Derivative of \( \displaystyle \left(x^{2} + 1\right)^{x} \)
Problem 2.396 · hard Video
Differentiate \( \displaystyle f(x) = \left(x^{2} + 1\right)^{x} \).
- \[ \frac{d}{d x} \left(x^{2} + 1\right)^{x} \]rewriteRewrite the base and exponent using the exponential and logarithm functions.✓ Proved
- \[ = \frac{d}{d x} e^{x \ln{\left(x^{2} + 1 \right)}} \]chainApply the chain rule to the exponential function.✓ Proved
- \[ = e^{x \ln{\left(x^{2} + 1 \right)}} \frac{d}{d x} x \ln{\left(x^{2} + 1 \right)} \]productApply the product rule to the exponent.✓ Proved
- \[ = \left(x \frac{d}{d x} \ln{\left(x^{2} + 1 \right)} + \ln{\left(x^{2} + 1 \right)} \frac{d}{d x} x\right) e^{x \ln{\left(x^{2} + 1 \right)}} \]derivativeDifferentiate the first part of the product.✓ Proved
- \[ = \left(x \frac{d}{d x} \ln{\left(x^{2} + 1 \right)} + \ln{\left(x^{2} + 1 \right)}\right) e^{x \ln{\left(x^{2} + 1 \right)}} \]derivativeDifferentiate the second part of the product.✓ Proved
- \[ = \left(\frac{x \frac{d}{d x} \left(x^{2} + 1\right)}{x^{2} + 1} + \ln{\left(x^{2} + 1 \right)}\right) e^{x \ln{\left(x^{2} + 1 \right)}} \]chainApply the chain rule to the logarithm.✓ Proved
- \[ = \left(\frac{2 x^{2}}{x^{2} + 1} + \ln{\left(x^{2} + 1 \right)}\right) e^{x \ln{\left(x^{2} + 1 \right)}} \]derivative algebraDifferentiate the inner function of the logarithm. Simplify the expression inside the parentheses.✓ Proved
- \[ = \left(x^{2} + 1\right)^{x} \left(\frac{2 x^{2}}{x^{2} + 1} + \ln{\left(x^{2} + 1 \right)}\right) \]rewrite simplifyConvert the exponential form back to the original power form. Final simplification.✓ Proved
Answer \( \left(x^{2} + 1\right)^{x} \left(\frac{2 x^{2}}{x^{2} + 1} + \log{\left(x^{2} + 1 \right)}\right) \)
✓ 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. 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 | ✓ 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 + 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 + 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 + 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 + 1 = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x**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 undefined where x**2 + 1 = 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated f directly and got the stated answer |
Reviewers
gpt-oss:20b: fail (style) — Step 3 incorrectly labels the application of the chain rule on the exponential as a product rule. The derivative of exp(u) is exp(u)*u', not a product rule application. The step itself is correct, but the rule name is wrong.qwen3.6:27b-mlx: pass — The solution correctly applies logarithmic differentiation, adhering to the one-change-per-step constraint and using valid rule labels from the fixed vocabulary.
Senior review claude-sonnet-5-5, 2026-10-08: pass — The derivation is correct and every line was proved by SymPy. The rule labels fit the steps, and the final answer matches the stated one.
gpt-oss:20b: dismiss — Step 2 to 3 is the chain-rule step, which is labeled chain in line 2; the step labeled product on line 3 is where the product rule is applied to the exponent, and the line 4 expression applies it, so the label is legitimate.
Every verdict on record (7)
claude-sonnet-5-5: pass 2026-10-08 — The derivation is correct and every line was proved by SymPy. The rule labels fit the steps, and the final answer matches the stated one.qwen3.6:27b-mlx: pass 2026-09-21 — The solution correctly applies logarithmic differentiation, adhering to the one-change-per-step constraint and using valid rule labels from the fixed vocabulary.gpt-oss:20b: fail (style) 2026-09-21 — Step 3 incorrectly labels the application of the chain rule on the exponential as a product rule. The derivative of exp(u) is exp(u)*u', not a product rule application. The step itself is correct, but the rule name is wrong.qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 4 is labeled 'derivative' but applies the product rule to the term x * log(x**2 + 1). The label 'derivative' is reserved for unfolding d/dx on known forms (like d/dx(x)), not for applying the product rule.gpt-oss:20b: fail (error) 2026-09-20 — Step 3 applies the product rule to the exponential function, but the correct rule is the chain rule. The step changes more than one thing (it rewrites the derivative of exp(u) as a product) and mislabels the rule.qwen3.6:27b-mlx: fail (error) 2026-09-20 — Step 4 is labeled 'derivative' but applies the product rule to the exponent. Step 5 is labeled 'derivative' but applies the chain rule to the logarithm. The labels do not match the rules applied.gpt-oss:20b: fail (style) 2026-09-20 — Step 3 incorrectly labels the application of the chain rule for the exponential as a product rule. The derivative of exp(u) is exp(u)*u', which is a chain rule step, not a product rule. This mislabeling could mislead a student about the correct rule to use.
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.