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 2.1330 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(\frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} - \frac{d}{d x} \frac{\ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \]sumApply the sum rule.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x - 1 \right)} + 1 \right)}}{4} - \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x - 1 \right)} \right)}}{2} \]constant-multipleFactor out the constants.✓ 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 terms.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan{\left(2 x - 1 \right)}}{2 \tan{\left(2 x - 1 \right)}} + \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x - 1 \right)}}{4 \left(\tan^{2}{\left(2 x - 1 \right)} + 1\right)} \]sumDifferentiate the sum inside the first term.✓ 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.✓ 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 tangent function. Simplify the constants and products. Distribute the constants into the fractions.≈ Checked numerically
- \[ = \tan{\left(2 x - 1 \right)} - \frac{\sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} \]algebra simplifyUse the identity tan(u)**2 + 1 = sec(u)**2. Cancel out the sec(2*x - 1)**2 term.≈ Checked numerically
- \[ = \frac{\tan^{2}{\left(2 x - 1 \right)} - \sec^{2}{\left(2 x - 1 \right)}}{\tan{\left(2 x - 1 \right)}} \]algebraCombine the terms over a common denominator.✓ Proved
- \[ = - \frac{1}{\tan{\left(2 x - 1 \right)}} \]algebra simplifySubstitute sec(2*x - 1)**2 with tan(2*x - 1)**2 + 1. Simplify the numerator.≈ Checked numerically
- \[ = - \cot{\left(2 x - 1 \right)} \]simplifyUse the identity 1/tan(u) = cot(u).✓ 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.
✓ Nihil obstat Lines: 13 proved, 3 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) = 0 undefined where tan(2*x - 1)**2 + 1 = 0 |
| 5 | ✓ 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 undefined where tan(2*x - 1)**2 + 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) = 0 undefined where tan(2*x - 1)**2 + 1 = 0 |
| 7 | ≈ 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) = 0 undefined where tan(2*x - 1)**2 + 1 = 0 sec has poles at odd multiples of pi/2 |
| 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 tan(2*x - 1) = 0 undefined where tan(2*x - 1)**2 + 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(2*x - 1) = 0 undefined where tan(2*x - 1)**2 + 1 = 0 |
| 10 | ≈ 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) = 0 undefined where tan(2*x - 1)**2 + 1 = 0 |
| 11 | ✓ 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 |
| 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 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: passqwen3.6:27b-mlx: fail (style) — Step 4 applies the chain rule to both terms simultaneously, violating the one-rule-per-step constraint. Step 6 applies the power rule to the outer square and the constant rule to the inner '1' simultaneously, also violating the constraint.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-10-03 — Step 4 applies the chain rule to both terms simultaneously, violating the one-rule-per-step constraint. Step 6 applies the power rule to the outer square and the constant rule to the inner '1' simultaneously, also violating the constraint.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-09-30gpt-oss:20b: pass 2026-09-30
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 2026-10-03 with SymPy 1.14.0.