∫Calc Practice

Derivative of \( \displaystyle - \frac{5 \ln{\left(\tan^{2}{\left(3 x + 1 \right)} + 1 \right)}}{6} + \frac{5 \ln{\left(\tan{\left(3 x + 1 \right)} \right)}}{3} \)

Problem 2.1120 · hard

Differentiate \( \displaystyle f(x) = - \frac{5 \ln{\left(\tan^{2}{\left(3 x + 1 \right)} + 1 \right)}}{6} + \frac{5 \ln{\left(\tan{\left(3 x + 1 \right)} \right)}}{3} \).
  1. \[ \frac{d}{d x} \left(- \frac{5 \ln{\left(\tan^{2}{\left(3 x + 1 \right)} + 1 \right)}}{6} + \frac{5 \ln{\left(\tan{\left(3 x + 1 \right)} \right)}}{3}\right) \]
    derivative constantDifferentiate the function.✓ Proved
  2. \[ = \frac{d}{d x} \left(- \frac{5 \ln{\left(\tan^{2}{\left(3 x + 1 \right)} + 1 \right)}}{6}\right) + \frac{d}{d x} \frac{5 \ln{\left(\tan{\left(3 x + 1 \right)} \right)}}{3} \]
    sum✓ Proved
  3. \[ = - \frac{5 \frac{d}{d x} \ln{\left(\tan^{2}{\left(3 x + 1 \right)} + 1 \right)}}{6} + \frac{5 \frac{d}{d x} \ln{\left(\tan{\left(3 x + 1 \right)} \right)}}{3} \]
    constant-multiple✓ Proved
  4. \[ = \frac{5 \frac{d}{d x} \tan{\left(3 x + 1 \right)}}{3 \tan{\left(3 x + 1 \right)}} - \frac{5 \frac{d}{d x} \left(\tan^{2}{\left(3 x + 1 \right)} + 1\right)}{6 \left(\tan^{2}{\left(3 x + 1 \right)} + 1\right)} \]
    chain✓ Proved
  5. \[ = \frac{5 \frac{d}{d x} \tan{\left(3 x + 1 \right)}}{3 \tan{\left(3 x + 1 \right)}} - \frac{5 \left(\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(3 x + 1 \right)}\right)}{6 \left(\tan^{2}{\left(3 x + 1 \right)} + 1\right)} \]
    sum✓ Proved
  6. \[ = \frac{5 \frac{d}{d x} \tan{\left(3 x + 1 \right)}}{3 \tan{\left(3 x + 1 \right)}} - \frac{5 \tan{\left(3 x + 1 \right)} \frac{d}{d x} \tan{\left(3 x + 1 \right)}}{3 \left(\tan^{2}{\left(3 x + 1 \right)} + 1\right)} \]
    power✓ Proved
  7. \[ = \frac{5 \sec^{2}{\left(3 x + 1 \right)}}{\tan{\left(3 x + 1 \right)}} - \frac{5 \tan{\left(3 x + 1 \right)} \sec^{2}{\left(3 x + 1 \right)}}{\tan^{2}{\left(3 x + 1 \right)} + 1} \]
    derivative algebra algebra≈ Checked numerically
  8. \[ = 5 \left(\frac{1}{\tan{\left(3 x + 1 \right)}} - \frac{\tan{\left(3 x + 1 \right)}}{\tan^{2}{\left(3 x + 1 \right)} + 1}\right) \sec^{2}{\left(3 x + 1 \right)} \]
    algebra✓ Proved
  9. \[ = \frac{5 \sec^{2}{\left(3 x + 1 \right)}}{\left(\tan^{2}{\left(3 x + 1 \right)} + 1\right) \tan{\left(3 x + 1 \right)}} \]
    algebra algebra simplify✓ Proved
Answer \( \frac{5}{\tan{\left(3 x + 1 \right)}} \)

✓ Nihil obstat Lines: 13 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

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
5✓ 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 0
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
tan has poles at odd multiples of pi/2
undefined where tan(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 0
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
tan has poles at odd multiples of pi/2
undefined where tan(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 0
8≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left (5*tan(3*x + 1)**2 - 5*sec(3*x + 1)**2 + 5)/(tan(3*x + 1)**3 + tan(3*x + 1)); numeric agreement only, at 24 of 24 sampled points
tan has poles at odd multiples of pi/2
undefined where tan(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 0
sec has poles at odd multiples of pi/2
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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 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(3*x + 1) = 0
undefined where tan(3*x + 1)**2 + 1 = 0
answer≈ Checked numericallysympy 1.14.0sympy 1.14.0: final line against the stated answer: simplify left (-5*tan(3*x + 1)**2 + 5*sec(3*x + 1)**2 - 5)/(tan(3*x + 1)**3 + tan(3*x + 1)); numeric agreement only, at 24 of 24 sampled points
tan has poles at odd multiples of pi/2
undefined where tan(3*x + 1) = 0
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — 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).
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-09-28 — 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 2026-09-28
  • qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies differentiation rules and algebraic simplifications. The final result is mathematically 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 2026-09-28

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 2026-09-28 with SymPy 1.14.0.