Derivative of \( \displaystyle - \frac{5 \ln{\left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)} \right)}}{2} \)
Problem 2.80 · hard
Differentiate \( \displaystyle f(x) = - \frac{5 \ln{\left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(- \frac{5 \ln{\left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)} \right)}}{2}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = - \frac{5 \frac{d}{d x} \ln{\left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)} \right)}}{2} \]constant-multiplePull out the constant factor.✓ Proved
- \[ = - \frac{5 \frac{d}{d x} \left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}\right)}{2 \left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}\right)} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = - \frac{5 \left(\frac{d}{d x} \cot{\left(2 x + 2 \right)} + \frac{d}{d x} \csc{\left(2 x + 2 \right)}\right)}{2 \left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}\right)} \]sumDifferentiate the sum inside the parentheses.✓ Proved
- \[ = - \frac{5 \left(- \cot{\left(2 x + 2 \right)} \csc{\left(2 x + 2 \right)} \frac{d}{d x} \left(2 x + 2\right) - \csc^{2}{\left(2 x + 2 \right)} \frac{d}{d x} \left(2 x + 2\right)\right)}{2 \left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}\right)} \]trigApply the chain rule to the trigonometric functions.✓ Proved
- \[ = - \frac{5 \left(- 2 \cot{\left(2 x + 2 \right)} \csc{\left(2 x + 2 \right)} - 2 \csc^{2}{\left(2 x + 2 \right)}\right)}{2 \left(\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}\right)} \]chain algebraEvaluate the derivative of the inner linear function. Factor out the common factor of 2.✓ Proved
- \[ = \frac{5 \cot{\left(2 x + 2 \right)} \csc{\left(2 x + 2 \right)} + 5 \csc^{2}{\left(2 x + 2 \right)}}{\cot{\left(2 x + 2 \right)} + \csc{\left(2 x + 2 \right)}} \]algebra algebraGroup the constants together. Multiply the constants (-5/2) * (-2) = 5.✓ Proved
- \[ = 5 \csc{\left(2 x + 2 \right)} \]algebra simplifyFactor out csc(2*x + 2) from the numerator. Cancel the common term in the numerator and denominator.✓ Proved
Answer \( \frac{5}{\sin{\left(2 x + 2 \right)}} \)
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.
✓ 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(2*x + 2) + csc(2*x + 2) = 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(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: passdeepseek-r1:70b: passqwen3.6:27b-mlx: pass
Every verdict on record (12)
qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression, and the labels accurately reflect the operations performed.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 2026-09-19gpt-oss:20b: fail 2026-09-17 — Step 5 applies both the chain rule and the derivative of trigonometric functions, yet it is labeled only as "trig" and combines two rules in one step, which misleads the student and violates the rule‑labeling guideline.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.