Derivative of \( \displaystyle \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} \)
Problem 2.1096 · hard
Differentiate \( \displaystyle f(x) = \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} \).
- \[ \frac{d}{d x} \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} \]productApply the product rule.✓ Proved
- \[ = \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) \frac{d}{d x} e^{2 x - 1}}{4} + \frac{e^{2 x - 1} \frac{d}{d x} \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right)}{4} \]chainApply the chain rule to both parts.✓ Proved
- \[ = \frac{\left(2 \sin{\left(2 x - 1 \right)} - 2 \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} + \frac{\left(2 \sin{\left(2 x - 1 \right)} + 2 \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{4} \]trig algebraDifferentiate the sine and cosine terms. Distribute the negative sign.✓ Proved
- \[ = e^{2 x - 1} \sin{\left(2 x - 1 \right)} \]algebra simplify simplifyDistribute the exponential term. Combine like terms. Simplify the final expression.✓ Proved
Answer \( e^{2 x - 1} \sin{\left(2 x - 1 \right)} \)
✓ 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.
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 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: fail (style) — Step 3 is labeled "chain" but it applies the product rule (and inside it the chain rule). The correct label for that step should be "product" (or "sum" if splitting the derivative).qwen3.6:27b-mlx: fail (style) — Step 3 applies the chain rule to both factors of the product, violating the 'one rule per step' constraint; the chain rule applications should be separated into distinct steps or labeled more granularly.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-09-28 — Step 3 applies the chain rule to both factors of the product, violating the 'one rule per step' constraint; the chain rule applications should be separated into distinct steps or labeled more granularly.gpt-oss:20b: fail (style) 2026-09-28 — Step 3 is labeled "chain" but it applies the product rule (and inside it the chain rule). The correct label for that step should be "product" (or "sum" if splitting the derivative).qwen3.6:27b-mlx: fail (style) 2026-09-28 — Step 3 applies the chain rule to both terms of the product simultaneously, violating the 'one change per step' constraint; it should be split into separate steps for each derivative.gpt-oss:20b: pass 2026-09-28
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 2026-09-28 with SymPy 1.14.0.