Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} \)
Problem 2.1013 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = - \frac{\frac{d}{d x} \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]logarithmicApply the chain rule for the logarithm.✓ Proved
- \[ = - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x + 1 \right)}}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan^{2}{\left(2 x + 1 \right)}}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]constantThe derivative of a constant is zero.✓ Proved
- \[ = - \frac{\tan{\left(2 x + 1 \right)} \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = - \frac{\tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right)}{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]trigApply the derivative of the tangent function.≈ Checked numerically
- \[ = - \frac{\tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1} \]chain algebraDifferentiate the inner linear function (2*x + 1). Multiply the constants together.✓ Proved
- \[ = - \frac{4 \tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)}}{4 \tan^{2}{\left(2 x + 1 \right)} + 4} \]algebraCombine the fraction into a single expression.✓ Proved
- \[ = - \frac{\tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1} \]simplifyCancel the common factor of 4.✓ Proved
- \[ = - \tan{\left(2 x + 1 \right)} \]algebra simplifyUse the trigonometric identity sec(u)**2 = 1 + tan(u)**2. Simplify the fraction by canceling the common term.≈ Checked numerically
Answer \( - \tan{\left(2 x + 1 \right)} \)
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: 12 proved, 2 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 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 4 | ✓ 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 |
| 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)**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)**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)/(tan(2*x + 1)**2 + 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 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)**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)**2 + 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 4*tan(2*x + 1)**2 + 4 = 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 4*tan(2*x + 1)**2 + 4 = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 12 | ≈ 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 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan 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 tan has poles at odd multiples of pi/2 |
| 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: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately reflect the operations performed, and the final simplification is correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately reflect the operations performed, and the final simplification is correct.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately reflect the operations performed, and the final simplification is correct.gpt-oss:20b: pass 2026-09-27
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.