Derivative of \( \displaystyle - \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{5} + \frac{2 \ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \)
Problem 2.1367 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{5} + \frac{2 \ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{5} + \frac{2 \ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5}\right) \]derivativeDifferentiate the function with respect to x.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(5 x - 1 \right)} + 1 \right)}}{5} + \frac{2 \frac{d}{d x} \ln{\left(\tan{\left(5 x - 1 \right)} \right)}}{5} \]sumApply the sum rule for derivatives.✓ Proved
- \[ = \frac{2 \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \tan{\left(5 x - 1 \right)}} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)}{5 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]chainApply the chain rule to both logarithmic terms.✓ Proved
- \[ = \frac{2 \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \tan{\left(5 x - 1 \right)}} - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(5 x - 1 \right)}}{5 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]sumApply the sum rule to the inner derivative.✓ Proved
- \[ = \frac{2 \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \tan{\left(5 x - 1 \right)}} - \frac{2 \tan{\left(5 x - 1 \right)} \frac{d}{d x} \tan{\left(5 x - 1 \right)}}{5 \left(\tan^{2}{\left(5 x - 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = \frac{2 \sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)}} - \frac{2 \tan{\left(5 x - 1 \right)} \sec^{2}{\left(5 x - 1 \right)}}{\tan^{2}{\left(5 x - 1 \right)} + 1} \]chain algebra algebraApply the chain rule to the tangent function. Simplify the products in the numerators. Distribute the constants and simplify terms.≈ Checked numerically
- \[ = - 2 \tan{\left(5 x - 1 \right)} + \frac{2 \sec^{2}{\left(5 x - 1 \right)}}{\tan{\left(5 x - 1 \right)}} \]algebra simplifyUse the identity tan(u)**2 + 1 = sec(u)**2. Cancel common terms in the first fraction.≈ Checked numerically
- \[ = \frac{2 \tan^{2}{\left(5 x - 1 \right)} + 2}{\tan{\left(5 x - 1 \right)}} - 2 \tan{\left(5 x - 1 \right)} \]algebraRewrite sec(u)**2 as tan(u)**2 + 1.≈ Checked numerically
- \[ = \frac{2}{\tan{\left(5 x - 1 \right)}} \]algebra algebra simplifySplit the fraction into two terms. Distribute the 2 into the parentheses. Combine like terms.✓ Proved
- \[ = 2 \cot{\left(5 x - 1 \right)} \]simplifyRewrite 1/tan(u) as cot(u).✓ Proved
Answer \( \frac{2}{\tan{\left(5 x - 1 \right)}} \)
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: 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 undefined where tan(5*x - 1)**2 + 1 = 0 undefined where tan(5*x - 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 undefined where tan(5*x - 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 undefined where tan(5*x - 1) = 0 |
| 6 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (2*tan(5*x - 1)**2 - 2*sec(5*x - 1)**2 + 2)/(tan(5*x - 1)**3 + 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)**2 + 1 = 0 undefined where tan(5*x - 1) = 0 sec has poles at odd multiples of pi/2 |
| 7 | ✓ 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 undefined where tan(5*x - 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)**2 + 1 = 0 undefined where tan(5*x - 1) = 0 |
| 9 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(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 undefined where tan(5*x - 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(5*x - 1) = 0 |
| 11 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 2*(-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 |
| 12 | ✓ 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 |
| 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: passqwen3.6:27b-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 2026-10-03 — 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: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-10-03gpt-oss:20b: pass 2026-10-03
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.