Derivative of \( \displaystyle \frac{\ln{\left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)} \right)}}{2} \)
Problem 2.1108 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)} \right)}}{2} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)} \right)}}{2} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)}\right)}{2 \left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)}\right)} \]chainApply the chain rule for the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(2 x + 1 \right)} + \frac{d}{d x} \sec{\left(2 x + 1 \right)}}{2 \left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)}\right)} \]sumApply the sum rule to the inner expression.✓ Proved
- \[ = \frac{2 \tan{\left(2 x + 1 \right)} \sec{\left(2 x + 1 \right)} + 2 \sec^{2}{\left(2 x + 1 \right)}}{2 \left(\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)}\right)} \]chainDifferentiate the tangent and secant terms using the chain rule.≈ Checked numerically
- \[ = \frac{\tan{\left(2 x + 1 \right)} \sec{\left(2 x + 1 \right)} + \sec^{2}{\left(2 x + 1 \right)}}{\tan{\left(2 x + 1 \right)} + \sec{\left(2 x + 1 \right)}} \]constant-multipleFactor out the common constant 2.✓ Proved
- \[ = \sec{\left(2 x + 1 \right)} \]algebra simplify simplifyFactor out sec(2*x + 1) from the numerator. Cancel the common terms in the numerator and denominator. Final simplified result.✓ Proved
Answer \( \frac{1}{\cos{\left(2 x + 1 \right)}} \)
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.
Lines: 9 proved, 1 checked numerically. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
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 sec 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 sec has poles at odd multiples of pi/2 undefined where tan(2*x + 1) + sec(2*x + 1) = 0 |
| 4 | ✓ 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) + sec(2*x + 1) = 0 |
| 5 | ≈ 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) + sec(2*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(2*x + 1) + sec(2*x + 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(2*x + 1) + sec(2*x + 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(2*x + 1) + sec(2*x + 1) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 sec has poles at odd multiples of pi/2 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 sec 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 undefined where cos(2*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 (error) — Step 5 applies the chain rule to both the tangent and secant terms simultaneously, changing two parts of the expression in one step. Each step must modify only one element, so this is a defect.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-oss:20b: fail (error) 2026-09-28 — Step 5 applies the chain rule to both the tangent and secant terms simultaneously, changing two parts of the expression in one step. Each step must modify only one element, so this is a defect.qwen3.6:27b-mlx: pass 2026-09-28gpt-oss:20b: pass 2026-09-28
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 2026-09-28 with SymPy 1.14.0.