Derivative of \( \displaystyle \frac{\ln{\left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)} \right)}}{3} \)
Problem 2.223 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)} \right)}}{3} \).
- \[ \frac{d}{d x} \frac{\ln{\left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)} \right)}}{3} \]constant-multiplePull out the constant factor 1/3.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)} \right)}}{3} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \frac{\frac{d}{d x} \left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)}\right)}{3 \left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)}\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = \frac{\frac{d}{d x} \cot{\left(3 x + 1 \right)} + \frac{d}{d x} \csc{\left(3 x + 1 \right)}}{3 \left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)}\right)} \]chainApply the chain rule to each trigonometric term.✓ Proved
- \[ = \frac{- 3 \cot{\left(3 x + 1 \right)} \csc{\left(3 x + 1 \right)} - 3 \csc^{2}{\left(3 x + 1 \right)}}{3 \left(\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)}\right)} \]trigDifferentiate the cotangent and cosecant functions.✓ Proved
- \[ = \frac{- \cot{\left(3 x + 1 \right)} \csc{\left(3 x + 1 \right)} - \csc^{2}{\left(3 x + 1 \right)}}{\cot{\left(3 x + 1 \right)} + \csc{\left(3 x + 1 \right)}} \]algebra simplifyFactor out the common constant 3. Cancel the 3 and the 1/3.✓ Proved
- \[ = - \csc{\left(3 x + 1 \right)} \]algebra simplifyFactor out -csc(3*x + 1) from the numerator. Simplify the fraction.✓ Proved
Answer \( - \frac{1}{\sin{\left(3 x + 1 \right)}} \)
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 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. 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 csc has poles at multiples of pi cot has poles at multiples of pi |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi cot has poles at multiples of pi undefined where cot(3*x + 1) + csc(3*x + 1) = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 csc has poles at multiples of pi |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 undefined where sin(3*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 (style) — Step 4 applies both the chain rule (to each trigonometric term) and the sum rule (to combine the derivatives) in a single step, violating the one‑rule‑per‑step requirement.deepseek-r1:70b: passqwen3.6:27b-mlx: pass
Every verdict on record (12)
qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: fail (style) 2026-09-20 — Step 4 applies both the chain rule (to each trigonometric term) and the sum rule (to combine the derivatives) in a single step, violating the one‑rule‑per‑step requirement.qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: fail (style) 2026-09-20 — Step 3 incorrectly labels the application of the chain rule for the logarithm as "sum"; step 4 incorrectly labels the sum rule as "chain". These mislabelings violate the rule‑granularity requirement.qwen3.6:27b-mlx: pass 2026-09-19deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: fail (style) 2026-09-19 — Step 3 incorrectly labels the application of the chain rule as "sum"; Step 4 incorrectly labels the application of the sum rule as "chain". These mislabelings violate the rule‑granularity requirement.qwen3.6:27b-mlx: pass 2026-09-18deepseek-r1:70b: pass 2026-09-18gpt-oss:20b: fail (style) 2026-09-18 — Step 3 incorrectly labels the chain rule as a sum; Step 4 incorrectly labels the sum rule as a chain rule. Both steps apply only one rule, but the labels do not match the rule actually used.gpt-oss:20b: fail 2026-09-17 — Step 3 incorrectly labels the rule; it applies both the chain rule for the logarithm and the sum rule for the inner function, but the rule label says "sum" only, which misleads the student.deepseek-r1:70b: pass 2026-09-17
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.