Derivative of \( \displaystyle \frac{5 \ln{\left(\ln{\left(2 x + 2 \right)} \right)}}{2} \)
Problem 2.872 · hard
Differentiate \( \displaystyle f(x) = \frac{5 \ln{\left(\ln{\left(2 x + 2 \right)} \right)}}{2} \).
- \[ \frac{d}{d x} \frac{5 \ln{\left(\ln{\left(2 x + 2 \right)} \right)}}{2} \]Start with the derivative of the function.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \ln{\left(\ln{\left(2 x + 2 \right)} \right)}}{2} \]constant-multiple rewrite algebraPull out the constant factor 5/2. Rewrite the inner log for clarity. Simplify the expression.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \ln{\left(2 x + 2 \right)}}{2 \ln{\left(2 x + 2 \right)}} \]chainApply the chain rule to the outer logarithm.✓ Proved
- \[ = \frac{5 \frac{d}{d x} \left(2 x + 2\right)}{2 \left(2 x + 2\right) \ln{\left(2 x + 2 \right)}} \]chainApply the chain rule to the inner logarithm.✓ Proved
- \[ = \frac{5}{\left(2 x + 2\right) \ln{\left(2 x + 2 \right)}} \]derivative algebra algebra simplifyDifferentiate the innermost linear function. Simplify the product. Simplify the fraction. Final simplified form.✓ Proved
Answer \( \frac{5}{\left(2 x + 2\right) \ln{\left(2 x + 2 \right)}} \)
Mind the domain. The answer is also defined on (-1, -1/2), where f(x) is not. Substituting there gives a number that is not a slope of f.
✓ Nihil obstat 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. 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 |
| 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 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2*x + 2) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(2*x + 2) = 0 undefined where 2*x + 2 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 undefined where log(2*x + 2) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 undefined where log(2*x + 2) = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 undefined where log(2*x + 2) = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x + 2 = 0 undefined where log(2*x + 2) = 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 2*x + 2 = 0 undefined where log(2*x + 2) = 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the chain rule and constant multiple rule. The intermediate steps involving rewriting log(log(2*x+2)) as log(log(exp(log(2*x+2)))) are redundant but algebraically valid and do not constitute a defect under the contract.gpt-oss:20b: pass 2026-09-26
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.