Home›Calculus 1›Derivatives of trigonometric functions›Problem 2.468

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

Problem 2.468 · hard Video

Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)} \right)}}{2} \).

\[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)} \right)}}{2}\right) \]
derivativeStart with the derivative of the function.✓ Proved
\[ = - \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)} \right)}}{2} \]
constant-multiplePull out the constant factor.✓ Proved
\[ = - \frac{\frac{d}{d x} \left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)}\right)}{2 \left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)}\right)} \]
logarithmicApply the derivative rule for the logarithm.✓ Proved
\[ = - \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)} + \frac{d}{d x} \sec{\left(2 x - 1 \right)}}{2 \left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)}\right)} \]
sumDifferentiate the sum inside the parentheses.✓ Proved
\[ = - \frac{2 \tan{\left(2 x - 1 \right)} \sec{\left(2 x - 1 \right)} + 2 \sec^{2}{\left(2 x - 1 \right)}}{2 \left(\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)}\right)} \]
chain algebraApply the chain rule to both terms. Rearrange the terms for clarity.✓ Checked independently
\[ = \frac{- 2 \tan{\left(2 x - 1 \right)} \sec{\left(2 x - 1 \right)} - 2 \sec^{2}{\left(2 x - 1 \right)}}{2 \tan{\left(2 x - 1 \right)} + 2 \sec{\left(2 x - 1 \right)}} \]
algebraCombine the fractions.✓ Proved
\[ = \frac{- \tan{\left(2 x - 1 \right)} \sec{\left(2 x - 1 \right)} - \sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)} + \sec{\left(2 x - 1 \right)}} \]
simplifyCancel the common factor of 2.✓ Proved
\[ = - \sec{\left(2 x - 1 \right)} \]
algebra simplifyFactor out sec(2*x - 1) from the numerator. Cancel the common term in the numerator and denominator.✓ Proved

Answer \( - \frac{1}{\cos{\left(2 x - 1 \right)}} \)

Top: the graph of f = -log(tan(2x - 1) + sec(2x - 1))/2. A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = -(2tan(2x - 1)² + 2tan(2x - 1)·sec(2x - 1) + 2)/(2(tan(2x - 1) + sec(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 Lines: 10 proved, 1 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	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2
3	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2x - 1) = 0
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2x - 1) = 0
5	✓ Checked independently	claude-sonnet-5-5 script, run by sympy 1.14.0, sympy 1.14.0	sympy 1.14.0: simplify left (-tan(2x - 1)2 + sec(2x - 1)*2 - 1)/(tan(2x - 1) + sec(2x - 1)); numeric agreement only, at 24 of 24 sampled points claude-sonnet-5-5 script, run by sympy 1.14.0: tan and sec derivatives match the chain-rule forms; line 4 evaluated equals line 5 after rewriting in cos (`checks/2.468/line5-157210ad.py`) tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2*x - 1) = 0
6	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2x - 1) = 0
7	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2x - 1) = 0 undefined where 2tan(2x - 1) + 2sec(2x - 1) = 0
8	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where 2tan(2x - 1) + 2sec(2x - 1) = 0 undefined where tan(2x - 1) + sec(2x - 1) = 0
9	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2x - 1) + sec(2x - 1) = 0
10	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 sec has poles at odd multiples of pi/2
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 undefined where cos(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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.

Senior review claude-sonnet-5-5, 2026-10-09: pass — Each derivative in line 5 is correct: d/dx tan(2x-1) = 2 sec^2(2x-1) and d/dx sec(2x-1) = 2 sec(2x-1) tan(2x-1). Labels and steps are legitimate, and the final answer -sec(2x-1) matches the stated -1/cos(2x-1). The panel listed no specific objections, so none are ruled on.

Every verdict on record (7)

claude-sonnet-5-5: pass 2026-10-09 — Each derivative in line 5 is correct: d/dx tan(2x-1) = 2 sec^2(2x-1) and d/dx sec(2x-1) = 2 sec(2x-1) tan(2x-1). Labels and steps are legitimate, and the final answer -sec(2x-1) matches the stated -1/cos(2x-1). The panel listed no specific objections, so none are ruled on.
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.
gpt-oss:20b: pass 2026-09-21
qwen3.6:27b-mlx: pass 2026-09-20
gpt-oss:20b: fail (style) 2026-09-20 — Step 5 applies the chain rule to two separate terms in a single line, violating the rule that each step must change only one thing. The step should be split into two separate chain‑rule applications.
qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels accurately reflect the operations performed, and the final result is correct.
gpt-oss:20b: fail (error) 2026-09-20 — Step 5 applies the chain rule to both the tan and sec terms in a single step, changing two terms at once. Each step should modify only one component, so this step violates the granularity rule.

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.