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

Problem 2.43 · hard

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

\[ \frac{d}{d x} \left(\frac{\ln{\left(\tan^{2}{\left(x + 1 \right)} + 1 \right)}}{2} - \ln{\left(\tan{\left(x + 1 \right)} \right)}\right) \]
sumStart with the derivative of the whole function.✓ Proved
\[ = \frac{d}{d x} \frac{\ln{\left(\tan^{2}{\left(x + 1 \right)} + 1 \right)}}{2} - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} \]
constant-multipleSeparate the terms.✓ Proved
\[ = \frac{\frac{d}{d x} \ln{\left(\tan^{2}{\left(x + 1 \right)} + 1 \right)}}{2} - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} \]
constant-multipleFactor out the constant 1/2.✓ Proved
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\frac{d}{d x} \left(\tan^{2}{\left(x + 1 \right)} + 1\right)}{2 \left(\tan^{2}{\left(x + 1 \right)} + 1\right)} \]
chainApply the chain rule to the first term.✓ Proved
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(x + 1 \right)}}{2 \left(\tan^{2}{\left(x + 1 \right)} + 1\right)} \]
sumApply the sum rule inside the derivative.✓ Proved
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\frac{d}{d x} \tan^{2}{\left(x + 1 \right)}}{2 \left(\tan^{2}{\left(x + 1 \right)} + 1\right)} \]
constantThe derivative of 1 is 0.✓ Proved
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\tan{\left(x + 1 \right)} \frac{d}{d x} \tan{\left(x + 1 \right)}}{\tan^{2}{\left(x + 1 \right)} + 1} \]
powerApply the power rule to tan(x + 1)**2.✓ Proved
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\tan{\left(x + 1 \right)} \sec^{2}{\left(x + 1 \right)} \frac{d}{d x} \left(x + 1\right)}{\tan^{2}{\left(x + 1 \right)} + 1} \]
chainApply the chain rule to tan(x + 1).≈ Checked numerically
\[ = - \frac{d}{d x} \ln{\left(\tan{\left(x + 1 \right)} \right)} + \frac{\tan{\left(x + 1 \right)} \sec^{2}{\left(x + 1 \right)}}{\tan^{2}{\left(x + 1 \right)} + 1} \]
chainThe derivative of x + 1 is 1.✓ Proved
\[ = - \frac{\frac{d}{d x} \tan{\left(x + 1 \right)}}{\tan{\left(x + 1 \right)}} + \frac{\tan{\left(x + 1 \right)} \sec^{2}{\left(x + 1 \right)}}{\tan^{2}{\left(x + 1 \right)} + 1} \]
chainApply the chain rule to the second term log(tan(x + 1)).✓ Proved
\[ = - \frac{\sec^{2}{\left(x + 1 \right)}}{\tan{\left(x + 1 \right)}} + \frac{\tan{\left(x + 1 \right)} \sec^{2}{\left(x + 1 \right)}}{\tan^{2}{\left(x + 1 \right)} + 1} \]
chain algebra algebraApply the chain rule to tan(x + 1) in the second term. Simplify the second term. Simplify the first term by canceling 2 and 1/2.≈ Checked numerically
\[ = \left(- \frac{1}{\tan{\left(x + 1 \right)}} + \frac{\tan{\left(x + 1 \right)}}{\tan^{2}{\left(x + 1 \right)} + 1}\right) \sec^{2}{\left(x + 1 \right)} \]
algebraFactor out sec(x + 1)**2.✓ Proved
\[ = - \frac{\sec^{2}{\left(x + 1 \right)}}{\left(\tan^{2}{\left(x + 1 \right)} + 1\right) \tan{\left(x + 1 \right)}} \]
algebra algebra simplifyFind a common denominator inside the parentheses. Simplify the numerator. Final simplified expression.✓ Proved

Answer \( - \frac{1}{\tan{\left(x + 1 \right)}} \)

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

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(x + 1)**2 + 1 = 0
5	✓ 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 + 1)**2 + 1 = 0
6	✓ 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 + 1)**2 + 1 = 0
7	✓ 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 + 1)**2 + 1 = 0
8	≈ Checked numerically	sympy 1.14.0	sympy 1.14.0: simplify left (tan(x + 1)2 - sec(x + 1)2 + 1)tan(x + 1)/(tan(x + 1)2 + 1); 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 + 1)*2 + 1 = 0 sec has poles at odd multiples of pi/2
9	✓ 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 sec has poles at odd multiples of pi/2 undefined where tan(x + 1)**2 + 1 = 0
10	✓ 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 sec has poles at odd multiples of pi/2 undefined where tan(x + 1)**2 + 1 = 0 undefined where tan(x + 1) = 0
11	≈ Checked numerically	sympy 1.14.0	sympy 1.14.0: simplify left (-tan(x + 1)2 + sec(x + 1)2 - 1)/tan(x + 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 + 1) = 0 undefined where tan(x + 1)**2 + 1 = 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(x + 1) = 0 undefined where tan(x + 1)**2 + 1 = 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(x + 1) = 0 undefined where tan(x + 1)**2 + 1 = 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(x + 1) = 0 undefined where tan(x + 1)**2 + 1 = 0
15	✓ 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 + 1) = 0 undefined where tan(x + 1)**2 + 1 = 0
16	✓ 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 + 1) = 0 undefined where tan(x + 1)**2 + 1 = 0
17	✓ 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 + 1) = 0 undefined where tan(x + 1)**2 + 1 = 0
answer	≈ Checked numerically	sympy 1.14.0	sympy 1.14.0: final line against the stated answer: simplify left (tan(x + 1)2 - sec(x + 1)2 + 1)/(tan(x + 1)**3 + tan(x + 1)); numeric agreement only, at 24 of 24 sampled points tan has poles at odd multiples of pi/2 undefined where tan(x + 1) = 0
answer, a second way	✓ Proved	sympy 1.14.0	SymPy differentiated f directly and got the stated answer

Reviewers

gpt-oss:20b: pass
deepseek-r1:70b: pass
qwen3.6:27b-mlx: fail (style) — Step 1 is labeled 'sum' but performs no operation; it merely restates the problem, violating the rule that each step must change one thing. Step 2 is labeled 'constant-multiple' but actually applies the sum/difference rule to separate the terms.

Every verdict on record (12)

qwen3.6:27b-mlx: fail (style) 2026-09-20 — Step 1 is labeled 'sum' but performs no operation; it merely restates the problem, violating the rule that each step must change one thing. Step 2 is labeled 'constant-multiple' but actually applies the sum/difference rule to separate the terms.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: fail (style) 2026-09-20 — Step 1 is labeled 'sum' but performs no differentiation; it merely restates the problem, violating the 'one change per step' rule. Step 2 is labeled 'constant-multiple' but performs the linearity of the derivative (splitting the difference), which should be labeled 'sum'.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: fail (error) 2026-09-19 — Step 2 applies the sum rule (differentiation of a difference) but is labeled 'constant-multiple'. Step 9 applies the derivative rule for a linear function (derivative of x+1 is 1) but is labeled 'chain'; while the chain rule is involved, the specific operation is evaluating a basic derivative, and labeling it 'chain' is imprecise compared to 'derivative' or 'simplify', but the primary error is Step 2's mislabeling.
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: pass 2026-09-19
qwen3.6:27b-mlx: fail (style) 2026-09-19 — Step 2 applies the sum rule to separate the terms but is labeled 'constant-multiple'. Step 9 computes the derivative of x+1 but is labeled 'chain' instead of 'derivative' or 'algebra'.
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: pass 2026-09-19
gpt-oss:20b: inconclusive 2026-09-17 — reviewer response could not be parsed:
deepseek-r1:70b: pass 2026-09-17

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.