Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \)
Problem 2.613 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{10} + \frac{\ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5}\right) \]derivativeDifferentiate the function with respect to x.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{10}\right) + \frac{d}{d x} \frac{\ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \]sumApply the sum rule for derivatives.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{10} + \frac{\frac{d}{d x} \ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \]constant-multipleFactor out the constant coefficients.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \tan{\left(5 x - 1 \right)}} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)}{10 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]chainApply the chain rule to both logarithmic terms.✓ Proved
- \[ = \frac{\sec^{2}{\left(5 x - 1 \right)} \frac{d}{d x} \left(5 x - 1\right)}{5 \tan{\left(5 x - 1 \right)}} - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(5 x - 1 \right)}}{10 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]sumDifferentiate the sum inside the first derivative.≈ Checked numerically
- \[ = \frac{\sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)}} - \frac{\tan{\left(5 x - 1 \right)} \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = \frac{\sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)}} - \frac{\tan{\left(5 x - 1 \right)} \sec^{2}{\left(5 x - 1 \right)}}{\tan^{2}{\left(5 x - 1 \right)} + 1} \]chain algebra algebraDifferentiate the inner tangent function. Simplify the expression by multiplying constants. Distribute the constant -1/10.≈ Checked numerically
- \[ = - \tan{\left(5 x - 1 \right)} + \frac{\sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)}} \]algebra algebraUse the identity 1 + tan(u)^2 = sec(u)^2. Cancel the sec(5*x - 1)**2 term in the first fraction.≈ Checked numerically
- \[ = \frac{\tan^{2}{\left(5 x - 1 \right)} + 1}{\tan{\left(5 x - 1 \right)}} - \tan{\left(5 x - 1 \right)} \]algebraRewrite sec(u)^2 as 1 + tan(u)^2.≈ Checked numerically
- \[ = \frac{1}{\tan{\left(5 x - 1 \right)}} \]algebra algebraSplit the fraction into two parts. Combine the tangent terms.✓ Proved
- \[ = \cot{\left(5 x - 1 \right)} \]rewriteUse the definition of cotangent.✓ Proved
Answer \( \frac{1}{\tan{\left(5 x - 1 \right)}} \)
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: 12 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(5*x - 1) = 0 undefined where tan(5*x - 1)**2 + 1 = 0 |
| 5 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (tan(5*x - 1)**2 - sec(5*x - 1)**2 + 1)/tan(5*x - 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) = 0 undefined where tan(5*x - 1)**2 + 1 = 0 sec has poles at odd multiples of pi/2 |
| 6 | ✓ 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) = 0 undefined where tan(5*x - 1)**2 + 1 = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-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) = 0 undefined where tan(5*x - 1)**2 + 1 = 0 |
| 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) = 0 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) = 0 undefined where tan(5*x - 1)**2 + 1 = 0 |
| 10 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (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) = 0 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 sec has poles at odd multiples of pi/2 undefined where tan(5*x - 1) = 0 |
| 12 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-tan(5*x - 1)**2 + sec(5*x - 1)**2 - 1)/tan(5*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(5*x - 1) = 0 |
| 13 | ✓ 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) = 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(5*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(5*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(5*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) — Step 4 applies the chain rule twice (once for each logarithmic term) in a single step, violating the one‑rule‑per‑step rule.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The algebraic simplifications are valid and clearly labeled.
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 algebraic simplifications are valid and clearly labeled.gpt-oss:20b: fail (style) 2026-09-21 — Step 4 applies the chain rule twice (once for each logarithmic term) in a single step, violating the one‑rule‑per‑step rule.qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single rule application, and the labels accurately reflect the operations performed.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.