Derivative of \( \displaystyle x \ln{\left(2 x - 1 \right)} - x - \frac{\ln{\left(2 x - 1 \right)}}{2} \)
Problem 2.1074 · hard Beautiful
Differentiate \( \displaystyle f(x) = x \ln{\left(2 x - 1 \right)} - x - \frac{\ln{\left(2 x - 1 \right)}}{2} \).
- \[ \frac{d}{d x} \left(x \ln{\left(2 x - 1 \right)} - x - \frac{\ln{\left(2 x - 1 \right)}}{2}\right) \]Start with the derivative of the function.✓ Proved
- \[ = - \frac{d}{d x} x + \frac{d}{d x} x \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{2} \]sumApply the sum rule.✓ Proved
- \[ = \frac{d}{d x} x \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{2} - 1 \]constantThe derivative of x is 1.✓ Proved
- \[ = x \frac{d}{d x} \ln{\left(2 x - 1 \right)} + \ln{\left(2 x - 1 \right)} \frac{d}{d x} x - \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{2} - 1 \]productApply the product rule to the first term.✓ Proved
- \[ = x \frac{d}{d x} \ln{\left(2 x - 1 \right)} + \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{2} - 1 \]constantThe derivative of x is 1.✓ Proved
- \[ = \frac{x \frac{d}{d x} \left(2 x - 1\right)}{2 x - 1} + \ln{\left(2 x - 1 \right)} - \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{2} - 1 \]chainApply the chain rule to the log terms.✓ Proved
- \[ = \frac{2 x}{2 x - 1} + \ln{\left(2 x - 1 \right)} - 1 - \frac{1}{2 x - 1} \]derivative algebraThe derivative of 2x - 1 is 2. Simplify the terms.✓ Proved
- \[ = \ln{\left(2 x - 1 \right)} \]algebra algebra simplifyCombine the fractions. Simplify the fraction. Final simplification.✓ Proved
Answer \( \ln{\left(2 x - 1 \right)} \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: fail (error) — Step 6 applies the chain rule to two separate terms in a single line, changing more than one thing at once, which violates the one‑change rule. The step should be split into two separate derivative applications.qwen3.6:27b-mlx: fail (error) — Step 6 applies both the chain rule and the constant-multiple rule simultaneously to the term Derivative(log(2*x - 1)/2, x), violating the one-rule-per-step constraint. Specifically, it factors out the 1/2 and applies the chain rule to the log in a single step.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — Step 6 applies both the chain rule and the constant-multiple rule simultaneously to the term Derivative(log(2*x - 1)/2, x), violating the one-rule-per-step constraint. Specifically, it factors out the 1/2 and applies the chain rule to the log in a single step.gpt-oss:20b: fail (error) 2026-09-28 — Step 6 applies the chain rule to two separate terms in a single line, changing more than one thing at once, which violates the one‑change rule. The step should be split into two separate derivative applications.qwen3.6:27b-mlx: fail (error) 2026-09-28 — Step 6 applies both the chain rule to the derivative of the log and the constant multiple rule to factor out the 1/2, violating the one-rule-per-step constraint. Additionally, the label 'chain' is insufficient to describe the simultaneous application of the constant multiple rule.gpt-oss:20b: fail (error) 2026-09-28 — Step 6 applies the chain rule twice in one line (to both log terms) while only one rule may be applied per step. This violates the single‑rule‑per‑step requirement.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-09-28 with SymPy 1.14.0.