Home›Calculus 1›Derivatives of trigonometric functions›Problem 2.411

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

Problem 2.411 · hard Beautiful

Differentiate \( \displaystyle f(x) = \frac{\ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} \).

\[ \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} \]
derivativeStart with the derivative of the function.✓ Proved
\[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 2 \right)} + 1 \right)}}{2} \]
constant-multiplePull out the constant factor.✓ Proved
\[ = \frac{\frac{d}{d x} \left(\tan^{2}{\left(x + 2 \right)} + 1\right)}{2 \left(\tan^{2}{\left(x + 2 \right)} + 1\right)} \]
chainApply the chain rule for the logarithm.✓ Proved
\[ = \frac{\frac{d}{d x} \tan^{2}{\left(x + 2 \right)}}{2 \left(\tan^{2}{\left(x + 2 \right)} + 1\right)} \]
sumDifferentiate the sum term by term.✓ Proved
\[ = \frac{\tan{\left(x + 2 \right)} \frac{d}{d x} \tan{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1} \]
powerApply the power rule to the squared term.✓ Proved
\[ = \frac{\tan{\left(x + 2 \right)} \sec^{2}{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1} \]
chain simplifyDifferentiate the tangent function. Cancel common factors and simplify the expression.≈ Checked numerically
\[ = \tan{\left(x + 2 \right)} \]
algebraWait, let's simplify using the identity sec(x)^2 = 1 + tan(x)^2.≈ Checked numerically
\[ = \frac{\tan{\left(x + 2 \right)} \sec^{2}{\left(x + 2 \right)}}{\tan^{2}{\left(x + 2 \right)} + 1} \]
simplifySubstitute the identity sec(x+2)**2 = tan(x+2)**2 + 1.≈ Checked numerically
\[ = \tan{\left(x + 2 \right)} \]
simplifyThe terms cancel out to leave the final simplified result.≈ Checked numerically

Answer \( \tan{\left(x + 2 \right)} \)

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: 7 proved, 4 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 undefined where tan(x + 2)**2 + 1 = 0
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(x + 2)**2 + 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(x + 2)**2 + 1 = 0
6	≈ Checked numerically	sympy 1.14.0	sympy 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 undefined where tan(x + 2)*2 + 1 = 0 sec has poles at odd multiples of pi/2
7	✓ 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(x + 2)**2 + 1 = 0
8	≈ Checked numerically	sympy 1.14.0	sympy 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
9	≈ Checked numerically	sympy 1.14.0	sympy 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
10	≈ Checked numerically	sympy 1.14.0	sympy 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
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2
answer, a second way	✓ Proved	sympy 1.14.0	SymPy differentiated f directly and got the stated answer

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (misleading) — Steps 8 and 9 are logically incoherent and redundant. Step 8 introduces the identity `sec^2 = 1 + tan^2` but writes the expression as `tan * (1 + tan^2) / (tan^2 + 1)`, which is just a rewrite of the previous line, not a simplification. Step 9 then substitutes `sec^2` back in, undoing the potential cancellation. The correct path from Step 7 is to recognize `sec^2 / (tan^2 + 1) = 1` directly. The current steps confuse the student by oscillating between forms without progressing toward the answer.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-09-21 — Steps 8 and 9 are logically incoherent and redundant. Step 8 introduces the identity `sec^2 = 1 + tan^2` but writes the expression as `tan * (1 + tan^2) / (tan^2 + 1)`, which is just a rewrite of the previous line, not a simplification. Step 9 then substitutes `sec^2` back in, undoing the potential cancellation. The correct path from Step 7 is to recognize `sec^2 / (tan^2 + 1) = 1` directly. The current steps confuse the student by oscillating between forms without progressing toward the answer.
gpt-oss:20b: pass 2026-09-21
qwen3.6:27b-mlx: fail (misleading) 2026-09-20 — Steps 8 and 9 are logically redundant and confusing; Step 8 rewrites the denominator using an identity, and Step 9 immediately rewrites it back, adding no value and obscuring the simplification in Step 10. Additionally, Step 6 is labeled 'chain' but primarily applies the 'derivative' rule for tangent, making the label imprecise.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: fail (misleading) 2026-09-20 — Steps 8 and 9 are logically redundant and confusing; Step 8 expands the denominator using an identity, and Step 9 immediately substitutes it back, creating a circular simplification that obscures the cancellation in Step 10.
gpt-oss:20b: pass 2026-09-20

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 2026-09-26 with SymPy 1.14.0.