Derivative of \( \displaystyle - x^{2} + 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} - x + \frac{\ln{\left(2 x + 1 \right)}}{2} \)
Problem 2.1208 · hard
Differentiate \( \displaystyle f(x) = - x^{2} + 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} - x + \frac{\ln{\left(2 x + 1 \right)}}{2} \).
- \[ \frac{d}{d x} \left(- x^{2} + 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} - x + \frac{\ln{\left(2 x + 1 \right)}}{2}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{d}{d x} 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} + \frac{d}{d x} \frac{\ln{\left(2 x + 1 \right)}}{2} \]sum constant-multipleApply the sum rule. Factor out the constant 1/2.✓ Proved
- \[ = \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{d}{d x} 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} + \frac{\frac{d}{d x} \ln{\left(2 x + 1 \right)}}{2} \]constantMove the constant outside the derivative.✓ Proved
- \[ = \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{d}{d x} 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} + \frac{\frac{d}{d x} \left(2 x + 1\right)}{2 \left(2 x + 1\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{d}{d x} 2 x \left(x + 1\right) \ln{\left(2 x + 1 \right)} + \frac{1}{2 x + 1} \]derivative algebraDifferentiate the inner function 2x + 1. Simplify the constant term.✓ Proved
- \[ = 2 x \left(x + 1\right) \frac{d}{d x} \ln{\left(2 x + 1 \right)} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} 2 x \left(x + 1\right) + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{1}{2 x + 1} \]productApply the product rule to the second term.✓ Proved
- \[ = 2 x \left(x + 1\right) \frac{d}{d x} \ln{\left(2 x + 1 \right)} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x^{2} + 2 x\right) + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{1}{2 x + 1} \]algebraExpand the polynomial inside the derivative.✓ Proved
- \[ = \frac{2 x \left(x + 1\right) \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x^{2} + 2 x\right) + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{1}{2 x + 1} \]chainApply the chain rule to the logarithm term again.✓ Proved
- \[ = \frac{4 x \left(x + 1\right)}{2 x + 1} + \ln{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x^{2} + 2 x\right) + \frac{d}{d x} \left(- x\right) + \frac{d}{d x} \left(- x^{2}\right) + \frac{1}{2 x + 1} \]derivative algebraDifferentiate the inner function 2x + 1. Simplify the product of constants and variables.✓ Proved
- \[ = \frac{4 x \left(x + 1\right)}{2 x + 1} - 2 x + \ln{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x^{2} + 2 x\right) - 1 + \frac{1}{2 x + 1} \]derivativeDifferentiate the remaining polynomial terms.✓ Proved
- \[ = \frac{4 x \left(x + 1\right)}{2 x + 1} - 2 x + \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} - 1 + \frac{1}{2 x + 1} \]derivativeDifferentiate the polynomial 2x^2 + 2x.✓ Proved
- \[ = - 2 x + \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} - 1 + \frac{4 x^{2} + 4 x + 1}{2 x + 1} \]algebraCombine the fractions with the same denominator.✓ Proved
- \[ = \left(4 x + 2\right) \ln{\left(2 x + 1 \right)} \]algebra simplify simplifyRecognize the numerator as a perfect square. Simplify the expression by canceling terms. Final simplified result.✓ 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 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 |
| 17 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 18 | ✓ 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 17 incorrectly drops the constant –1 from the expression, leading to an incorrect final result. The simplification should cancel –2*x with +2*x and –1 with +1, yielding (4*x+2)*log(2*x+1), but the omitted –1 makes the step algebraically wrong.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. All labels are appropriate for the operations performed.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. All labels are appropriate for the operations performed.gpt-oss:20b: fail (error) 2026-09-29 — Step 17 incorrectly drops the constant –1 from the expression, leading to an incorrect final result. The simplification should cancel –2*x with +2*x and –1 with +1, yielding (4*x+2)*log(2*x+1), but the omitted –1 makes the step algebraically wrong.qwen3.6:27b-mlx: pass 2026-09-29gpt-oss:20b: pass 2026-09-29
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 2026-09-29 with SymPy 1.14.0.