Home›Calculus 1›Derivatives of trigonometric functions›Problem 2.706

Derivative of \( \displaystyle - \frac{\ln{\left(\cot{\left(2 x + 1 \right)} + \csc{\left(2 x + 1 \right)} \right)}}{2} \)

Problem 2.706 · hard Video

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 \( \frac{1}{\sin{\left(2 x + 1 \right)}} \)

Top: the graph of f = -log(cot(2x + 1) + csc(2x + 1))/2. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = -(-2cot(2x + 1)² - 2cot(2x + 1)·csc(2x + 1) - 2)/(2(cot(2x + 1) + csc(2x + 1))).

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. 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	✓ 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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(2x + 1) + csc(2x + 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: fail (error) — Step 5 applies the trig derivative rule to both the cotangent and cosecant terms simultaneously, violating the rule‑by‑rule granularity requirement.
qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single operation, and the labels accurately reflect the rules applied (e.g., 'chain' for the logarithm and inner trig functions, 'sum' for splitting the derivative of the sum, 'algebra' for simplification).

Senior review claude-sonnet-5-5, 2026-10-12: pass — Every line is algebraically equivalent to the previous one (SymPy proved each), and the rule labels are acceptable. The derivative is correct: d/du log(csc u + cot u) = -csc u, so the result is csc(2x+1), matching the stated answer.

gpt-oss:20b: dismiss — Step 5 applies a single rule (the trig derivative rules, with the inner derivative left as Derivative(2*x+1, x)) to the two terms of a sum already split in step 4. Applying one rule across the terms of a sum is one change, and the 'trig' label is legitimate.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-12 — Every line is algebraically equivalent to the previous one (SymPy proved each), and the rule labels are acceptable. The derivative is correct: d/du log(csc u + cot u) = -csc u, so the result is csc(2x+1), matching the stated answer.
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single operation, and the labels accurately reflect the rules applied (e.g., 'chain' for the logarithm and inner trig functions, 'sum' for splitting the derivative of the sum, 'algebra' for simplification).
gpt-oss:20b: fail (error) 2026-09-26 — Step 5 applies the trig derivative rule to both the cotangent and cosecant terms simultaneously, violating the rule‑by‑rule granularity requirement.
qwen3.6:27b-mlx: pass 2026-09-21
gpt-oss:20b: pass 2026-09-21
qwen3.6:27b-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.