Derivative of \( \displaystyle \frac{3 \ln{\left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)} \right)}}{2} \)
Problem 2.360 · hard
Differentiate \( \displaystyle f(x) = \frac{3 \ln{\left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \frac{3 \ln{\left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)} \right)}}{2} \]derivativeDifferentiate the function.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \ln{\left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)} \right)}}{2} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)}\right)}{2 \left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)}\right)} \]logarithmicApply the chain rule for the logarithm.✓ Proved
- \[ = \frac{3 \left(\frac{d}{d x} \tan{\left(2 x + 2 \right)} + \frac{d}{d x} \sec{\left(2 x + 2 \right)}\right)}{2 \left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)}\right)} \]sumDifferentiate the sum inside the parenthesis.✓ Proved
- \[ = \frac{3 \left(2 \tan{\left(2 x + 2 \right)} \sec{\left(2 x + 2 \right)} + 2 \sec^{2}{\left(2 x + 2 \right)}\right)}{2 \left(\tan{\left(2 x + 2 \right)} + \sec{\left(2 x + 2 \right)}\right)} \]derivativeApply the chain rule to the tangent and secant terms.✓ Proved
- \[ = 3 \sec{\left(2 x + 2 \right)} \]algebra simplifyFactor out the common term 2*sec(2*x + 2). Cancel the common term in the numerator and denominator.✓ Proved
Answer \( \frac{3}{\cos{\left(2 x + 2 \right)}} \)
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.
Every line of this solution was proved by the computer algebra system SymPy. 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 + 2) + sec(2*x + 2) = 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 + 2) + sec(2*x + 2) = 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(2*x + 2) + sec(2*x + 2) = 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 + 2) + sec(2*x + 2) = 0 |
| 7 | ✓ 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 + 2) = 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 derivative rule twice (for tan and sec) in a single step, violating the rule that each step must change only one thing.deepseek-r1:70b: passqwen3.6:27b-mlx: pass
Every verdict on record (10)
qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: fail (error) 2026-09-20 — Step 5 applies the derivative rule twice (for tan and sec) in a single step, violating the rule that each step must change only one thing.qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies the constant multiple, logarithmic, sum, and derivative rules in sequence. The final simplification is algebraically sound and correctly labeled.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-19deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: pass 2026-09-19deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19
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.