Derivative of ln(2x)
ln(2x) is calculus's favorite trick question: the chain rule fires, the 2s cancel, and the answer is identical to the derivative of plain ln(x).
Step-by-step solution
Outer function ln(u) with derivative 1/u; inner function u = 2x with derivative 2. So f′(x) = [1/(2x)] · 2.
The 2 from the inner derivative cancels the 2 in the denominator: f′(x) = 1/x.
ln(2x) = ln 2 + ln(x). The constant ln 2 differentiates to zero, and ln(x) differentiates to 1/x — same answer, no chain rule needed.
Why it works
The cancellation isn't a coincidence — it's a logarithm property wearing a chain-rule costume. Multiplying a positive input of a log by a positive constant just shifts the graph vertically (by ln of that constant), and vertical shifts never change slope. So ln(2x), ln(5x), and ln(x/7) all share the derivative 1/x.
This example is worth internalizing because it teaches a general lesson: before differentiating, ask whether an algebraic rewrite makes the problem trivial. Splitting logs, expanding products, or simplifying fractions first often eliminates entire rule applications — and eliminates the opportunities for error that come with them.
Domain and logarithm properties
The real-valued function ln(2x) requires 2x > 0, so both this function and its derivative are considered on x > 0. The expression 1/x exists at negative inputs, but it is not the derivative of ln(2x) there because ln(2x) is undefined.
More generally, for a nonzero real constant k, ln(kx) has derivative k/(kx) = 1/x wherever kx > 0. The rewrite ln(kx) = ln(k) + ln(x) is valid over the reals when k and x are both positive. If both are negative, use ln|k| + ln|x| instead.
For ln(x²), the input is positive for every x ≠ 0. The correct identity on this whole domain is ln(x²) = 2 ln|x|. Its derivative is 2/x on both intervals x < 0 and x > 0. For ln(2x + 1), the derivative is 2/(2x + 1) only for x > −1/2.
Common mistakes
- Answering 1/(2x): applying the outer derivative but forgetting to multiply by the inner derivative 2.
- Answering 2/x: multiplying by the inner derivative but forgetting the 2 already sitting in the denominator.
- Overgeneralizing the cancellation: it works for ln(kx) with constant k, but not for ln(x²), whose derivative is 2/x.
Practice problems
Differentiate ln(5x)
Answer: 1/x — the same cancellation.
Differentiate ln(x²)
Answer: 2/x for x ≠ 0 — write it as 2 ln|x| first. The identity 2 ln(x) covers only x > 0.
Differentiate ln(2x + 1)
Answer: 2/(2x + 1) — no cancellation now, because the inside isn't a pure multiple of x.