Home›Calculus 1›Chain rule›Problem 2.264

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

Problem 2.264 · hard Beautiful

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

\[ \frac{d}{d x} \left(\frac{\ln{\left(\cos{\left(x \right)} - 1 \right)}}{2} - \frac{\ln{\left(\cos{\left(x \right)} + 1 \right)}}{2}\right) \]
constant-multipleStart with the derivative of the function. Factor out the constant 1/2 from both terms.✓ Proved
\[ = \frac{\frac{d}{d x} \ln{\left(\cos{\left(x \right)} - 1 \right)}}{2} - \frac{\frac{d}{d x} \ln{\left(\cos{\left(x \right)} + 1 \right)}}{2} \]
sumApply the linearity of the derivative.✓ Proved
\[ = - \frac{\frac{d}{d x} \left(\cos{\left(x \right)} + 1\right)}{2 \left(\cos{\left(x \right)} + 1\right)} + \frac{\frac{d}{d x} \left(\cos{\left(x \right)} - 1\right)}{2 \left(\cos{\left(x \right)} - 1\right)} \]
chainApply the chain rule to both logarithmic terms.✓ Proved
\[ = \frac{\sin{\left(x \right)}}{2 \left(\cos{\left(x \right)} + 1\right)} - \frac{\sin{\left(x \right)}}{2 \left(\cos{\left(x \right)} - 1\right)} \]
trig algebraDifferentiate the inner functions cos(x) - 1 and cos(x) + 1. Simplify the signs and distribute the terms.✓ Proved
\[ = \frac{\left(\frac{1}{\cos{\left(x \right)} + 1} - \frac{1}{\cos{\left(x \right)} - 1}\right) \sin{\left(x \right)}}{2} \]
algebraFactor out the common term (1/2)sin(x).✓ Proved
\[ = - \frac{\sin{\left(x \right)}}{\left(\cos{\left(x \right)} - 1\right) \left(\cos{\left(x \right)} + 1\right)} \]
algebraCombine the fractions using a common denominator.✓ Proved
\[ = - \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)} - 1} \]
algebra algebraSimplify the numerator and the denominator. Cancel the 2 and simplify the expression.✓ Proved
\[ = \frac{\sin{\left(x \right)}}{1 - \cos^{2}{\left(x \right)}} \]
algebraMultiply the numerator and denominator by -1.✓ Proved
\[ = \frac{1}{\sin{\left(x \right)}} \]
trig simplifyUse the trigonometric identity 1 - cos(x)**2 = sin(x)**2. Simplify the fraction by canceling sin(x).✓ Proved

Answer \( \frac{1}{\sin{\left(x \right)}} \)

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.

Every line of this solution was proved by the computer algebra system SymPy. 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
3	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0
5	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0
6	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0
7	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0
8	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0
9	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x) + 1 = 0 undefined where cos(x) - 1 = 0 undefined where cos(x)**2 - 1 = 0
10	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x)**2 - 1 = 0
11	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where cos(x)2 - 1 = 0 undefined where 1 - cos(x)2 = 0
12	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where 1 - cos(x)**2 = 0 undefined where sin(x) = 0
13	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 undefined where sin(x) = 0
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 undefined where sin(x) = 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: pass

Every verdict on record (12)

qwen3.6:27b-mlx: pass 2026-09-20
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: pass 2026-09-20
gpt-oss:20b: fail (style) 2026-09-20 — Step 4 applies the chain rule to both logarithmic terms in a single step, violating the rule that each step must change only one thing. It should be split into two separate chain-rule applications.
qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies differentiation rules and algebraic simplifications. Each step changes only one aspect of the expression, and the labels accurately describe the operations performed.
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: pass 2026-09-19
qwen3.6:27b-mlx: pass 2026-09-19
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: fail (error) 2026-09-19 — Step 4 applies the chain rule to both logarithmic terms in a single step, violating the rule that each step must change only one thing.
gpt-oss:20b: fail 2026-09-17 — The function f(x) = log(cos(x)-1)/2 - log(cos(x)+1)/2 is not real‑valued for any real x because cos(x)-1 ≤ 0, so the derivative is undefined. The solution ignores this domain restriction and later cancels sin(x), which would be zero at the points where the function is already undefined.
deepseek-r1:70b: fail 2026-09-17 — The solution ignores the domain issue where the function is undefined for all real x, making the derivative invalid.

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.