Derivative of \( \displaystyle \ln{\left(x + 1 \right)}^{x + 1} \)
Problem 2.841 · hard
Differentiate \( \displaystyle f(x) = \ln{\left(x + 1 \right)}^{x + 1} \).
- \[ \frac{d}{d x} \ln{\left(x + 1 \right)}^{x + 1} \]derivativeStart with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \]rewriteRewrite the base-exponent power using the exponential identity.≈ Checked numerically
- \[ = e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \frac{d}{d x} \left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)} \]chainApply the chain rule to the exponential function.✓ Proved
- \[ = \left(\left(x + 1\right) \frac{d}{d x} \ln{\left(\ln{\left(x + 1 \right)} \right)} + \ln{\left(\ln{\left(x + 1 \right)} \right)} \frac{d}{d x} \left(x + 1\right)\right) e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \]productApply the product rule to the exponent.✓ Proved
- \[ = \left(\left(x + 1\right) \frac{d}{d x} \ln{\left(\ln{\left(x + 1 \right)} \right)} + \ln{\left(\ln{\left(x + 1 \right)} \right)}\right) e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \]constantDifferentiate the linear term x + 1.✓ Proved
- \[ = \left(\frac{\left(x + 1\right) \frac{d}{d x} \ln{\left(x + 1 \right)}}{\ln{\left(x + 1 \right)}} + \ln{\left(\ln{\left(x + 1 \right)} \right)}\right) e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \]chainApply the chain rule to the log(log(x + 1)) term.✓ Proved
- \[ = \left(\ln{\left(\ln{\left(x + 1 \right)} \right)} + \frac{1}{\ln{\left(x + 1 \right)}}\right) e^{\left(x + 1\right) \ln{\left(\ln{\left(x + 1 \right)} \right)}} \]chain algebraDifferentiate the inner log(x + 1) term. Simplify the expression by canceling (x + 1).✓ Proved
- \[ = \left(\ln{\left(\ln{\left(x + 1 \right)} \right)} + \frac{1}{\ln{\left(x + 1 \right)}}\right) \ln{\left(x + 1 \right)}^{x + 1} \]simplifyConvert the exponential form back to the original power form.≈ Checked numerically
Answer \( \left(\ln{\left(x + 1 \right)} \ln{\left(\ln{\left(x + 1 \right)} \right)} + 1\right) \ln{\left(x + 1 \right)}^{x} \)
✓ Nihil obstat Lines: 8 proved, 2 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 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (log(x + 1)*log(log(x + 1)) + 1)*(-exp((x + 1)*log(log(x + 1))) + log(x + 1)**(x + 1))/log(x + 1); numeric agreement only, at 24 of 24 sampled points 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 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(x + 1) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(x + 1) = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where log(x + 1) = 0 |
| 9 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (log(x + 1)*log(log(x + 1)) + 1)*(exp((x + 1)*log(log(x + 1))) - log(x + 1)**(x + 1))/log(x + 1); numeric agreement only, at 24 of 24 sampled points log is undefined for non-positive arguments undefined where log(x + 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 |
| 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 — The solution correctly applies logarithmic differentiation by rewriting the power function as an exponential, then systematically applies the chain and product rules. Each step isolates a single rule application, and the labels are accurate.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies logarithmic differentiation by rewriting the power function as an exponential, then systematically applies the chain and product rules. Each step isolates a single rule application, and the labels are accurate.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the chain rule, product rule, and logarithmic differentiation steps. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.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.