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Home›Calculus 1›Logarithmic differentiation›Problem 2.327

Derivative of \( \displaystyle \sin^{2 x + 1}{\left(2 x + 1 \right)} \)

Problem 2.327 · hard

Differentiate \( \displaystyle f(x) = \sin^{2 x + 1}{\left(2 x + 1 \right)} \).
  1. \[ \frac{d}{d x} \sin^{2 x + 1}{\left(2 x + 1 \right)} \]
    derivativeStart with the derivative of the function.✓ Proved
  2. \[ = \frac{d}{d x} e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \]
    rewriteRewrite the power using the exponential and logarithm.≈ Checked numerically
  3. \[ = e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \frac{d}{d x} \left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)} \]
    chainApply the chain rule to the exponential function.✓ Proved
  4. \[ = \left(\left(2 x + 1\right) \frac{d}{d x} \ln{\left(\sin{\left(2 x + 1 \right)} \right)} + \ln{\left(\sin{\left(2 x + 1 \right)} \right)} \frac{d}{d x} \left(2 x + 1\right)\right) e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \]
    productApply the product rule to the exponent.✓ Proved
  5. \[ = \left(\left(2 x + 1\right) \frac{d}{d x} \ln{\left(\sin{\left(2 x + 1 \right)} \right)} + 2 \ln{\left(\sin{\left(2 x + 1 \right)} \right)}\right) e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \]
    derivativeDifferentiate the linear term 2*x + 1.✓ Proved
  6. \[ = \left(\frac{\left(2 x + 1\right) \frac{d}{d x} \sin{\left(2 x + 1 \right)}}{\sin{\left(2 x + 1 \right)}} + 2 \ln{\left(\sin{\left(2 x + 1 \right)} \right)}\right) e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \]
    chainApply the chain rule to the logarithm.✓ Proved
  7. \[ = \left(\frac{2 \left(2 x + 1\right) \cos{\left(2 x + 1 \right)}}{\sin{\left(2 x + 1 \right)}} + 2 \ln{\left(\sin{\left(2 x + 1 \right)} \right)}\right) e^{\left(2 x + 1\right) \ln{\left(\sin{\left(2 x + 1 \right)} \right)}} \]
    trigDifferentiate the sine function.✓ Proved
  8. \[ = \left(\frac{\left(4 x + 2\right) \cos{\left(2 x + 1 \right)}}{\sin{\left(2 x + 1 \right)}} + 2 \ln{\left(\sin{\left(2 x + 1 \right)} \right)}\right) \sin^{2 x + 1}{\left(2 x + 1 \right)} \]
    rewriteConvert the exponential back to the power form.≈ Checked numerically
  9. \[ = \left(\left(4 x + 2\right) \cot{\left(2 x + 1 \right)} + 2 \ln{\left(\sin{\left(2 x + 1 \right)} \right)}\right) \sin^{2 x + 1}{\left(2 x + 1 \right)} \]
    simplifySimplify the expression using the cotangent identity.≈ Checked numerically
Answer \( \left(\frac{4 x + 2}{\tan{\left(2 x + 1 \right)}} + 2 \log{\left(\sin{\left(2 x + 1 \right)} \right)}\right) \sin^{2 x + 1}{\left(2 x + 1 \right)} \)

✓ Nihil obstat Lines: 7 proved, 3 checked numerically. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left 2*((2*x + 1)*cos(2*x + 1) + log(sin(2*x + 1))*sin(2*x + 1))*(-exp((2*x + 1)*log(sin(2*x + 1))) + sin(2*x + 1)**(2*x + 1))/sin(2*x + 1); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where sin(2*x + 1) = 0
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
undefined where sin(2*x + 1) = 0
8≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left 2*((2*x + 1)*cos(2*x + 1) + log(sin(2*x + 1))*sin(2*x + 1))*(exp((2*x + 1)*log(sin(2*x + 1))) - sin(2*x + 1)**(2*x + 1))/sin(2*x + 1); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
undefined where sin(2*x + 1) = 0
9≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left 2*(((2*x + 1)*cos(2*x + 1) + log(sin(2*x + 1))*sin(2*x + 1))*sin(2*x + 1)**(2*x + 1) - ((2*x + 1)*cot(2*x + 1) + log(sin(2*x + 1)))*sin(2*x + 1)**(2*x + 2))/sin(2*x + 1); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
undefined where sin(2*x + 1) = 0
cot has poles at multiples of pi
answer✓ Provedsympy 1.14.0final line against the stated answer: simplify(a - b) reduced to 0
log is undefined for non-positive arguments
tan has poles at odd multiples of pi/2
undefined where tan(2*x + 1) = 0
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: pass
  • deepseek-r1:70b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly applies logarithmic differentiation via rewriting as an exponential, and each step isolates a single rule application with appropriate labels.
Every verdict on record (13)
  • qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies logarithmic differentiation via rewriting as an exponential, and each step isolates a single rule application with appropriate labels.
  • gpt-oss:20b: pass 2026-09-20
  • qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies logarithmic differentiation by rewriting the power function as an exponential, then systematically applying the chain and product rules. Each step isolates a single rule application, and the labels accurately reflect the operations performed.
  • gpt-oss:20b: pass 2026-09-20
  • qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies logarithmic differentiation via rewriting, and each step isolates a single rule application with appropriate labels.
  • deepseek-r1:70b: pass 2026-09-19
  • gpt-oss:20b: pass 2026-09-19
  • qwen3.6:27b-mlx: pass 2026-09-19
  • deepseek-r1:70b: pass 2026-09-19
  • gpt-oss:20b: pass 2026-09-19
  • qwen3.6:27b-mlx: pass 2026-09-18
  • deepseek-r1:70b: pass 2026-09-18
  • gpt-oss:20b: pass 2026-09-18

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.

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