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The multivariable chain rule

Problem 10.544 · easy

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

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

Answer \( 1 \)

Lines: 2 proved, 1 not checked. 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	Not checked	—	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: fail (error) — The solution fails to evaluate the partial derivatives at the specified point (s=1, t=1) and does not compute the partials of x and y with respect to s. It simply asserts '1 = 1' without showing the necessary substitution or calculation steps.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to evaluate the partial derivatives at the specified point (s=1, t=1) and does not compute the partials of x and y with respect to s. It simply asserts '1 = 1' without showing the necessary substitution or calculation steps.
gpt-oss:20b: fail (misleading) 2026-10-12 — The solution skips the actual substitution and evaluation of the partial derivatives; it merely states "1 = 1" without showing the calculation, which could mislead a student into thinking the result follows automatically.
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to compute the partial derivatives of x and y with respect to s, nor does it evaluate the partial derivatives of z with respect to x and y at the specific point (s=1, t=1). It merely states the chain rule formula and asserts the final answer without showing the necessary intermediate calculations.

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 2026-10-12 with SymPy 1.14.0.