∫Calc Practice

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

Problem 2.1226 · hard

Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \ln{\left(\tan{\left(x + 2 \right)} \right)} \).
  1. \[ \frac{d}{d x} \left(\frac{\ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \ln{\left(\tan{\left(x + 2 \right)} \right)}\right) \]
    derivativeStart with the derivative of the function.✓ Proved
  2. \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \frac{d}{d x} \ln{\left(\tan{\left(x + 2 \right)} \right)} \]
    sumApply the difference rule.✓ Proved
  3. \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \frac{\frac{d}{d x} \tan{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} \]
    logarithmicApply the chain rule for the second term.✓ Proved
  4. \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \frac{\sec^{2}{\left(x + 2 \right)} \frac{d}{d x} \left(x + 2\right)}{\tan{\left(x + 2 \right)}} \]
    trigDifferentiate the tangent function.≈ Checked numerically
  5. \[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} \]
    derivative algebraDifferentiate the inner linear function. Simplify the expression.✓ Proved
  6. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} + \frac{\frac{d}{d x} \left(\tan^{2}{\left(x + 2 \right)} + 1\right)}{2 \left(\tan^{2}{\left(x + 2 \right)} + 1\right)} \]
    logarithmicApply the chain rule to the first term.✓ Proved
  7. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} + \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(x + 2 \right)}}{2 \left(\tan^{2}{\left(x + 2 \right)} + 1\right)} \]
    derivativeDifferentiate the sum inside the derivative.✓ Proved
  8. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} + \frac{\frac{d}{d x} \tan^{2}{\left(x + 2 \right)}}{2 \left(\tan^{2}{\left(x + 2 \right)} + 1\right)} \]
    constantThe derivative of a constant is zero.✓ Proved
  9. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} + \frac{\tan{\left(x + 2 \right)} \frac{d}{d x} \tan{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1} \]
    powerApply the power rule and chain rule.✓ Proved
  10. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan{\left(x + 2 \right)}} + \frac{\tan{\left(x + 2 \right)} \sec^{2}{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1} \]
    trig algebraDifferentiate the tangent function again. Simplify the coefficients.≈ Checked numerically
  11. \[ = \left(- \frac{1}{\tan{\left(x + 2 \right)}} + \frac{\tan{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1}\right) \sec^{2}{\left(x + 2 \right)} \]
    algebraFactor out the common secant term.✓ Proved
  12. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\left(\tan^{2}{\left(x + 2 \right)} + 1\right) \tan{\left(x + 2 \right)}} \]
    algebra algebra algebraFind a common denominator. Simplify the numerator. Final simplification.✓ Proved
  13. \[ = - \frac{\sec^{2}{\left(x + 2 \right)}}{\tan^{3}{\left(x + 2 \right)} + \tan{\left(x + 2 \right)}} \]
    algebraExpand the denominator.✓ Proved
Answer \( - \frac{1}{\tan{\left(x + 2 \right)}} \)
Mind the domain. The answer is also defined at points where f(x) is not. Substituting there gives a number that is not a slope of f.

Lines: 15 proved, 3 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
undefined where tan(x + 2) = 0
4≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left (-tan(x + 2)**2 + sec(x + 2)**2 - 1)/tan(x + 2); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
tan has poles at odd multiples of pi/2
undefined where tan(x + 2) = 0
sec 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
sec has poles at odd multiples of pi/2
undefined where tan(x + 2) = 0
6✓ 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
sec has poles at odd multiples of pi/2
undefined where tan(x + 2) = 0
7✓ 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
sec has poles at odd multiples of pi/2
undefined where tan(x + 2) = 0
undefined where tan(x + 2)**2 + 1 = 0
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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 0
11≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left (tan(x + 2)**2 - sec(x + 2)**2 + 1)*tan(x + 2)/(tan(x + 2)**2 + 1); numeric agreement only, at 24 of 24 sampled points
tan has poles at odd multiples of pi/2
sec has poles at odd multiples of pi/2
undefined where tan(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 0
16✓ 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 0
17✓ 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(x + 2)**2 + 1 = 0
undefined where tan(x + 2) = 0
undefined where tan(x + 2)**3 + tan(x + 2) = 0
answer≈ Checked numericallysympy 1.14.0sympy 1.14.0: final line against the stated answer: simplify left (tan(x + 2)**2 - sec(x + 2)**2 + 1)/(tan(x + 2)**3 + tan(x + 2)); numeric agreement only, at 24 of 24 sampled points
tan has poles at odd multiples of pi/2
undefined where tan(x + 2) = 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: fail (style) — Step 10 applies both the power rule and the chain rule simultaneously, violating the one-rule-per-step constraint. Step 2 labels the application of linearity (constant multiple and difference) as 'sum', which is imprecise given the available vocabulary.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (style) 2026-09-29 — Step 10 applies both the power rule and the chain rule simultaneously, violating the one-rule-per-step constraint. Step 2 labels the application of linearity (constant multiple and difference) as 'sum', which is imprecise given the available vocabulary.
  • gpt-oss:20b: pass 2026-09-29
  • qwen3.6:27b-mlx: fail (error) 2026-09-29 — Step 10 applies both the power rule and the chain rule simultaneously, violating the one-rule-per-step constraint. Additionally, the final answer in step 17 does not match the stated answer of -1/tan(x + 2); the solution fails to simplify sec^2(u)/(tan(u)(tan^2(u)+1)) to 1/tan(u) using the identity tan^2(u)+1 = sec^2(u).
  • gpt-oss:20b: fail (error) 2026-09-29 — Step 10 applies both the power rule and the chain rule in a single step, violating the rule that each step must change only one thing. The label "power" also fails to acknowledge the chain rule applied to tan(x+2).

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.