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 2.1779 · hard
Differentiate \( \displaystyle f(x) = \frac{x^{2}}{4} + \frac{x}{2} - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)} \).
- \[ \frac{d}{d x} \left(\frac{x^{2}}{4} + \frac{x}{2} - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)}\right) \]sumStart with the derivative of the entire expression.✓ Proved
- \[ = \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} - \frac{d}{d x} \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \ln{\left(x + 1 \right)} \]sumApply the sum rule to separate the terms.✓ Proved
- \[ = - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} - \ln{\left(x + 1 \right)} \frac{d}{d x} \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} \]productApply the product rule to the third term.✓ Proved
- \[ = - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} - \left(\frac{d}{d x} \frac{1}{2} + \frac{d}{d x} x + \frac{d}{d x} \frac{x^{2}}{2}\right) \ln{\left(x + 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} \]sumApply the sum rule to the derivative of the polynomial.✓ Proved
- \[ = - \left(\frac{d}{d x} x + \frac{d}{d x} \frac{x^{2}}{2}\right) \ln{\left(x + 1 \right)} - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} \]constantThe derivative of the constant 1/2 is 0.✓ Proved
- \[ = - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} - \ln{\left(x + 1 \right)} \frac{d}{d x} x - \ln{\left(x + 1 \right)} \frac{d}{d x} \frac{x^{2}}{2} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} \]algebra algebraDistribute the log(x + 1) term. Remove the zero term.✓ Proved
- \[ = - \left(\frac{d}{d x} x + \frac{d}{d x} \frac{x^{2}}{2}\right) \ln{\left(x + 1 \right)} - \left(\frac{x^{2}}{2} + x + \frac{1}{2}\right) \frac{d}{d x} \ln{\left(x + 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} \]algebraFactor out log(x + 1) from the parenthesis.✓ Proved
- \[ = - \left(\frac{d}{d x} x + \frac{d}{d x} \frac{x^{2}}{2}\right) \ln{\left(x + 1 \right)} + \frac{d}{d x} \frac{x}{2} + \frac{d}{d x} \frac{x^{2}}{4} - \frac{\frac{x^{2}}{2} + x + \frac{1}{2}}{x + 1} \]logarithmicDifferentiate the log(x + 1) term.✓ Proved
- \[ = \frac{x}{2} - \left(x + 1\right) \ln{\left(x + 1 \right)} + \frac{1}{2} - \frac{\frac{x^{2}}{2} + x + \frac{1}{2}}{x + 1} \]power algebra algebraDifferentiate the power terms. Simplify the coefficients. Factor out 1/2 from the numerator.✓ Proved
- \[ = - \left(x + 1\right) \ln{\left(x + 1 \right)} \]algebra algebra algebraRecognize the perfect square trinomial. Cancel the common factor (x + 1). Distribute the division by 2.✓ Proved
- \[ = \left(- x - 1\right) \ln{\left(x + 1 \right)} \]simplifyCombine all remaining terms.✓ Proved
Answer \( \left(- x - 1\right) \ln{\left(x + 1 \right)} \)
✓ 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.
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 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 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 undefined where x + 1 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| 16 | ✓ 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)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies differentiation rules and algebraic simplifications in a step-by-step manner. Each step adheres to the single-rule constraint and uses valid labels from the provided vocabulary.
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 2026-10-07 with SymPy 1.14.0.