Derivative of \( \displaystyle \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \)
Problem 2.554 · hard Video
Differentiate \( \displaystyle f(x) = \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \).
- \[ \frac{d}{d x} \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \]rewriteRewrite the base and exponent using the exponential and logarithm.✓ Checked independently
- \[ = e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \frac{d}{d x} \ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)} \]chainApply the chain rule for the exponential function.✓ Proved
- \[ = \left(\ln{\left(2 x + 1 \right)} \frac{d}{d x} \cos{\left(2 x + 1 \right)} + \cos{\left(2 x + 1 \right)} \frac{d}{d x} \ln{\left(2 x + 1 \right)}\right) e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \]productApply the product rule to the inner expression.✓ Proved
- \[ = \left(- \ln{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right) + \frac{\cos{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right)}{2 x + 1}\right) e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \]trigDifferentiate the trigonometric function.✓ Proved
- \[ = \left(- 2 \ln{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{2 \cos{\left(2 x + 1 \right)}}{2 x + 1}\right) e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \]derivative algebraDifferentiate the inner linear function. Simplify the expression algebraically.✓ Proved
- \[ = \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \left(- 2 \ln{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{2 \cos{\left(2 x + 1 \right)}}{2 x + 1}\right) \]rewriteConvert the exponential form back to the original power form.✓ Checked independently
- \[ = 2 \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \left(- \ln{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{\cos{\left(2 x + 1 \right)}}{2 x + 1}\right) \]simplifyFactor out the common 2 and simplify the expression.✓ Proved
Answer \( \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \left(- 2 \log{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{2 \cos{\left(2 x + 1 \right)}}{2 x + 1}\right) \)
✓ Nihil obstat Lines: 8 proved, 2 checked by independent solution. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Checked independently | claude-sonnet-5-5 script, run by sympy 1.14.0, sympy 1.14.0 | sympy 1.14.0: simplify left 2*((2*x + 1)**cos(2*x + 1)*(-(2*x + 1)*log(2*x + 1)*sin(2*x + 1) + cos(2*x + 1)) + ((2*x + 1)*log(2*x + 1)*sin(2*x + 1) - cos(2*x + 1))*exp(log(2*x + 1)*cos(2*x + 1)))/(2*x + 1); numeric agreement only, at 24 of 24 sampled points claude-sonnet-5-5 script, run by sympy 1.14.0: F rewritten as exp(cos*log(base)) equals line-2 inner expression, so derivatives agree ( checks/2.554/line2-33bb6570.py)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 | ✓ Checked independently | claude-sonnet-5-5 script, run by sympy 1.14.0, sympy 1.14.0 | sympy 1.14.0: simplify left 2*((2*x + 1)**cos(2*x + 1)*((2*x + 1)*log(2*x + 1)*sin(2*x + 1) - cos(2*x + 1)) + (-(2*x + 1)*log(2*x + 1)*sin(2*x + 1) + cos(2*x + 1))*exp(log(2*x + 1)*cos(2*x + 1)))/(2*x + 1); numeric agreement only, at 24 of 24 sampled points claude-sonnet-5-5 script, run by sympy 1.14.0: replacing (2x+1)**cos(2x+1) by exp(cos*log(2x+1)) in line 8 gives line 7 exactly ( checks/2.554/line8-61b5e76b.py)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 |
| 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 undefined where 2*x + 1 = 0 |
| 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
Senior review claude-sonnet-5-5, 2026-10-10: pass — The derivation is correct: rewrite as exp, chain, product, differentiate the trig and linear pieces, simplify, convert back, and factor. Every step has a legitimate rule label and the final line matches the stated answer. The two weak lines only rewrite u**c as exp(c*log(u)) or back, which is valid where 2x+1>0 (a domain matter, not a defect).
Every verdict on record (7)
claude-sonnet-5-5: pass 2026-10-10 — The derivation is correct: rewrite as exp, chain, product, differentiate the trig and linear pieces, simplify, convert back, and factor. Every step has a legitimate rule label and the final line matches the stated answer. The two weak lines only rewrite u**c as exp(c*log(u)) or back, which is valid where 2x+1>0 (a domain matter, not a defect).qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: fail (error) 2026-09-21 — Step 5 applies two derivative rules at once (derivative of the cosine and derivative of the logarithm) and labels the step only as "trig", violating the one‑rule‑per‑step rule and mislabeling the operation.
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.