Derivative of \( \displaystyle - \frac{x^{2}}{2} + x \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{x}{2} + \frac{\ln{\left(2 x - 1 \right)}}{4} \)
Problem 2.49 · hard
Differentiate \( \displaystyle f(x) = - \frac{x^{2}}{2} + x \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{x}{2} + \frac{\ln{\left(2 x - 1 \right)}}{4} \).
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{2} + x \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{x}{2} + \frac{\ln{\left(2 x - 1 \right)}}{4}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{d}{d x} x \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{4} \]sumApply the sum rule.✓ Proved
- \[ = \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{d}{d x} x \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]constantPull out the constant factor from the last term.✓ Proved
- \[ = x \left(x - 1\right) \frac{d}{d x} \ln{\left(2 x - 1 \right)} + x \ln{\left(2 x - 1 \right)} \frac{d}{d x} \left(x - 1\right) + \left(x - 1\right) \ln{\left(2 x - 1 \right)} \frac{d}{d x} x + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]productApply the product rule to the second term.✓ Proved
- \[ = x \left(x - 1\right) \frac{d}{d x} \ln{\left(2 x - 1 \right)} + x \ln{\left(2 x - 1 \right)} + \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]derivative algebraDifferentiate the individual components within the product rule. Simplify the expression by removing unnecessary parentheses.✓ Proved
- \[ = \frac{2 x \left(x - 1\right)}{2 x - 1} + x \ln{\left(2 x - 1 \right)} + \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{1}{2 \left(2 x - 1\right)} \]chain algebraApply the chain rule to the logarithmic terms. Simplify the products and fractions.✓ Proved
- \[ = x \ln{\left(2 x - 1 \right)} + \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{2 x^{2} - 2 x}{2 x - 1} + \frac{1}{2 \left(2 x - 1\right)} \]algebraExpand the numerator in the third term.✓ Proved
- \[ = x \ln{\left(2 x - 1 \right)} + \left(x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{1}{2} + \frac{2 x^{2} - 2 x}{2 x - 1} + \frac{1}{2 \left(2 x - 1\right)} \]derivativeDifferentiate the term x/2.✓ Proved
- \[ = \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{d}{d x} \left(- \frac{x^{2}}{2}\right) + \frac{1}{2} + \frac{2 x^{2} - 2 x}{2 x - 1} + \frac{1}{2 \left(2 x - 1\right)} \]algebraCombine the log terms using factoring.✓ Proved
- \[ = - x + \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{1}{2} + \frac{2 x^{2} - 2 x}{2 x - 1} + \frac{1}{2 \left(2 x - 1\right)} \]derivativeDifferentiate the first term -x**2/2.✓ Proved
- \[ = - x + \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{2 x^{2} - x - \frac{1}{2}}{2 x - 1} + \frac{1}{2 \left(2 x - 1\right)} \]algebra algebra algebraCombine the fraction terms over a common denominator. Expand the numerator. Combine like terms in the numerator.✓ Proved
- \[ = - x + \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} + \frac{2 x^{2} - x}{2 x - 1} \]simplifyCombine all rational terms into one single fraction.✓ Proved
Answer \( \left(2 x - 1\right) \log{\left(2 x - 1 \right)} \)
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. The reviewers disagree about how one step is explained; every verdict is in 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 |
| 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 undefined where 2*x - 1 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 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 4 applies the product rule to the triple product x*(x-1)*log(2*x-1) as if it were a product of only two factors. The derivative of (x-1)*log(2*x-1) itself requires another product rule, so the step combines two rules at once and is mathematically incorrect.deepseek-r1:70b: passqwen3.6:27b-mlx: fail (error) — Step 13 is mathematically incorrect: it attempts to combine `(2*x**2 - 2*x)/(2*x - 1)` and `1/2` into a single fraction but incorrectly calculates the numerator as `2*x**2 - 2*x + 1/2*(2*x - 1)` while simultaneously retaining the separate term `(1/2)/(2*x - 1)` on the RHS, leading to a mismatch in value and logic. Specifically, `A + B` cannot become `A_combined + B` where `A_combined` already includes `B`'s contribution.
Every verdict on record (12)
qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 13 is mathematically incorrect: it attempts to combine `(2*x**2 - 2*x)/(2*x - 1)` and `1/2` into a single fraction but incorrectly calculates the numerator as `2*x**2 - 2*x + 1/2*(2*x - 1)` while simultaneously retaining the separate term `(1/2)/(2*x - 1)` on the RHS, leading to a mismatch in value and logic. Specifically, `A + B` cannot become `A_combined + B` where `A_combined` already includes `B`'s contribution.gpt-oss:20b: fail (error) 2026-09-20 — Step 4 applies the product rule to the triple product x*(x-1)*log(2*x-1) as if it were a product of only two factors. The derivative of (x-1)*log(2*x-1) itself requires another product rule, so the step combines two rules at once and is mathematically incorrect.qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 13 contains a mathematical error: the term `(1/2)/(2*x - 1)` is present in Step 12 but is incorrectly omitted in the final combined fraction in Step 13 (the numerator `2*x**2 - 2*x + 1/2*(2*x - 1)` accounts for the previous fraction but drops the separate `1/2` term). Consequently, Step 16 simplifies to an incorrect result.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: fail (error) 2026-09-19 — Step 13 is algebraically incorrect: it combines (2*x**2 - 2*x)/(2*x - 1) and 1/2 into a single fraction but fails to include the remaining term (1/2)/(2*x - 1) in that combination, effectively dropping it or misplacing it. Step 16 then incorrectly simplifies the result to (2*x**2 - x)/(2*x - 1), which does not equal the sum of the rational terms present in step 15. The final answer is missing the rational part of the derivative.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: fail (error) 2026-09-19 — Step 4 applies a product rule to the entire term x*(x-1)*log(2*x-1) as if it were a single product, but it is actually a product of three factors; the subsequent differentiation in step 5 treats each factor separately, which is inconsistent with the rule applied. Additionally, step 13 incorrectly combines the constant 1/2 with the fraction (2*x**2-2*x)/(2*x-1), treating 1/2 as if it had the same denominator, leading to an algebraic error.qwen3.6:27b-mlx: fail (error) 2026-09-19 — Step 13 is mathematically incorrect; the term (1/2)/(2*x - 1) is not combined into the fraction in the way shown, and the resulting numerator arithmetic is wrong. Specifically, the previous step had separate terms (2*x**2 - 2*x)/(2*x - 1) and (1/2)/(2*x - 1) and 1/2. Step 13 attempts to combine the first fraction with the 1/2 term but ignores the last term or combines them incorrectly, leading to a wrong intermediate expression that does not algebraically equal the previous line.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: inconclusive 2026-09-19 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The final simplification step (step 16) claims to combine all rational terms into a single fraction, but it fails to cancel the remaining terms. The expregpt-oss:20b: inconclusive 2026-09-17 — reviewer response could not be parsed:deepseek-r1:70b: pass 2026-09-17
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 2026-09-26 with SymPy 1.14.0.