Derivative of \( \displaystyle - \frac{3 \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} + \frac{3 \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \)
Problem 2.1318 · hard
Differentiate \( \displaystyle f(x) = - \frac{3 \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} + \frac{3 \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(- \frac{3 \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} + \frac{3 \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2}\right) \]constant-multipleStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{3 \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4}\right) + \frac{d}{d x} \frac{3 \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \]sumSplit the derivative into two parts.✓ Proved
- \[ = - \frac{3 \frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x + 1 \right)} + 1 \right)}}{4} + \frac{3 \frac{d}{d x} \ln{\left(\tan{\left(2 x + 1 \right)} \right)}}{2} \]constant-multiplePull out the constant factors.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} - \frac{3 \frac{d}{d x} \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]chainApply the chain rule to the logarithmic terms.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} - \frac{3 \left(\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x + 1 \right)}\right)}{4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]sumDifferentiate the sum inside the first parenthesis.✓ Proved
- \[ = \frac{3 \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \tan{\left(2 x + 1 \right)}} - \frac{3 \tan{\left(2 x + 1 \right)} \frac{d}{d x} \tan{\left(2 x + 1 \right)}}{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)} \]powerApply the power rule to the squared tangent term.✓ Proved
- \[ = \frac{3 \sec^{2}{\left(2 x + 1 \right)}}{\tan{\left(2 x + 1 \right)}} - \frac{3 \tan{\left(2 x + 1 \right)} \sec^{2}{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1} \]chain algebra algebraApply the chain rule to the tangent term. Simplify the products and fractions. Simplify the constant multiplication.≈ Checked numerically
- \[ = 3 \left(\frac{1}{\tan{\left(2 x + 1 \right)}} - \frac{\tan{\left(2 x + 1 \right)}}{\tan^{2}{\left(2 x + 1 \right)} + 1}\right) \sec^{2}{\left(2 x + 1 \right)} \]algebraFactor out the common term.✓ Proved
- \[ = \frac{3 \sec^{2}{\left(2 x + 1 \right)}}{\left(\tan^{2}{\left(2 x + 1 \right)} + 1\right) \tan{\left(2 x + 1 \right)}} \]algebra algebraFind a common denominator for the terms in the parenthesis. Simplify the numerator of the fraction.✓ Proved
- \[ = \frac{3 \sec^{2}{\left(2 x + 1 \right)}}{\tan^{3}{\left(2 x + 1 \right)} + \tan{\left(2 x + 1 \right)}} \]simplifyDistribute the denominator and finalize the expression.✓ Proved
Answer \( \frac{3}{\tan{\left(2 x + 1 \right)}} \)
Lines: 12 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 1) = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (3*tan(2*x + 1)**2 - 3*sec(2*x + 1)**2 + 3)/(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)**2 + 1 = 0 undefined where tan(2*x + 1) = 0 sec has poles at odd multiples of pi/2 |
| 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(2*x + 1)**2 + 1 = 0 undefined where tan(2*x + 1) = 0 |
| 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 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)**2 + 1 = 0 undefined where tan(2*x + 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 (-3*tan(2*x + 1)**2 + 3*sec(2*x + 1)**2 - 3)/(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: passqwen3.6:27b-mlx: fail (error) — The final answer provided (3/tan(2*x + 1)) does not match the result of the derivation steps (3*sec(2*x + 1)**2 / (tan(2*x + 1)**3 + tan(2*x + 1))). The solution fails to simplify the derived expression to the stated answer, and the stated answer is mathematically incorrect for the given function.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — The final answer provided (3/tan(2*x + 1)) does not match the result of the derivation steps (3*sec(2*x + 1)**2 / (tan(2*x + 1)**3 + tan(2*x + 1))). The solution fails to simplify the derived expression to the stated answer, and the stated answer is mathematically incorrect for the given function.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-09-30 — The solution correctly applies differentiation rules and algebraic simplifications. The final result is algebraically equivalent to the stated answer, as sec^2(u)/(tan(u)*(tan^2(u)+1)) simplifies to 1/tan(u) using the identity tan^2(u)+1 = sec^2(u).gpt-oss:20b: pass 2026-09-30
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 2026-10-03 with SymPy 1.14.0.