Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} + \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \)
Problem 2.112 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} + \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} + \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2}\right) \]derivative constant-multipleStart with the derivative of the function. Distribute the constant factors.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4}\right) + \frac{d}{d x} \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \]sumApply the sum rule for derivatives.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4}\right) + \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \]constant-multipleFactor out the constant from the second term.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)}}{2 \tan{\left(2 x - 1 \right)}} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]chainApply the chain rule to both logarithmic terms.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)}}{2 \tan{\left(2 x - 1 \right)}} - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x - 1 \right)}}{4 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]sumApply the sum rule to the inner derivative.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)}}{2 \tan{\left(2 x - 1 \right)}} - \frac{\tan{\left(2 x - 1 \right)} \frac{d}{d x} \tan{\left(2 x - 1 \right)}}{2 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = \frac{\sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} - \frac{\tan{\left(2 x - 1 \right)} \sec^{2}{\left(2 x - 1 \right)}}{\tan^{2}{\left(2 x - 1 \right)} + 1} \]chain algebra algebraApply the chain rule to the derivative of tan(2*x - 1). Simplify the products in the numerators. Simplify the fractions by combining constants.≈ Checked numerically
- \[ = - \tan{\left(2 x - 1 \right)} + \frac{\sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} \]rewrite simplifyUse the identity tan(u)**2 + 1 = sec(u)**2. Cancel the sec(2*x - 1)**2 term in the first fraction.≈ Checked numerically
- \[ = \frac{\tan^{2}{\left(2 x - 1 \right)} + 1}{\tan{\left(2 x - 1 \right)}} - \tan{\left(2 x - 1 \right)} \]rewriteRewrite sec(2*x - 1)**2 in terms of tangent.≈ Checked numerically
- \[ = \frac{1}{\tan{\left(2 x - 1 \right)}} \]algebra simplifySplit the fraction into two parts. Combine the tangent terms.✓ Proved
- \[ = \cot{\left(2 x - 1 \right)} \]simplifyUse the definition of the cotangent function.✓ Proved
Answer \( \frac{1}{\tan{\left(2 x - 1 \right)}} \)
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.
Lines: 14 proved, 3 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 | ✓ 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 |
| 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 |
| 4 | ✓ 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 |
| 5 | ✓ 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 undefined where tan(2*x - 1)**2 + 1 = 0 undefined where tan(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 undefined where tan(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 |
| 8 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(2*x - 1)**2 - sec(2*x - 1)**2 + 1)/(tan(2*x - 1)**3 + tan(2*x - 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 sec has poles at odd multiples of pi/2 |
| 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(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 |
| 10 | ✓ 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(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 |
| 11 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(2*x - 1)**2 - sec(2*x - 1)**2 + 1)*tan(2*x - 1)/(tan(2*x - 1)**2 + 1); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2*x - 1)**2 + 1 = 0 undefined where tan(2*x - 1) = 0 |
| 12 | ✓ 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(2*x - 1) = 0 |
| 13 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-tan(2*x - 1)**2 + sec(2*x - 1)**2 - 1)/tan(2*x - 1); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where tan(2*x - 1) = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x - 1) = 0 |
| 15 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x - 1) = 0 |
| 16 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x - 1) = 0 cot has poles at multiples of pi |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(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 chain rule to two logarithmic terms in a single step, violating the rule that each step must change only one thing. This is a multi‑rule application and should be split into two separate steps.deepseek-r1:70b: passqwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications. Each step adheres to the single-change constraint and uses valid labels from the fixed vocabulary.
Every verdict on record (15)
qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications. Each step adheres to the single-change constraint and uses valid labels from the fixed vocabulary.gpt-oss:20b: fail (error) 2026-09-20 — Step 5 applies the chain rule to two logarithmic terms in a single step, violating the rule that each step must change only one thing. This is a multi‑rule application and should be split into two separate steps.qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications. Each step adheres to the single-change constraint and uses valid labels from the fixed vocabulary.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules and algebraic simplifications. Each step adheres to the single-change constraint and uses valid labels from the fixed vocabulary.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications and trigonometric identities are applied correctly to reach the final result.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: pass 2026-09-18 — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single transformation, and the labels accurately reflect the operations performed.deepseek-r1:70b: pass 2026-09-18gpt-oss:20b: pass 2026-09-18gpt-oss:20b: inconclusive 2026-09-17 — reviewer response could not be parsed:deepseek-r1:70b: pass 2026-09-17
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.