Derivative of \( \displaystyle \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{2} \)
Problem 2.319 · hard
Differentiate \( \displaystyle f(x) = \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{2} \).
- \[ \frac{d}{d x} \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{2} \]Start with the derivative of the function.✓ Proved
- \[ = \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) \frac{d}{d x} \frac{e^{2 x - 1}}{2} + \frac{e^{2 x - 1} \frac{d}{d x} \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right)}{2} \]productApply the product rule.✓ Proved
- \[ = \frac{\left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} \frac{d}{d x} \left(2 x - 1\right)}{2} + \frac{e^{2 x - 1} \frac{d}{d x} \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right)}{2} \]chainApply the chain rule to the exponential part.✓ Proved
- \[ = \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} + \frac{e^{2 x - 1} \frac{d}{d x} \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right)}{2} \]derivative algebraDifferentiate the inner function 2*x - 1. Simplify the constant factor 2/2.✓ Proved
- \[ = \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} + \frac{\left(\frac{d}{d x} \sin{\left(2 x - 1 \right)} - \frac{d}{d x} \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{2} \]sumDistribute the derivative over the subtraction.✓ Proved
- \[ = \frac{\left(\sin{\left(2 x - 1 \right)} \frac{d}{d x} \left(2 x - 1\right) + \cos{\left(2 x - 1 \right)} \frac{d}{d x} \left(2 x - 1\right)\right) e^{2 x - 1}}{2} + \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} \]trigDifferentiate the sine and cosine terms using the chain rule.✓ Proved
- \[ = \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} + \frac{\left(2 \sin{\left(2 x - 1 \right)} + 2 \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1}}{2} \]derivative algebraDifferentiate the inner function 2*x - 1. Factor out the 2.✓ Proved
- \[ = \left(\sin{\left(2 x - 1 \right)} - \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} + \left(\sin{\left(2 x - 1 \right)} + \cos{\left(2 x - 1 \right)}\right) e^{2 x - 1} \]algebraCancel the 2 in the numerator and denominator.✓ Proved
- \[ = 2 e^{2 x - 1} \sin{\left(2 x - 1 \right)} \]algebra simplify algebraFactor out the common exponential term. Combine like terms inside the parentheses. Rearrange the terms for the final answer.✓ Proved
Answer \( 2 e^{2 x - 1} \sin{\left(2 x - 1 \right)} \)
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 |
| 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 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 13 | ✓ 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: passdeepseek-r1:70b: passqwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed.
Every verdict on record (12)
qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules and algebraic simplifications in a step-by-step manner. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: fail (error) 2026-09-19 — Step 7 applies both the trig derivative rule and the chain rule in a single step, violating the rule‑by‑step granularity requirement.qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately describe the operations performed, and the final result is correct.deepseek-r1:70b: fail (style) 2026-09-19 — Step 1 is missing a rule label.gpt-oss:20b: pass 2026-09-19gpt-oss:20b: pass 2026-09-17deepseek-r1:70b: pass 2026-09-17
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 2026-09-26 with SymPy 1.14.0.