Derivative of \( \displaystyle - x^{2} - x + \left(2 x^{2} + 2 x + \frac{1}{2}\right) \ln{\left(2 x + 1 \right)} \)
Problem 2.1984 · hard
Differentiate \( \displaystyle f(x) = - x^{2} - x + \left(2 x^{2} + 2 x + \frac{1}{2}\right) \ln{\left(2 x + 1 \right)} \).
- \[ \frac{d}{d x} \left(- x^{2} - x + \left(2 x^{2} + 2 x + \frac{1}{2}\right) \ln{\left(2 x + 1 \right)}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{d}{d x} \left(2 x^{2} + 2 x + \frac{1}{2}\right) \ln{\left(2 x + 1 \right)} \]sumApply the sum rule.✓ Proved
- \[ = \left(2 x^{2} + 2 x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(2 x + 1 \right)} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x^{2} + 2 x + \frac{1}{2}\right) + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) \]productApply the product rule to the third term.✓ Proved
- \[ = \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} + \left(2 x^{2} + 2 x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) \]derivativeDifferentiate the polynomial factor.✓ Proved
- \[ = \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{\left(2 x^{2} + 2 x + \frac{1}{2}\right) \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{2 \left(2 x^{2} + 2 x + \frac{1}{2}\right)}{2 x + 1} \]derivativeDifferentiate the inner function 2x + 1.✓ Proved
- \[ = - 2 x + \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} - 1 + \frac{2 \left(2 x^{2} + 2 x + \frac{1}{2}\right)}{2 x + 1} \]derivativeDifferentiate the remaining polynomial terms.✓ Proved
- \[ = - 2 x + \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} - 1 + \frac{4 x^{2} + 4 x + 1}{2 x + 1} \]algebraDistribute the 2 in the last term.✓ Proved
- \[ = \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} \]algebra simplify simplifyRecognize the numerator as a perfect square. Cancel the common factor (2*x + 1). Combine like terms to get the final answer.✓ Proved
Answer \( 2 \left(2 x + 1\right) \ln{\left(2 x + 1 \right)} \)
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. 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 undefined where 2*x + 1 = 0 |
| 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 7 replaces both Derivative(-x**2, x) and Derivative(-x, x) in a single step, applying the derivative rule twice at once. Each step must change only one thing; this violates the one‑change rule.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications in the final steps are valid and clearly labeled.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-10 — Step 7 replaces both Derivative(-x**2, x) and Derivative(-x, x) in a single step, applying the derivative rule twice at once. Each step must change only one thing; this violates the one‑change rule.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications in the final steps are valid and clearly labeled.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately reflect the operations performed, and the final simplification is correct.gpt-oss:20b: pass 2026-10-10
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 2026-10-10 with SymPy 1.14.0.