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The multivariable chain rule

Problem 10.391 · easy

Let \( \displaystyle z = \ln{\left(x^{2} + y^{2} + 1 \right)} \) with \( \displaystyle x = s^{2} - t^{2} \), \( \displaystyle y = 2 s t \). Find \( \displaystyle \frac{\partial z}{\partial s} \) at \( \displaystyle s = 1 \), \( \displaystyle t = 0 \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)}\\\frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)}\end{matrix}\right] = \left[\begin{matrix}\frac{2 x}{x^{2} + y^{2} + 1}\\\frac{2 y}{x^{2} + y^{2} + 1}\end{matrix}\right] \]
∂z/∂x and ∂z/∂y.✓ Proved
∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s).
Reviewed
\[ 2 \]
Substitute.✓ Proved

Answer \( 2 \)

Lines: 2 proved, 1 reviewed. 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

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	substituted first, then differenced numerically

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule and the algebraic verification confirms the result.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies the chain rule and the algebraic verification confirms the result.
gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution skips the calculation of the partial derivatives of x and y with respect to s, and fails to evaluate the partials of z with respect to x and y at the specific point (s=1, t=0) before substituting. It presents a final numerical equality without showing the intermediate values or the chain rule application steps, making it impossible to verify if the logic was sound or if it was just guessing the answer.

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 generator:structured/multivariable_chain_rule, checked 2026-10-06 with SymPy 1.14.0.