ln(2x + 1)$ the derivative is $\frac{2}{2x + 1}$.

The **derivative of $\ln(2x)$ is $\dfrac{1}{x}$**. This surprises many students: the $2$ disappears, and the answer matches the [derivative of ln x](/derivatives/ln-x).

$$\frac{d}{dx}\,\ln(2x) = \frac{1}{x}, \qquad x > 0$$

## Proof 1: the chain rule

Let $u = 2x$, so $u' = 2$. The [chain rule](/rules/chain-rule) for logs says "derivative of the inside over the inside":

$$\frac{d}{dx}\ln(2x) = \frac{u'}{u} = \frac{2}{2x} = \frac{1}{x}$$

The $2$ on top cancels the $2$ on the bottom.

## Proof 2: split the log first

The product rule for logarithms says $\ln(ab) = \ln a + \ln b$. So

$$\ln(2x) = \ln 2 + \ln x$$

Now $\ln 2 \approx 0.693$ is just a number, and the derivative of a constant is $0$:

$$\frac{d}{dx}\ln(2x) = 0 + \frac{1}{x} = \frac{1}{x}$$

This second proof explains *why* the answer is the same as for $\ln x$.

## Graph: two curves, same slopes

```graph
title: 'Graph of y = ln(2x) and y = ln x'
caption: 'The curve $\ln(2x)$ is $\ln x$ shifted up by $\ln 2 \approx 0.693$. Shifting a graph up does not change its slope, so both have derivative $\frac{1}{x}$.'
x: [-0.5, 6]
y: [-3, 3]
functions:
  - expr: ln(2x)
    label: y = ln(2x)
    domain: [0.02, 6]
  - expr: ln(x)
    label: y = ln x
    dashed: true
    domain: [0.04, 6]
points:
  - [0.5, 0, '(1/2, 0)']
  - [1, 0, '(1, 0)']
```

At any $x$, the two curves are the same vertical distance apart, so their tangent lines are parallel. At $x = 2$, for example, both have slope $\frac{1}{2}$.

## When the constant does not cancel

The trick only works when the inside is a constant *times* $x$. Compare:

| Function | Derivative | Does the constant cancel? |
| --- | --- | --- |
| $\ln(2x)$ | $\frac{2}{2x} = \frac{1}{x}$ | yes |
| $\ln(5x)$ | $\frac{5}{5x} = \frac{1}{x}$ | yes |
| $\ln(2x + 1)$ | $\frac{2}{2x + 1}$ | no |
| $\ln(x + 2)$ | $\frac{1}{x + 2}$ | no |
| $\ln(2x^3)$ | $\frac{6x^2}{2x^3} = \frac{3}{x}$ | the $2$ cancels, the power stays |

The last row works because $\ln(2x^3) = \ln 2 + 3\ln x$.

## Worked examples

**Example 1: a tangent line.** At $x = \frac{1}{2}$, $\ln(2x) = \ln 1 = 0$ and the slope is $\frac{1}{1/2} = 2$. The tangent line is

$$y = 2\left(x - \tfrac{1}{2}\right) = 2x - 1$$

**Example 2: product rule.** For $f(x) = x\ln(2x)$, the [product rule](/rules/product-rule) gives

$$f'(x) = \ln(2x) + x\cdot\frac{1}{x} = \ln(2x) + 1$$

**Example 3: a square.** For $g(x) = \left[\ln(2x)\right]^2$, the outer function is a square:

$$g'(x) = 2\ln(2x)\cdot\frac{1}{x} = \frac{2\ln(2x)}{x}$$

## Common mistakes

1. **Writing $\frac{1}{2x}$.** You divided by the inside but forgot to multiply by its derivative, $2$.
2. **Writing $\frac{2}{x}$.** You multiplied by $2$ but did not keep the $2$ in the denominator.
3. **Cancelling in $\ln(2x + 1)$.** The $2$ only cancels when the inside is exactly $2x$.

See also the [derivative of ln(x²)](/derivatives/ln-x-squared), where a power instead of a multiple sits inside the log, and practise more with [derivatives of exponential and logarithmic functions](/practice-problems/derivatives-of-exponential-and-logarithmic-functions).

## Frequently asked questions

### What is the derivative of ln(2x)?

The derivative of $\ln(2x)$ is $\frac{1}{x}$.

### Why is the derivative of ln(2x) the same as ln x?

Because $\ln(2x) = \ln 2 + \ln x$. The two functions differ only by the constant $\ln 2$, so their graphs are vertical shifts of each other and have the same slope everywhere.

### What is the derivative of ln(3x)?

Also $\frac{1}{x}$. For any positive constant $k$, $\frac{d}{dx}\ln(kx) = \frac{k}{kx} = \frac{1}{x}$.

### What is the derivative of ln(2x + 1)?

It is $\frac{2}{2x + 1}$. Adding $1$ inside means the constant no longer splits off as a separate log, so nothing cancels.