Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(2 x - 1 \right)} \right)} \)
Problem 2.968 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(2 x - 1 \right)} \right)} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(2 x - 1 \right)} \right)}\right) \]sumDifferentiate the sum of two terms.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{2}\right) + \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} \]constantPull out the constant factor.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{2} + \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} \]chainApply the chain rule to the first term.✓ Proved
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)}{2 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]sumDifferentiate the argument of the logarithm.✓ Proved
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x - 1 \right)}}{2 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]constantDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{\frac{d}{d x} \tan^{2}{\left(2 x - 1 \right)}}{2 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]constantThe derivative of a constant is zero.✓ Proved
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{\tan{\left(2 x - 1 \right)} \frac{d}{d x} \tan{\left(2 x - 1 \right)}}{\tan^{2}{\left(2 x - 1 \right)} + 1} \]powerApply the power rule to tan(2*x - 1)**2.✓ Proved
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{\tan{\left(2 x - 1 \right)} \sec^{2}{\left(2 x - 1 \right)} \frac{d}{d x} \left(2 x - 1\right)}{\tan^{2}{\left(2 x - 1 \right)} + 1} \]chainApply the chain rule to tan(2*x - 1).≈ Checked numerically
- \[ = \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} - \frac{2 \tan{\left(2 x - 1 \right)} \sec^{2}{\left(2 x - 1 \right)}}{\tan^{2}{\left(2 x - 1 \right)} + 1} \]chain algebra algebraDifferentiate the inner linear function 2*x - 1. Simplify the product of constants. Combine the fraction and the constant.✓ Proved
- \[ = - 2 \tan{\left(2 x - 1 \right)} + \frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)} \]trig algebraUse the identity tan(u)**2 + 1 = sec(u)**2. Cancel the sec(2*x - 1)**2 term.≈ Checked numerically
- \[ = - 2 \tan{\left(2 x - 1 \right)} + \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} \]chainApply the chain rule to the second term's logarithm.✓ Proved
- \[ = - 2 \tan{\left(2 x - 1 \right)} + \frac{2 \sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} \]chain algebraDifferentiate tan(2*x - 1). Simplify the product.≈ Checked numerically
- \[ = \frac{2 \tan^{2}{\left(2 x - 1 \right)} + 2}{\tan{\left(2 x - 1 \right)}} - 2 \tan{\left(2 x - 1 \right)} \]trigSubstitute sec(2*x - 1)**2 with 1 + tan(2*x - 1)**2.≈ Checked numerically
- \[ = \frac{2}{\tan{\left(2 x - 1 \right)}} \]algebra algebraDistribute the division by tan(2*x - 1). Combine like terms.✓ Proved
Answer \( \frac{2}{\tan{\left(2 x - 1 \right)}} \)
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.
✓ Nihil obstat Lines: 16 proved, 4 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
| 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 undefined where tan(2*x - 1)**2 + 1 = 0 |
| 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 |
| 6 | ✓ 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 |
| 7 | ✓ 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 |
| 8 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(-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 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 sec has poles at odd multiples of pi/2 |
| 9 | ✓ 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(2*x - 1)**2 + 1 = 0 |
| 10 | ✓ 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(2*x - 1)**2 + 1 = 0 |
| 11 | ✓ 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(2*x - 1)**2 + 1 = 0 |
| 12 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(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 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(2*x - 1)**2 + 1 = 0 |
| 13 | ✓ 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 |
| 14 | ✓ 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) = 0 |
| 15 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(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 undefined where tan(2*x - 1) = 0 sec has poles at odd multiples of pi/2 |
| 16 | ✓ 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 |
| 17 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(-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 |
| 18 | ✓ 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 |
| 19 | ✓ 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 |
| 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 (style) — Several steps apply more than one rule or use an incorrect label. For example, step 5 is a sum‑rule application but is labeled "constant"; step 2 pulls out a constant factor and should be labeled "constant‑multiple"; step 4 applies the chain rule to the logarithm but is labeled "sum". These mislabelings violate the contract’s granularity rules.qwen3.6:27b-mlx: fail (style) — Step 2 is labeled 'constant' but performs the linearity of the derivative (splitting the sum), which should be labeled 'sum'. Step 3 is labeled 'chain' but performs pulling out a constant factor, which should be labeled 'constant-multiple'. The labels do not match the operations performed.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-09-27 — Step 2 is labeled 'constant' but performs the linearity of the derivative (splitting the sum), which should be labeled 'sum'. Step 3 is labeled 'chain' but performs pulling out a constant factor, which should be labeled 'constant-multiple'. The labels do not match the operations performed.gpt-oss:20b: fail (style) 2026-09-27 — Several steps apply more than one rule or use an incorrect label. For example, step 5 is a sum‑rule application but is labeled "constant"; step 2 pulls out a constant factor and should be labeled "constant‑multiple"; step 4 applies the chain rule to the logarithm but is labeled "sum". These mislabelings violate the contract’s granularity rules.qwen3.6:27b-mlx: fail (style) 2026-09-27 — Step 2 is labeled 'constant' but performs the linearity of the derivative (splitting the sum), which should be labeled 'sum'. Step 3 is labeled 'chain' but performs pulling out a constant factor, which should be labeled 'constant-multiple'. The labels do not match the operations performed.gpt-oss:20b: fail (style) 2026-09-27 — Step 5 incorrectly labels the operation as "constant"; it is actually a "sum" rule applied to the derivative of the argument of the logarithm.
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 2026-09-27 with SymPy 1.14.0.