∫Calc Practice

Derivative of \( \displaystyle \frac{\ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \)

Problem 2.1241 · hard

Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \).
  1. \[ \frac{d}{d x} \left(\frac{\ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} - \frac{\ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2}\right) \]
    derivativeStart with the derivative of the function.✓ Proved
  2. \[ = \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} - \frac{d}{d x} \frac{\ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \]
    sumApply the difference rule.✓ Proved
  3. \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(2 x - 3 \right)} + 1 \right)}}{4} - \frac{\frac{d}{d x} \ln{\left(\tan{\left(2 x - 3 \right)} \right)}}{2} \]
    constant-multipleFactor out the constants.✓ Proved
  4. \[ = - \frac{\frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} + \frac{\frac{d}{d x} \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)}{4 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]
    chainApply the chain rule to both terms.✓ Proved
  5. \[ = - \frac{\frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} + \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(2 x - 3 \right)}}{4 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]
    sumDifferentiate the sum inside the first term.✓ Proved
  6. \[ = - \frac{\frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \tan{\left(2 x - 3 \right)}} + \frac{\tan{\left(2 x - 3 \right)} \frac{d}{d x} \tan{\left(2 x - 3 \right)}}{2 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]
    powerApply the power rule to tan(2*x - 3)**2.✓ Proved
  7. \[ = - \frac{\sec^{2}{\left(2 x - 3 \right)} \frac{d}{d x} \left(2 x - 3\right)}{2 \tan{\left(2 x - 3 \right)}} + \frac{\tan{\left(2 x - 3 \right)} \sec^{2}{\left(2 x - 3 \right)} \frac{d}{d x} \left(2 x - 3\right)}{2 \left(\tan^{2}{\left(2 x - 3 \right)} + 1\right)} \]
    chainApply the chain rule to tan(2*x - 3).≈ Checked numerically
  8. \[ = - \frac{\sec^{2}{\left(2 x - 3 \right)}}{\tan{\left(2 x - 3 \right)}} + \frac{\tan{\left(2 x - 3 \right)} \sec^{2}{\left(2 x - 3 \right)}}{\tan^{2}{\left(2 x - 3 \right)} + 1} \]
    derivative algebra algebraDifferentiate the innermost function 2*x - 3. Simplify the constants and products. Distribute 1/4 and 1/2 into the terms.✓ Proved
  9. \[ = \left(- \frac{1}{\tan{\left(2 x - 3 \right)}} + \frac{\tan{\left(2 x - 3 \right)}}{\tan^{2}{\left(2 x - 3 \right)} + 1}\right) \sec^{2}{\left(2 x - 3 \right)} \]
    algebraFactor out sec(2*x - 3)**2.✓ Proved
  10. \[ = - \frac{\sec^{2}{\left(2 x - 3 \right)}}{\left(\tan^{2}{\left(2 x - 3 \right)} + 1\right) \tan{\left(2 x - 3 \right)}} \]
    algebra algebraFind a common denominator for the terms in the parentheses. Simplify the numerator.✓ Proved
  11. \[ = - \frac{\sec^{2}{\left(2 x - 3 \right)}}{\tan^{3}{\left(2 x - 3 \right)} + \tan{\left(2 x - 3 \right)}} \]
    algebraDistribute tan(2*x - 3) in the denominator.✓ Proved
  12. \[ = - \frac{1}{\left(\tan^{2}{\left(2 x - 3 \right)} + 1\right) \cos^{2}{\left(2 x - 3 \right)} \tan{\left(2 x - 3 \right)}} \]
    algebraRewrite sec(2*x - 3)**2 as 1/cos(2*x - 3)**2.✓ Proved
  13. \[ = - \frac{1}{\cos^{2}{\left(2 x - 3 \right)} \tan^{3}{\left(2 x - 3 \right)} + \cos^{2}{\left(2 x - 3 \right)} \tan{\left(2 x - 3 \right)}} \]
    algebraDistribute the cosine term in the denominator.✓ Proved
  14. \[ = - \frac{1}{\left(\tan^{2}{\left(2 x - 3 \right)} + 1\right) \cos^{2}{\left(2 x - 3 \right)} \tan{\left(2 x - 3 \right)}} \]
    algebraFactor out cos(2*x - 3)**2 from the denominator.✓ Proved
  15. \[ = - \frac{\sec^{2}{\left(2 x - 3 \right)}}{\tan^{3}{\left(2 x - 3 \right)} + \tan{\left(2 x - 3 \right)}} \]
    simplifyFinal simplification.✓ Proved
Answer \( - \frac{1}{\tan{\left(2 x - 3 \right)}} \)

Lines: 17 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
tan has poles at odd multiples of pi/2
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
tan has poles at odd multiples of pi/2
4✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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 numericallysympy 1.14.0sympy 1.14.0: simplify left (-tan(2*x - 3)**2 + sec(2*x - 3)**2 - 1)/(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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✓ Provedsympy 1.14.0simplify(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
undefined where tan(2*x - 3)**3 + tan(2*x - 3) = 0
15✓ Provedsympy 1.14.0simplify(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)**3 + tan(2*x - 3) = 0
undefined where cos(2*x - 3) = 0
undefined where tan(2*x - 3)**2 + 1 = 0
undefined where tan(2*x - 3) = 0
16✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
tan has poles at odd multiples of pi/2
undefined where cos(2*x - 3) = 0
undefined where tan(2*x - 3)**2 + 1 = 0
undefined where tan(2*x - 3) = 0
undefined where cos(2*x - 3)**2*tan(2*x - 3)**3 + cos(2*x - 3)**2*tan(2*x - 3) = 0
17✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
tan has poles at odd multiples of pi/2
undefined where cos(2*x - 3)**2*tan(2*x - 3)**3 + cos(2*x - 3)**2*tan(2*x - 3) = 0
undefined where cos(2*x - 3) = 0
undefined where tan(2*x - 3)**2 + 1 = 0
undefined where tan(2*x - 3) = 0
18✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
tan has poles at odd multiples of pi/2
undefined where cos(2*x - 3) = 0
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
undefined where tan(2*x - 3)**3 + tan(2*x - 3) = 0
answer≈ Checked numericallysympy 1.14.0sympy 1.14.0: final line against the stated answer: simplify left (tan(2*x - 3)**2 - sec(2*x - 3)**2 + 1)/(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✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: fail (error) — The final simplified derivative is not “-1/tan(2*x - 3)”. The correct derivative, as shown by the steps, is “- sec(2*x - 3)**2/(tan(2*x - 3)**3 + tan(2*x - 3))” (or equivalently “-1/(tan(2*x - 3)*(tan(2*x - 3)**2 + 1)*cos(2*x - 3)**2)”).
  • qwen3.6:27b-mlx: fail (error) — The final answer is mathematically incorrect; the derivative simplifies to -2/cot(2x-3) or -2*sin(2x-3)*cos(2x-3), not -1/tan(2x-3). The error occurs in step 15-18 where the simplification logic is flawed, specifically failing to cancel the sec^2 term with the (tan^2+1) term correctly to reach the stated answer.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-09-29 — The final answer is mathematically incorrect; the derivative simplifies to -2/cot(2x-3) or -2*sin(2x-3)*cos(2x-3), not -1/tan(2x-3). The error occurs in step 15-18 where the simplification logic is flawed, specifically failing to cancel the sec^2 term with the (tan^2+1) term correctly to reach the stated answer.
  • gpt-oss:20b: fail (error) 2026-09-29 — The final simplified derivative is not “-1/tan(2*x - 3)”. The correct derivative, as shown by the steps, is “- sec(2*x - 3)**2/(tan(2*x - 3)**3 + tan(2*x - 3))” (or equivalently “-1/(tan(2*x - 3)*(tan(2*x - 3)**2 + 1)*cos(2*x - 3)**2)”).
  • qwen3.6:27b-mlx: fail (error) 2026-09-29 — The final answer is incorrect. The solution simplifies to -sec(2x-3)^2 / (tan(2x-3)(tan(2x-3)^2+1)). Since tan^2+1 = sec^2, this simplifies to -1/tan(2x-3), but the derivative of the original function f(x) = log(sec^2(2x-3))/4 - log(tan(2x-3))/2 simplifies to 0 because log(sec^2) = -2log(cos) and the terms cancel out or rather, let's re-evaluate. f(x) = 1/4 log(tan^2+1) - 1/2 log(tan) = 1/4 log(sec^2) - 1/2 log(tan) = 1/2 log(sec) - 1/2 log(tan) = 1/2 log(1/cos) - 1/2 log(sin/cos) = -1/2 log(cos) - 1/2 log(sin) + 1/2 log(cos) = -1/2 log(sin). The derivative of -1/2 log(sin(2x-3)) is -1/2 * cot(2x-3) * 2 = -cot(2x-3) = -1/tan(2x-3). Wait, the stated answer IS -1/tan(2x-3). Let me re-read the steps. Step 18 returns to the form in Step 14. Step 14 is -sec^2 / (tan^3 + tan). This is equal to -1/tan. So the math is correct. Why did I think it was wrong? Let's check the contract. Step 4 applies chain rule to BOTH terms. This is two applications of the chain rule in one step. The contract says 'Each step changes ONE thing. A step that applies two rules at once is a defect.' Step 4 applies the chain rule to the first term AND the second term simultaneously. This is a defect.
  • gpt-oss:20b: pass 2026-09-29

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.