Derivative of \( \displaystyle \frac{\ln{\left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)} \right)}}{5} \)
Problem 2.1010 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)} \right)}}{5} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)} \right)}}{5} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)}\right)}{5 \left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)}\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} \tan{\left(5 x + 1 \right)} + \frac{d}{d x} \sec{\left(5 x + 1 \right)}}{5 \left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)}\right)} \]sumDifferentiate the sum inside the parenthesis.✓ Proved
- \[ = \frac{5 \tan{\left(5 x + 1 \right)} \sec{\left(5 x + 1 \right)} + 5 \sec^{2}{\left(5 x + 1 \right)}}{5 \left(\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)}\right)} \]chainApply the chain rule to the tangent and secant terms.≈ Checked numerically
- \[ = \frac{\tan{\left(5 x + 1 \right)} \sec{\left(5 x + 1 \right)} + \sec^{2}{\left(5 x + 1 \right)}}{\tan{\left(5 x + 1 \right)} + \sec{\left(5 x + 1 \right)}} \]constant-multiple simplifyFactor out the constant 5. Simplify the expression by canceling the 5 and the 1/5.✓ Proved
- \[ = \sec{\left(5 x + 1 \right)} \]algebra simplifyFactor out sec(5*x + 1) from the numerator. Cancel the common term in the numerator and denominator.✓ Proved
Answer \( \frac{1}{\cos{\left(5 x + 1 \right)}} \)
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.
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(5*x + 1) + sec(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 sec has poles at odd multiples of pi/2 undefined where tan(5*x + 1) + sec(5*x + 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) + sec(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) + sec(5*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(5*x + 1) + sec(5*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(5*x + 1) + sec(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) + sec(5*x + 1) = 0 |
| 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(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 (error) — Step 5 applies the chain rule twice (to both tan and sec) in a single step, violating the rule that each step must change only one thing.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and simplifies the expression to the stated answer. 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 2026-09-27 — The solution correctly applies differentiation rules and simplifies the expression to the stated answer. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-oss:20b: fail (error) 2026-09-27 — Step 5 applies the chain rule twice (to both tan and sec) in a single step, violating the rule that each step must change only one thing.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules and algebraic simplifications in single steps. The labels used are appropriate for the operations performed.gpt-oss:20b: fail (style) 2026-09-27 — Step 5 applies the chain rule to both tan and sec terms in a single line, effectively combining two rule applications (chain for each term) without a separate sum step. This violates the one‑rule‑per‑step requirement.
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 2026-09-27 with SymPy 1.14.0.