Derivative of \( \displaystyle - \frac{\ln{\left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)} \right)}}{5} \)
Problem 2.1083 · hard
Differentiate \( \displaystyle f(x) = - \frac{\ln{\left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)} \right)}}{5} \).
- \[ \frac{d}{d x} \left(- \frac{\ln{\left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)} \right)}}{5}\right) \]derivativeStart with the derivative of the given function.✓ Proved
- \[ = - \frac{\frac{d}{d x} \ln{\left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)} \right)}}{5} \]constantPull out the constant factor.✓ Proved
- \[ = - \frac{\frac{d}{d x} \left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}\right)}{5 \left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = - \frac{\frac{d}{d x} \cot{\left(5 x - 1 \right)} + \frac{d}{d x} \csc{\left(5 x - 1 \right)}}{5 \left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = - \frac{- \cot{\left(5 x - 1 \right)} \csc{\left(5 x - 1 \right)} \frac{d}{d x} \left(5 x - 1\right) - \csc^{2}{\left(5 x - 1 \right)} \frac{d}{d x} \left(5 x - 1\right)}{5 \left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}\right)} \]trigApply the chain rule to the cotangent and cosecant terms.✓ Proved
- \[ = - \frac{- 5 \cot{\left(5 x - 1 \right)} \csc{\left(5 x - 1 \right)} - 5 \csc^{2}{\left(5 x - 1 \right)}}{5 \left(\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}\right)} \]chainEvaluate the derivative of the inner linear function 5*x - 1.✓ Proved
- \[ = - \frac{- \cot{\left(5 x - 1 \right)} \csc{\left(5 x - 1 \right)} - \csc^{2}{\left(5 x - 1 \right)}}{\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}} \]constant-multiple constant-multipleFactor out the common 5 from the derivative terms. Simplify the constant factor -1/5 * 5.✓ Proved
- \[ = \frac{\cot{\left(5 x - 1 \right)} \csc{\left(5 x - 1 \right)} + \csc^{2}{\left(5 x - 1 \right)}}{\cot{\left(5 x - 1 \right)} + \csc{\left(5 x - 1 \right)}} \]algebraDistribute the negative sign and multiply the terms.✓ Proved
- \[ = \csc{\left(5 x - 1 \right)} \]algebra simplifyFactor out csc(5*x - 1) from the numerator. Cancel the common factor in the numerator and denominator.✓ Proved
Answer \( \frac{1}{\sin{\left(5 x - 1 \right)}} \)
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.
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 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(5*x - 1) + csc(5*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(5*x - 1) + csc(5*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(5*x - 1) + csc(5*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(5*x - 1) + csc(5*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(5*x - 1) + csc(5*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(5*x - 1) + csc(5*x - 1) = 0 |
| 9 | ✓ 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(5*x - 1) + csc(5*x - 1) = 0 |
| 10 | ✓ 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(5*x - 1) + csc(5*x - 1) = 0 |
| 11 | ✓ 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(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 (style) — Step 5 applies the trig rule to both the cotangent and cosecant derivatives in one line, violating the rule‑by‑rule granularity requirement.qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels used are consistent with the provided vocabulary and accurately describe the operations performed.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies differentiation rules and algebraic simplifications in single-step increments. The labels used are consistent with the provided vocabulary and accurately describe the operations performed.gpt-oss:20b: fail (style) 2026-09-28 — Step 5 applies the trig rule to both the cotangent and cosecant derivatives in one line, violating the rule‑by‑rule granularity requirement.qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies differentiation rules and simplifies the expression. Each step adheres to the single-rule constraint and uses valid labels from the fixed vocabulary.gpt-oss:20b: fail (error) 2026-09-28 — Step 5 applies both the trigonometric derivative rule and the chain rule in one step, violating the one‑rule‑per‑step rule.
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.