Derivative of \( \displaystyle - \frac{x^{2}}{4} - \frac{x}{2} + \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)} \)
Problem 2.1589 · hard
Differentiate \( \displaystyle f(x) = - \frac{x^{2}}{4} - \frac{x}{2} + \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)} \).
- \[ \frac{d}{d x} \left(- \frac{x^{2}}{4} - \frac{x}{2} + \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)}\right) \]Start with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{x}{2}\right) + \frac{d}{d x} \left(- \frac{x^{2}}{4}\right) + \frac{d}{d x} \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)} \]sumApply the sum rule.✓ Proved
- \[ = \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} + \ln{\left(x + 1 \right)} \frac{d}{d x} \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) + \frac{d}{d x} \left(- \frac{x}{2}\right) + \frac{d}{d x} \left(- \frac{x^{2}}{4}\right) \]productApply the product rule to the third term.✓ Proved
- \[ = \left(x + 1\right) \ln{\left(x + 1 \right)} + \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} + \frac{d}{d x} \left(- \frac{x}{2}\right) + \frac{d}{d x} \left(- \frac{x^{2}}{4}\right) \]derivative algebraDifferentiate the polynomial part of the product. Simplify the polynomial expression.✓ Proved
- \[ = \left(x + 1\right) \ln{\left(x + 1 \right)} + \frac{d}{d x} \left(- \frac{x}{2}\right) + \frac{d}{d x} \left(- \frac{x^{2}}{4}\right) + \frac{\frac{x^{2}}{2} + x + \frac{1}{2}}{x + 1} \]derivativeDifferentiate the logarithm.✓ Proved
- \[ = - \frac{x}{2} + \left(x + 1\right) \ln{\left(x + 1 \right)} - \frac{1}{2} + \frac{\frac{x^{2}}{2} + x + \frac{1}{2}}{x + 1} \]derivative algebra algebraDifferentiate the remaining power and constant terms. Simplify the first term. Factor out 1/2 from the numerator.✓ Proved
- \[ = \left(x + 1\right) \ln{\left(x + 1 \right)} \]algebra algebra algebra simplifyRecognize the perfect square trinomial. Cancel the common factor (x + 1). Distribute the 1/2. Combine like terms.✓ Proved
Answer \( \left(x + 1\right) \ln{\left(x + 1 \right)} \)
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 undefined where x + 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 13 | ✓ 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: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: fail (error) 2026-10-05 — Step 7 applies two derivative rules in a single step, violating the one‑rule‑per‑step rule.
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 2026-10-05 with SymPy 1.14.0.