Home›Calculus 1›Derivatives of trigonometric functions›Problem 2.996

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

Problem 2.996 · hard Beautiful

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

\[ \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(x - 1 \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(x - 1 \right)} \right)}\right) \]
derivativeStart with the derivative of the function.✓ Proved
\[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(x - 1 \right)} + 1 \right)}}{2}\right) + \frac{d}{d x} \ln{\left(\tan{\left(x - 1 \right)} \right)} \]
sumApply the sum rule.✓ Proved
\[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(x - 1 \right)} + 1 \right)}}{2}\right) + \frac{\frac{d}{d x} \tan{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} \]
logarithmicApply the chain rule for the second logarithm.✓ Proved
\[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(x - 1 \right)} + 1 \right)}}{2}\right) + \frac{\sec^{2}{\left(x - 1 \right)} \frac{d}{d x} \left(x - 1\right)}{\tan{\left(x - 1 \right)}} \]
trigDifferentiate the tangent function.≈ Checked numerically
\[ = \frac{d}{d x} \left(- \frac{\ln{\left(\tan^{2}{\left(x - 1 \right)} + 1 \right)}}{2}\right) + \frac{\sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} \]
derivativeDifferentiate the inner function x - 1.✓ Proved
\[ = \frac{\sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} - \frac{\frac{d}{d x} \left(\tan^{2}{\left(x - 1 \right)} + 1\right)}{2 \left(\tan^{2}{\left(x - 1 \right)} + 1\right)} \]
logarithmicApply the chain rule for the first logarithm.✓ Proved
\[ = \frac{\sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} - \frac{\frac{d}{d x} 1 + \frac{d}{d x} \tan^{2}{\left(x - 1 \right)}}{2 \left(\tan^{2}{\left(x - 1 \right)} + 1\right)} \]
sumApply the sum rule inside the derivative.✓ Proved
\[ = \frac{\sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} - \frac{\tan{\left(x - 1 \right)} \frac{d}{d x} \tan{\left(x - 1 \right)}}{\tan^{2}{\left(x - 1 \right)} + 1} \]
powerApply the power rule to tan(x - 1)**2.✓ Proved
\[ = \frac{\sec^{2}{\left(x - 1 \right)}}{\tan{\left(x - 1 \right)}} - \frac{\tan{\left(x - 1 \right)} \sec^{2}{\left(x - 1 \right)}}{\tan^{2}{\left(x - 1 \right)} + 1} \]
trig algebraDifferentiate the tangent function again. Simplify the constants and terms.≈ 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 the common sec(x - 1)**2 term.✓ Proved
\[ = \frac{\sec^{2}{\left(x - 1 \right)}}{\left(\tan^{2}{\left(x - 1 \right)} + 1\right) \tan{\left(x - 1 \right)}} \]
algebra algebra simplify algebraFind a common denominator for the terms in the parentheses. Expand the numerator. Cancel the tan(x - 1)**2 terms. Rewrite the expression as a single fraction.✓ Proved
\[ = \frac{1}{\tan{\left(x - 1 \right)}} \]
algebra algebraSubstitute tan(x - 1)**2 + 1 with sec(x - 1)**2. Cancel the sec(x - 1)**2 terms.≈ Checked numerically
\[ = \cot{\left(x - 1 \right)} \]
trigRewrite 1/tan(u) as cot(u).✓ Proved

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

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.

✓ Nihil obstat Lines: 16 proved, 3 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

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 - 1) = 0
4	≈ 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 log is undefined for non-positive arguments tan has poles at odd multiples of pi/2 undefined where tan(x - 1) = 0 sec has poles at odd multiples of pi/2
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 sec has poles at odd multiples of pi/2 undefined where tan(x - 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 sec has poles at odd multiples of pi/2 undefined where tan(x - 1) = 0 undefined where tan(x - 1)**2 + 1 = 0
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 - 1)**2 + 1 = 0 undefined where tan(x - 1) = 0
8	✓ 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)**2 + 1 = 0 undefined where tan(x - 1) = 0
9	≈ 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 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
10	✓ 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)**2 + 1 = 0 undefined where tan(x - 1) = 0
11	✓ 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)**2 + 1 = 0 undefined where tan(x - 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)**2 + 1 = 0 undefined where tan(x - 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)**2 + 1 = 0 undefined where tan(x - 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)**2 + 1 = 0 undefined where tan(x - 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)**2 + 1 = 0 undefined where tan(x - 1) = 0
16	≈ Checked numerically	sympy 1.14.0	sympy 1.14.0: 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 sec has poles at odd multiples of pi/2 undefined where tan(x - 1)2 + 1 = 0 undefined where tan(x - 1) = 0
17	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(x - 1) = 0
18	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 tan has poles at odd multiples of pi/2 undefined where tan(x - 1) = 0 cot has poles at multiples of pi
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 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
qwen3.6:27b-mlx: pass — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed, and the algebraic simplifications are valid.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed, and the algebraic simplifications are valid.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies differentiation rules step-by-step, adhering to the one-change-per-step constraint. The labels used are appropriate for the operations performed, and the final simplification is mathematically sound.
gpt-oss:20b: pass 2026-09-27

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 2026-09-27 with SymPy 1.14.0.