Derivative of \( \displaystyle - \frac{5 \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} + \frac{5 \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \)
Problem 2.1204 · hard
Differentiate \( \displaystyle f(x) = - \frac{5 \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} + \frac{5 \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \left(- \frac{5 \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} + \frac{5 \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2}\right) \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} \left(- \frac{5 \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4}\right) + \frac{d}{d x} \frac{5 \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \]sumApply the sum rule.✓ Proved
- \[ = - \frac{5 \frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} + \frac{5 \frac{d}{d x} \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \]constant-multipleFactor out the constants.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} - \frac{5 \frac{d}{d x} \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]chainApply the chain rule to both logarithmic terms.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} - \frac{5 \left(\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x - 3 \right)}\right)}{4 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]sumApply the sum rule to the first inner derivative.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} - \frac{5 \tan{\left(2 x - 3 \right)} \frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]powerApply the power rule to the tan squared term.✓ Proved
- \[ = \frac{5 \sec^{2}{\left(2 x - 3 \right)}}{\tan{\left(2 x - 3 \right)}} - \frac{5 \tan{\left(2 x - 3 \right)} \sec^{2}{\left(2 x - 3 \right)}}{\tan^{2}{\left(2 x - 3 \right)} + 1} \]chain algebra algebraApply the chain rule to the tan term. Simplify the products within the terms. Simplify the coefficients.≈ Checked numerically
- \[ = 5 \left(\frac{1}{\tan{\left(2 x - 3 \right)}} - \frac{\tan{\left(2 x - 3 \right)}}{\tan^{2}{\left(2 x - 3 \right)} + 1}\right) \sec^{2}{\left(2 x - 3 \right)} \]algebraFactor out the common secant term.✓ Proved
- \[ = \frac{5 \sec^{2}{\left(2 x - 3 \right)}}{\left(\tan^{2}{\left(2 x - 3 \right)} + 1\right) \tan{\left(2 x - 3 \right)}} \]algebra algebra simplify simplifyFind a common denominator for the terms in parentheses. Simplify the numerator. Combine the terms into a single fraction. Final simplified form.✓ Proved
Answer \( \frac{5}{\tan{\left(2 x - 3 \right)}} \)
Lines: 13 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (5*tan(2*x - 3)**2 - 5*sec(2*x - 3)**2 + 5)/(tan(2*x - 3)**3 + tan(2*x - 3)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(2*x - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 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 - 3)**2 + 1 = 0 undefined where tan(2*x - 3) = 0 |
| answer | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: final line against the stated answer: simplify left (-5*tan(2*x - 3)**2 + 5*sec(2*x - 3)**2 - 5)/(tan(2*x - 3)**3 + tan(2*x - 3)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(2*x - 3) = 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: pass — The solution correctly applies differentiation rules and algebraic simplifications. The use of the identity tan^2(u) + 1 = sec^2(u) allows the terms to cancel, leading to the correct final result. Each step adheres to the one-change-per-step constraint and uses valid labels.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies differentiation rules and algebraic simplifications. The use of the identity tan^2(u) + 1 = sec^2(u) allows the terms to cancel, leading to the correct final result. Each step adheres to the one-change-per-step constraint and uses valid labels.gpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: inconclusive 2026-09-29 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "Step 7 applies the chain rule to differentiate tan(2*x - 3), but it fails to include the derivative of the inner function (2*x - 3), which is 2. Thegpt-oss:20b: fail (error) 2026-09-29 — The final expression 5*sec(2*x - 3)**2/(tan(2*x - 3)*(tan(2*x - 3)**2 + 1)) is algebraically equivalent to 5/tan(2*x - 3), but the solution stops short of simplifying to the stated answer. Since the stated answer is 5/tan(2*x - 3), the solution does not fully match the required result.
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 2026-09-29 with SymPy 1.14.0.