Derivative of \( \displaystyle \operatorname{atan}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)} \right)} \)
Problem 2.530 · hard Beautiful
Differentiate \( \displaystyle f(x) = \operatorname{atan}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)} \right)} \).
- \[ \frac{d}{d x} \operatorname{atan}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)} \right)} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)^{2} + 1} \]chainApply the chain rule for the arctangent function.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(3 x + 1 \right)} + \frac{d}{d x} \sec{\left(3 x + 1 \right)}}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)^{2} + 1} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{\tan{\left(3 x + 1 \right)} \sec{\left(3 x + 1 \right)} \frac{d}{d x} \left(3 x + 1\right) + \sec^{2}{\left(3 x + 1 \right)} \frac{d}{d x} \left(3 x + 1\right)}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)^{2} + 1} \]trigApply the derivatives for tangent and secant.≈ Checked numerically
- \[ = \frac{3 \tan{\left(3 x + 1 \right)} \sec{\left(3 x + 1 \right)} + 3 \sec^{2}{\left(3 x + 1 \right)}}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)^{2} + 1} \]chain constant-multipleDifferentiate the inner linear functions. Factor out the constant 3.✓ Proved
- \[ = \frac{3 \left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right) \sec{\left(3 x + 1 \right)}}{\left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right)^{2} + 1} \]algebraFactor out sec(3*x + 1) from the numerator.✓ Proved
- \[ = \frac{3 \left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right) \sec{\left(3 x + 1 \right)}}{\tan^{2}{\left(3 x + 1 \right)} + 2 \tan{\left(3 x + 1 \right)} \sec{\left(3 x + 1 \right)} + \sec^{2}{\left(3 x + 1 \right)} + 1} \]algebraExpand the square in the denominator.✓ Proved
- \[ = \frac{3 \left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right) \sec{\left(3 x + 1 \right)}}{2 \tan^{2}{\left(3 x + 1 \right)} + 2 \tan{\left(3 x + 1 \right)} \sec{\left(3 x + 1 \right)} + 2} \]algebra algebra algebraSubstitute sec(3*x + 1)**2 with 1 + tan(3*x + 1)**2. Combine like terms in the denominator. Factor out 2 from the denominator.≈ Checked numerically
- \[ = \frac{3 \left(\tan{\left(3 x + 1 \right)} + \sec{\left(3 x + 1 \right)}\right) \sec{\left(3 x + 1 \right)}}{2 \tan{\left(3 x + 1 \right)} \sec{\left(3 x + 1 \right)} + 2 \sec^{2}{\left(3 x + 1 \right)}} \]algebraSubstitute tan(3*x + 1)**2 with sec(3*x + 1)**2 - 1.≈ Checked numerically
- \[ = \frac{3}{2} \]algebra simplifyFactor out sec(3*x + 1) from the denominator. Cancel common terms in the numerator and denominator.✓ Proved
Answer \( \frac{3}{2} \)
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, 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 tan has poles at odd multiples of pi/2 sec has poles at odd multiples of pi/2 undefined where (tan(3*x + 1) + sec(3*x + 1))**2 + 1 = 0 |
| 3 | ✓ 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(3*x + 1) + sec(3*x + 1))**2 + 1 = 0 |
| 4 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 3*(tan(3*x + 1)**2 - sec(3*x + 1)**2 + 1)/((tan(3*x + 1) + sec(3*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(3*x + 1) + sec(3*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 sec has poles at odd multiples of pi/2 undefined where (tan(3*x + 1) + sec(3*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 sec has poles at odd multiples of pi/2 undefined where (tan(3*x + 1) + sec(3*x + 1))**2 + 1 = 0 |
| 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(3*x + 1) + sec(3*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(3*x + 1) + sec(3*x + 1))**2 + 1 = 0 undefined where tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + sec(3*x + 1)**2 + 1 = 0 |
| 9 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 3*(tan(3*x + 1) + sec(3*x + 1))*(tan(3*x + 1)**2 - sec(3*x + 1)**2 + 1)*sec(3*x + 1)/(2*(tan(3*x + 1)**2 + tan(3*x + 1)*sec(3*x + 1) + 1)*(tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + sec(3*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(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + sec(3*x + 1)**2 + 1 = 0 undefined where 2*tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + 2 = 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 2*tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + 2 = 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 2*tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + 2 = 0 |
| 12 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left 3*(-tan(3*x + 1)**2 + sec(3*x + 1)**2 - 1)/(2*(tan(3*x + 1)**2 + tan(3*x + 1)*sec(3*x + 1) + 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 2*tan(3*x + 1)**2 + 2*tan(3*x + 1)*sec(3*x + 1) + 2 = 0 undefined where 2*tan(3*x + 1)*sec(3*x + 1) + 2*sec(3*x + 1)**2 = 0 |
| 13 | ✓ 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 2*tan(3*x + 1)*sec(3*x + 1) + 2*sec(3*x + 1)**2 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 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 and algebraic simplifications. Each step changes only one aspect of the expression, and the labels accurately reflect the operations performed.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression, and the labels accurately reflect the operations performed.gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications. Each step isolates a single transformation, and the labels accurately reflect the operations performed.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels accurately reflect the operations performed at each stage.gpt-oss:20b: fail (style) 2026-09-20 — Step 4 applies multiple rules (trig derivative for tan and sec plus chain rule for the inner linear function) in a single line but labels it only as "trig". Each step should change only one rule, so this is a labeling defect.
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.