Home›Calculus 1›Logarithmic differentiation›Problem 2.841

Derivative of \( \displaystyle \ln{\left(x + 1 \right)}^{x + 1} \)

Problem 2.841 · hard Video

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 \( \left(\ln{\left(x + 1 \right)} \ln{\left(\ln{\left(x + 1 \right)} \right)} + 1\right) \ln{\left(x + 1 \right)}^{x} \)

Top: the graph of f = log(x + 1)^(x + 1). A tangent line slides along it, and its slope is plotted below as it goes, tracing f' = (log(log(x + 1)) + 1/log(x + 1))·log(x + 1)^(x + 1).

✓ 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. How we verify

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: pass
qwen3.6:27b-mlx: pass — 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.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-27 — 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 2026-09-27
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 2026-09-26
qwen3.6:27b-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 2026-09-27 with SymPy 1.14.0.