Derivative of \( \displaystyle \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \)
Problem 2.554 · hard
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 numerically
- \[ = e^{\ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}} \frac{d}{d x} \ln{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)} \]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 numerically
- \[ = 2 \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \left(- \ln{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{\cos{\left(2 x + 1 \right)}}{2 x + 1}\right) \]simplifyFactor out the common 2 and simplify the expression.✓ Proved
Answer \( \left(2 x + 1\right)^{\cos{\left(2 x + 1 \right)}} \left(- 2 \log{\left(2 x + 1 \right)} \sin{\left(2 x + 1 \right)} + \frac{2 \cos{\left(2 x + 1 \right)}}{2 x + 1}\right) \)
Lines: 8 proved, 2 checked numerically. 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 | ≈ Checked numerically | 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 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 numerically | 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 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
Every verdict on record (4)
qwen3.6:27b-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.