Derivative of \( \displaystyle \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{2} \)
Problem 2.637 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{2} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{2} \]derivative✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{2} \]constant-multiple✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)}{2 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]chain✓ Proved
- \[ = \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(5 x - 1 \right)}}{2 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]sum✓ Proved
- \[ = \frac{\frac{d}{d x} \tan^{2}{\left(5 x - 1 \right)}}{2 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]constant✓ Proved
- \[ = \frac{\tan{\left(5 x - 1 \right)} \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{\tan^{2}{\left(5 x - 1 \right)} + 1} \]power✓ Proved
- \[ = \frac{\tan{\left(5 x - 1 \right)} \sec^{2}{\left(5 x - 1 \right)} \frac{d}{d x} \left(5 x - 1\right)}{\tan^{2}{\left(5 x - 1 \right)} + 1} \]trig≈ Checked numerically
- \[ = \frac{5 \tan{\left(5 x - 1 \right)} \sec^{2}{\left(5 x - 1 \right)}}{\tan^{2}{\left(5 x - 1 \right)} + 1} \]chain simplify✓ Proved
- \[ = 5 \tan{\left(5 x - 1 \right)} \]algebra simplifyUse the identity sec(x)**2 = 1 + tan(x)**2.≈ Checked numerically
Answer \( 5 \tan{\left(5 x - 1 \right)} \)
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: 10 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(5*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(5*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(5*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(5*x - 1)**2 + 1 = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 5*(tan(5*x - 1)**2 - sec(5*x - 1)**2 + 1)*tan(5*x - 1)/(tan(5*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(5*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(5*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(5*x - 1)**2 + 1 = 0 |
| 10 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 5*(-tan(5*x - 1)**2 + sec(5*x - 1)**2 - 1)*tan(5*x - 1)/(tan(5*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(5*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 |
| 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, including the correct use of trigonometric identities for simplification.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels accurately reflect the operations performed, including the correct use of trigonometric identities for simplification.gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-21 — 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 mathematically sound.gpt-oss:20b: pass 2026-09-21
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.