Derivative of \( \displaystyle - \frac{\ln{\left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)} \right)}}{2} \)
Problem 2.706 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)} \right)}}{2}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)} \right)}}{2} \]constantPull out the constant factor.✓ Proved
- \[ = - \frac{\frac{d}{d x} \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)}{2 \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = - \frac{\frac{d}{d x} \cot{\left(2 x + 1 \right)} + \frac{d}{d x} \csc{\left(2 x + 1 \right)}}{2 \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = - \frac{- \cot{\left(2 x + 1 \right)} \csc{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right) - \csc^{2}{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right)}{2 \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)} \]trigApply the chain rule to the cotangent and cosecant functions.✓ Proved
- \[ = - \frac{- 2 \cot{\left(2 x + 1 \right)} \csc{\left(2 x + 1 \right)} - 2 \csc^{2}{\left(2 x + 1 \right)}}{2 \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)} \]chain constant-multipleDifferentiate the inner linear function 2*x + 1. Factor out the common constant 2.✓ Proved
- \[ = \frac{2 \cot{\left(2 x + 1 \right)} \csc{\left(2 x + 1 \right)} + 2 \csc^{2}{\left(2 x + 1 \right)}}{2 \left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}\right)} \]algebraDistribute the negative sign and simplify the product of constants.✓ Proved
- \[ = \frac{\cot{\left(2 x + 1 \right)} \csc{\left(2 x + 1 \right)} + \csc^{2}{\left(2 x + 1 \right)}}{\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)}} \]algebraSimplify the expression by canceling the 2s and the 1/2.✓ Proved
- \[ = \csc{\left(2 x + 1 \right)} \]algebra simplifyFactor out csc(2*x + 1) from the numerator. Cancel the common factor in the numerator and denominator.✓ Proved
Answer \( \frac{1}{\sin{\left(2 x + 1 \right)}} \)
Mind the domain. The answer is also defined at points where f(x) is not. Substituting there gives a number that is not a slope of f.
✓ 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 log is undefined for non-positive arguments csc has poles at multiples of pi cot has poles at multiples of pi |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(2*x + 1) + csc(2*x + 1) = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 undefined where sin(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 2026-09-21 — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single operation, and the labels accurately reflect the rules applied.gpt-oss:20b: pass 2026-09-21
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.