∫Calc Practice

Derivative of \( \displaystyle x^{x^{2}} \)

Problem 2.941 · hard

Differentiate \( \displaystyle f(x) = x^{x^{2}} \).
  1. \[ \frac{d}{d x} x^{x^{2}} \]
    rewriteRewrite the base and exponent using the exponential identity.✓ Proved
  2. \[ = \frac{d}{d x} e^{x^{2} \ln{\left(x \right)}} \]
    chainApply the chain rule to the exponential function.≈ Checked numerically
  3. \[ = e^{x^{2} \ln{\left(x \right)}} \frac{d}{d x} x^{2} \ln{\left(x \right)} \]
    productApply the product rule to the exponent.✓ Proved
  4. \[ = \left(x^{2} \frac{d}{d x} \ln{\left(x \right)} + \ln{\left(x \right)} \frac{d}{d x} x^{2}\right) e^{x^{2} \ln{\left(x \right)}} \]
    derivativeDifferentiate the two parts of the product.✓ Proved
  5. \[ = \left(2 x \ln{\left(x \right)} + x\right) e^{x^{2} \ln{\left(x \right)}} \]
    derivative simplifyCompute the derivatives of x**2 and log(x). Simplify the expression inside the parentheses.✓ Proved
  6. \[ = x^{x^{2}} \left(2 x \ln{\left(x \right)} + x\right) \]
    simplifySubstitute back the original form of the exponential term.≈ Checked numerically
Answer \( x^{x^{2}} \left(2 x \ln{\left(x \right)} + x\right) \)

Lines: 6 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left x*(x**(x**2) - exp(x**2*log(x)))*(2*log(x) + 1); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
7≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left x*(-x**(x**2) + exp(x**2*log(x)))*(2*log(x) + 1); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
answer✓ Provedsympy 1.14.0final line against the stated answer: simplify(a - b) reduced to 0
log is undefined for non-positive arguments
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated f directly and got the stated answer

Reviewers

  • gpt-oss:20b: fail (style) — Step 3 incorrectly labels the chain rule result as a product rule, and step 4 incorrectly labels a product differentiation as a simple derivative. The labels do not match the operations performed.
  • qwen3.6:27b-mlx: fail (error) — Step 1 applies the exponential identity rewrite but also attempts to differentiate, violating the 'one change per step' rule. Step 2 is labeled 'chain' but performs no differentiation; it is merely a presentation of the rewritten form, making the label incorrect for the operation shown.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — Step 1 applies the exponential identity rewrite but also attempts to differentiate, violating the 'one change per step' rule. Step 2 is labeled 'chain' but performs no differentiation; it is merely a presentation of the rewritten form, making the label incorrect for the operation shown.
  • gpt-oss:20b: fail (style) 2026-09-27 — Step 3 incorrectly labels the chain rule result as a product rule, and step 4 incorrectly labels a product differentiation as a simple derivative. The labels do not match the operations performed.
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — Step 1 is labeled 'rewrite' but performs a differentiation (Derivative(...)), which is not a presentational step. Step 2 is labeled 'chain' but the expression is still wrapped in a Derivative operator; it should be labeled 'derivative' to unfold the outer exponential derivative, or the step is logically incomplete.
  • gpt-oss:20b: fail (style) 2026-09-27 — Step 3 incorrectly labels the application of the chain rule as a product rule; the derivative of exp(u) is exp(u)*u', not a product rule on the exponent. The subsequent steps misuse the product rule terminology, leading to a labeling defect.

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.