Derivative of \( \displaystyle \left(x + 2\right)^{x + 1} \)
Problem 2.371 · medium
Differentiate \( \displaystyle f(x) = \left(x + 2\right)^{x + 1} \).
- \[ \frac{d}{d x} \left(x + 2\right)^{x + 1} \]derivative✓ Proved
- \[ = \frac{d}{d x} e^{\left(x + 1\right) \ln{\left(x + 2 \right)}} \]rewriteRewrite the base and exponent using the exponential and logarithm.≈ Checked numerically
- \[ = e^{\left(x + 1\right) \ln{\left(x + 2 \right)}} \frac{d}{d x} \left(x + 1\right) \ln{\left(x + 2 \right)} \]chainApply the chain rule.✓ Proved
- \[ = \left(\left(x + 1\right) \frac{d}{d x} \ln{\left(x + 2 \right)} + \ln{\left(x + 2 \right)} \frac{d}{d x} \left(x + 1\right)\right) e^{\left(x + 1\right) \ln{\left(x + 2 \right)}} \]productApply the product rule to the inner expression.✓ Proved
- \[ = \left(\left(x + 1\right) \frac{d}{d x} \ln{\left(x + 2 \right)} + \ln{\left(x + 2 \right)}\right) e^{\left(x + 1\right) \ln{\left(x + 2 \right)}} \]derivativeDifferentiate x + 1.✓ Proved
- \[ = \left(\frac{x + 1}{x + 2} + \ln{\left(x + 2 \right)}\right) e^{\left(x + 1\right) \ln{\left(x + 2 \right)}} \]derivative simplifyDifferentiate log(x + 2). Simplify the expression.✓ Proved
- \[ = \left(x + 2\right)^{x + 1} \left(\frac{x + 1}{x + 2} + \ln{\left(x + 2 \right)}\right) \]simplifyConvert the exponential form back to the original power form.≈ Checked numerically
Answer \( \left(x + 2\right)^{x} \left(x + \left(x + 2\right) \log{\left(x + 2 \right)} + 1\right) \)
✓ Nihil obstat Lines: 7 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 ((x + 2)**(x + 1) - exp((x + 1)*log(x + 2)))*(x + (x + 2)*log(x + 2) + 1)/(x + 2); 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 x + 2 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 2 = 0 |
| 8 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left (-(x + 2)**(x + 1) + exp((x + 1)*log(x + 2)))*(x + (x + 2)*log(x + 2) + 1)/(x + 2); numeric agreement only, at 24 of 24 sampled points log is undefined for non-positive arguments undefined where 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 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: passdeepseek-r1:70b: passqwen3.6:27b-mlx: pass
Every verdict on record (10)
qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-20gpt-oss:20b: pass 2026-09-20qwen3.6:27b-mlx: pass 2026-09-19deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19qwen3.6:27b-mlx: pass 2026-09-19 — The solution correctly applies logarithmic differentiation by rewriting the power function, applying the chain rule, and then the product rule. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.deepseek-r1:70b: pass 2026-09-19gpt-oss:20b: pass 2026-09-19
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.