Derivative of \( \displaystyle \left(\left(x + 2\right)^{2} + 1\right)^{x + 2} \)
Problem 2.651 · hard
Differentiate \( \displaystyle f(x) = \left(\left(x + 2\right)^{2} + 1\right)^{x + 2} \).
- \[ \frac{d}{d x} \left(\left(x + 2\right)^{2} + 1\right)^{x + 2} \]Start with the derivative of the function.✓ Proved
- \[ = \frac{d}{d x} e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]rewriteRewrite the function using the exponential and logarithm identity.✓ Proved
- \[ = e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \frac{d}{d x} \left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)} \]chainApply the chain rule for the exponential function.✓ Proved
- \[ = \left(\left(x + 2\right) \frac{d}{d x} \ln{\left(\left(x + 2\right)^{2} + 1 \right)} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)} \frac{d}{d x} \left(x + 2\right)\right) e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]productApply the product rule to the exponent's derivative.✓ Proved
- \[ = \left(\left(x + 2\right) \frac{d}{d x} \ln{\left(\left(x + 2\right)^{2} + 1 \right)} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)}\right) e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]derivativeDifferentiate the first part of the product.✓ Proved
- \[ = \left(\frac{\left(x + 2\right) \frac{d}{d x} \left(\left(x + 2\right)^{2} + 1\right)}{\left(x + 2\right)^{2} + 1} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)}\right) e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \left(\frac{\left(x + 2\right) \left(2 x + 4\right)}{\left(x + 2\right)^{2} + 1} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)}\right) e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]chainDifferentiate the inner function of the logarithm.✓ Proved
- \[ = \left(\frac{2 \left(x + 2\right)^{2}}{\left(x + 2\right)^{2} + 1} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)}\right) e^{\left(x + 2\right) \ln{\left(\left(x + 2\right)^{2} + 1 \right)}} \]algebraSimplify the expression algebraically.✓ Proved
- \[ = \left(\frac{2 \left(x + 2\right)^{2}}{\left(x + 2\right)^{2} + 1} + \ln{\left(\left(x + 2\right)^{2} + 1 \right)}\right) \left(\left(x + 2\right)^{2} + 1\right)^{x + 2} \]simplifySubstitute back the original base for the final simplified form.✓ Proved
Answer \( \left(\left(x + 2\right)^{2} + 1\right)^{x + 1} \left(2 \left(x + 2\right)^{2} + \left(\left(x + 2\right)^{2} + 1\right) \log{\left(\left(x + 2\right)^{2} + 1 \right)}\right) \)
✓ 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 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where (x + 2)**2 + 1 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where (x + 2)**2 + 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where (x + 2)**2 + 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where (x + 2)**2 + 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
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-21gpt-oss:20b: pass 2026-09-21qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies logarithmic differentiation. Each step isolates a single rule application, and the labels accurately reflect the operations performed.gpt-oss:20b: pass 2026-09-21
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.