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Exponential and log derivatives

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.

Exponential and log derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\ln(2x) = \frac{1}{x}
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}
On this page
  1. Proof 1: the chain rule
  2. Proof 2: split the log first
  3. Graph: two curves, same slopes
  4. When the constant does not cancel
  5. Worked examples
  6. Common mistakes

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.

\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 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 of y = ln(2x) and y = ln x
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}.

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 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), where a power instead of a multiple sits inside the log, and practise more with derivatives of exponential and logarithmic functions.

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Chain rule: any function

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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.

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