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 2.705 · hard Video
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
- \[ = - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{d}{d x} x \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{\ln{\left(2 x - 1 \right)}}{4} \]sumApply the sum rule.✓ Proved
- \[ = - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{d}{d x} x \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]constantFactor out the constant from the last term.✓ Proved
- \[ = - x \left(x - 1\right) \frac{d}{d x} \ln{\left(2 x - 1 \right)} - x \ln{\left(2 x - 1 \right)} \frac{d}{d x} \left(x - 1\right) - \left(x - 1\right) \ln{\left(2 x - 1 \right)} \frac{d}{d x} x - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]productApply the product rule to the second term.✓ Proved
- \[ = - x \left(x - 1\right) \frac{d}{d x} \ln{\left(2 x - 1 \right)} - x \ln{\left(2 x - 1 \right)} - \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]derivativeDifferentiate the individual parts of the product rule.✓ Proved
- \[ = - \frac{2 x \left(x - 1\right)}{2 x - 1} - x \ln{\left(2 x - 1 \right)} - \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{4} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = - \frac{2 x \left(x - 1\right)}{2 x - 1} - x \ln{\left(2 x - 1 \right)} - \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{2} - \frac{1}{2 \left(2 x - 1\right)} \]algebraSimplify the derivative of the log term.✓ Proved
- \[ = - \frac{2 x \left(x - 1\right)}{2 x - 1} - x \ln{\left(2 x - 1 \right)} + x - \left(x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{1}{2} - \frac{1}{4 x - 2} \]derivativeEvaluate the remaining derivatives.✓ Proved
- \[ = x - \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{1}{2} - \frac{1}{4 x - 2} - \frac{2 x^{2} - 2 x}{2 x - 1} \]algebra algebra algebraCombine the log terms. Group the rational terms.✓ Proved
- \[ = x - \frac{2 x}{4 x - 2} - \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{2 x^{2} - 2 x}{2 x - 1} \]algebra algebraFind a common denominator for the rational part. Simplify the numerator of the rational part.✓ Proved
- \[ = x - \frac{x}{2 x - 1} - \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{2 x^{2} - 2 x}{2 x - 1} \]algebraSimplify the fraction.✓ Proved
- \[ = x - \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} - \frac{2 x^{2} - x}{2 x - 1} \]algebra algebraCombine the rational terms over the common denominator. Simplify the numerator.✓ Proved
- \[ = - \left(2 x - 1\right) \ln{\left(2 x - 1 \right)} \]algebra algebraFactor out x from the numerator. Simplify the fraction.✓ Proved
- \[ = \left(1 - 2 x\right) \ln{\left(2 x - 1 \right)} \]algebraFinal simplification.✓ Proved
Answer \( \left(1 - 2 x\right) \log{\left(2 x - 1 \right)} \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation. How we verify
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 undefined where 4*x - 2 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 4*x - 2 = 0 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 4*x - 2 = 0 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 4*x - 2 = 0 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 4*x - 2 = 0 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 4*x - 2 = 0 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 4*x - 2 = 0 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 |
| 17 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 18 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 19 | ✓ 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: passqwen3.6:27b-mlx: fail (error) — Step 3 is labeled 'constant' but applies the constant-multiple rule (factoring 1/4). The vocabulary requires 'constant-multiple' for this operation; 'constant' is reserved for the derivative of a constant being zero. Additionally, Step 7 is labeled 'algebra' but performs the differentiation of the log term (applying the chain rule result), which should be labeled 'chain' or 'derivative', not 'algebra'.
Senior review claude-sonnet-5-5, 2026-10-11: fail — The math is correct throughout and every line was proved by SymPy, so the final answer (1-2x)*log(2x-1) is right. Two steps carry the wrong rule label: step 3 should be 'constant-multiple', and step 7 should be 'chain' or 'derivative' instead of 'algebra'.
qwen3.6:27b-mlx: uphold — Step 3 pulls the factor 1/4 out of the derivative, which is the constant-multiple rule, not 'constant', and step 7 substitutes the chain-rule derivative 2/(2x-1) for Derivative(log(2x-1),x), which is a derivative/chain step rather than 'algebra'; both labels name a rule the step did not apply.
Every verdict on record (7)
claude-sonnet-5-5: fail (style) 2026-10-11 — The math is correct throughout and every line was proved by SymPy, so the final answer (1-2x)*log(2x-1) is right. Two steps carry the wrong rule label: step 3 should be 'constant-multiple', and step 7 should be 'chain' or 'derivative' instead of 'algebra'.qwen3.6:27b-mlx: fail (error) 2026-09-26 — Step 3 is labeled 'constant' but applies the constant-multiple rule (factoring 1/4). The vocabulary requires 'constant-multiple' for this operation; 'constant' is reserved for the derivative of a constant being zero. Additionally, Step 7 is labeled 'algebra' but performs the differentiation of the log term (applying the chain rule result), which should be labeled 'chain' or 'derivative', not 'algebra'.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: fail (error) 2026-09-21 — Step 4 applies the product rule twice in a single line, combining two separate rule applications into one. This violates the rule that each step must change only one thing. The correct approach would split the product rule application into two separate steps or use a single product rule on the two outer factors and then expand the inner product in a subsequent step.qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21
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.