Derivative of \( \displaystyle \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \)
Problem 2.183 · hard
Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(\frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2}\right) \]sumDifferentiate the terms separately.✓ Proved
- \[ = \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} - \frac{d}{d x} \frac{\ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \]constant-multiplePull out the constant factors.✓ Proved
- \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} - \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \]logarithmicApply the derivative rule for logarithms.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} + \frac{\frac{d}{d x} \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]chainApply the chain rule to the logarithmic arguments.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} + \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x + 1 \right)}}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]sumDifferentiate the sum inside the first term.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} + \frac{\frac{d}{d x} \tan^{2}{\left(2 x + 1 \right)}}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]constantThe derivative of the constant 1 is 0.✓ Proved
- \[ = - \frac{\frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} + \frac{\tan{\left(2 x + 1 \right)} \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = - \frac{\sec^{2}{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right)}{2 \tan{\left(2 x + 1 \right)}} + \frac{\tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)} \frac{d}{d x} \left(2 x + 1\right)}{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]trigApply the derivative rule for tangent.≈ Checked numerically
- \[ = - \frac{\sec^{2}{\left(2 x + 1 \right)}}{\tan{\left(2 x + 1 \right)}} + \frac{\tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1} \]chain algebra algebraDifferentiate the innermost linear function 2*x + 1. Simplify the products and quotients in each term. Simplify the coefficients 1/4 * 4 and 1/2 * 2.✓ Proved
- \[ = \left(- \frac{1}{\tan{\left(2 x + 1 \right)}} + \frac{\tan{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1}\right) \sec^{2}{\left(2 x + 1 \right)} \]algebraFactor out the common secant squared term.✓ Proved
- \[ = - \frac{\sec^{2}{\left(2 x + 1 \right)}}{\left(\tan^{2}{\left(2 x + 1 \right)} + 1\right) \tan{\left(2 x + 1 \right)}} \]algebra algebra simplify simplifyFind a common denominator for the terms in the parentheses. Distribute the negative sign in the numerator. Cancel the tan(2*x + 1)**2 terms in the numerator. Remove the unnecessary parentheses.✓ Proved
- \[ = - \frac{\sec^{2}{\left(2 x + 1 \right)}}{\tan^{3}{\left(2 x + 1 \right)} + \tan{\left(2 x + 1 \right)}} \]simplifyExpand the denominator.✓ Proved
Answer \( - \frac{1}{\tan{\left(2 x + 1 \right)}} \)
Lines: 16 proved, 2 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 |
| 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 |
| 4 | ✓ 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 undefined where tan(2*x + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(2*x + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 8 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-tan(2*x + 1)**2 + sec(2*x + 1)**2 - 1)/(tan(2*x + 1)**3 + tan(2*x + 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(2*x + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 sec has poles at odd multiples of pi/2 |
| 9 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 10 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 11 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 12 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 13 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 14 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 15 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 16 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 |
| 17 | ✓ 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 + 1) = 0 undefined where tan(2*x + 1)**2 + 1 = 0 undefined where tan(2*x + 1)**3 + tan(2*x + 1) = 0 |
| answer | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: final line against the stated answer: simplify left (tan(2*x + 1)**2 - sec(2*x + 1)**2 + 1)/(tan(2*x + 1)**3 + tan(2*x + 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(2*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: passdeepseek-r1:70b: passqwen3.6:27b-mlx: fail (style) — Step 2 applies the linearity of the derivative (splitting the difference) and pulling out constant factors simultaneously, violating the one-rule-per-step constraint. It should be split into a 'sum' step followed by 'constant-multiple' steps.
Every verdict on record (15)
qwen3.6:27b-mlx: fail (style) 2026-09-20 — Step 2 applies the linearity of the derivative (splitting the difference) and pulling out constant factors simultaneously, violating the one-rule-per-step constraint. It should be split into a 'sum' step followed by 'constant-multiple' steps.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: fail (error) 2026-09-20 — The final answer provided in the task description is -1/tan(2*x + 1), but the derivation correctly simplifies to -1/(tan(2*x + 1) * (tan(2*x + 1)**2 + 1)). Since tan^2 + 1 = sec^2, the correct derivative is -cos^2(2*x+1)/sin(2*x+1) or -cot(2*x+1)*cos(2*x+1), which is not equal to -cot(2*x+1) unless cos(2*x+1)=1. The stated answer is mathematically incorrect relative to the derived steps.gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: fail (error) 2026-09-19 — The final answer is mathematically incorrect; the solution fails to simplify the expression using the identity tan^2 + 1 = sec^2, leaving an extra factor of 1/sec^2 in the denominator. Additionally, Step 2 applies both the sum rule and constant-multiple rule simultaneously, violating the one-rule-per-step constraint.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: fail (error) 2026-09-19 — The solution fails to simplify the expression to the stated answer. Step 17 leaves the result as -sec(2*x + 1)**2 / (tan(2*x + 1)**3 + tan(2*x + 1)), but using the identity tan^2 + 1 = sec^2, this simplifies to -1/tan(2*x + 1). The final line does not match the stated answer.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: inconclusive 2026-09-18 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The final answer provided in the task description is -1/tan(2*x + 1), but the derivation correctly simplifies to -1/tan(2*x + 1) only if sec(2*x + 1deepseek-r1:70b: pass 2026-09-18gpt-oss:20b: pass 2026-09-18gpt-oss:20b: fail 2026-09-17 — The final derivative obtained in the solution is - sec(2*x+1)**2/(tan(2*x+1)*(tan(2*x+1)**2+1)), which simplifies to -sec(2*x+1)**2/(tan(2*x+1)**3+tan(2*x+1)). This does not equal the stated answer -1/tan(2*x+1). The solution is therefore incorrect.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.