Home›Calculus 1›Chain rule›Problem 2.138

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

Problem 2.138 · medium

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

\[ \frac{d}{d x} \frac{\ln{\left(\ln{\left(2 x - 1 \right)} \right)}}{2} \]
constantStart with the derivative of the function.✓ Proved
\[ = \frac{\frac{d}{d x} \ln{\left(\ln{\left(2 x - 1 \right)} \right)}}{2} \]
constant-multiplePull out the constant factor.✓ Proved
\[ = \frac{\frac{d}{d x} \ln{\left(2 x - 1 \right)}}{2 \ln{\left(2 x - 1 \right)}} \]
chainApply the chain rule to the outer logarithm.✓ Proved
\[ = \frac{\frac{d}{d x} \left(2 x - 1\right)}{2 \left(2 x - 1\right) \ln{\left(2 x - 1 \right)}} \]
chainApply the chain rule to the inner logarithm.✓ Proved
\[ = \frac{1}{\left(2 x - 1\right) \ln{\left(2 x - 1 \right)}} \]
derivative simplifyDifferentiate the innermost linear expression. Simplify the expression by canceling the 2.✓ Proved
\[ = \frac{1}{2 x \ln{\left(2 x - 1 \right)} - \ln{\left(2 x - 1 \right)}} \]
algebraDistribute the term in the denominator.✓ Proved

Answer \( \frac{1}{\left(2 x - 1\right) \log{\left(2 x - 1 \right)}} \)

Mind the domain. The answer is also defined on (1/2, 1), 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 undefined where log(2*x - 1) = 0
4	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2x - 1) = 0 undefined where 2x - 1 = 0
5	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2x - 1) = 0 undefined where 2x - 1 = 0
6	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2x - 1) = 0 undefined where 2x - 1 = 0
7	✓ Proved	sympy 1.14.0	simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2x - 1) = 0 undefined where 2x - 1 = 0 undefined where 2xlog(2x - 1) - log(2x - 1) = 0
answer	✓ Proved	sympy 1.14.0	final line against the stated answer: simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2x - 1) = 0 undefined where 2x - 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: fail (error) — Step 1 incorrectly labels the rule as 'constant' instead of 'constant-multiple'.
qwen3.6:27b-mlx: pass — The solution correctly applies the constant multiple rule, chain rule, and derivative rules in a step-by-step manner. The final algebraic simplification is valid and the labels are appropriate.

Every verdict on record (15)

qwen3.6:27b-mlx: pass 2026-09-20 — The solution correctly applies the constant multiple rule, chain rule, and derivative rules in a step-by-step manner. The final algebraic simplification is valid and the labels are appropriate.
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: pass 2026-09-20
gpt-oss:20b: pass 2026-09-20
qwen3.6:27b-mlx: pass 2026-09-19
deepseek-r1:70b: fail (error) 2026-09-19 — Step 1 incorrectly labels the rule as 'constant' instead of 'constant-multiple'.
gpt-oss:20b: fail (style) 2026-09-19 — Step 1 incorrectly labels the initial derivative as a "constant" rule; the constant‑multiple rule should be applied in step 2. The rest of the steps correctly apply single rules and are algebraically valid.
qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies the chain rule and constant multiple rule in distinct steps. The final algebraic expansion is valid, though not strictly necessary, it does not constitute a defect.
deepseek-r1:70b: pass 2026-09-19
gpt-oss:20b: fail (error) 2026-09-19 — Step 1 incorrectly labels the derivative of a constant‑multiple as a ‘constant’ rule; step 3 applies the chain rule to the outer log but uses 1/log(2*x‑1) instead of 1/log(log(2*x‑1)), and step 7 distributes the denominator incorrectly, turning (2*x‑1)*log(2*x‑1) into 2*x*log(2*x‑1)‑log(2*x‑1).
qwen3.6:27b-mlx: pass 2026-09-18 — The solution correctly applies the constant multiple rule, chain rule, and derivative rules in a step-by-step manner. The final algebraic expansion is valid, though not strictly necessary for the simplest form, but it is not a defect.
deepseek-r1:70b: fail (error) 2026-09-18 — Step 7 incorrectly applies algebra by distributing, which is invalid for the product in the denominator.
gpt-oss:20b: fail (error) 2026-09-18 — Step 1 incorrectly labels the derivative of the whole function as a constant rule, and step 7 incorrectly distributes the denominator, turning a product into a difference. Both steps violate the one‑rule‑per‑step rule and produce an algebraically incorrect final expression.
gpt-oss:20b: fail 2026-09-17 — The solution omits the domain restriction for the logarithms. Since log(log(2*x-1)) requires 2*x-1>1 (i.e., x>1), the derivative is only valid for x>1, but this is not mentioned.
deepseek-r1:70b: pass 2026-09-17 — The solution correctly applies the chain rule and handles constants appropriately. The algebraic manipulation in step 7 is unnecessary but correct.

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.