The multivariable chain rule
Problem 10.546 · easy
Let \( \displaystyle z = x y^{2} + 3 x \) with \( \displaystyle x = s t \), \( \displaystyle y = s + t \). Find \( \displaystyle \frac{\partial z}{\partial s} \) at \( \displaystyle s = 2 \), \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x y^{2} + 3 x\right)\\\frac{\partial}{\partial y} \left(x y^{2} + 3 x\right)\end{matrix}\right] = \left[\begin{matrix}y^{2} + 3\\2 x y\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- ∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s).
- \[ 0 \]Substitute.✓ Proved
Answer \( 0 \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to compute the partial derivatives of x and y with respect to s, nor does it evaluate the terms at the specified point (s=2, t=0). Instead, it provides a tautology '0=0' as the substitution step, which is mathematically unjustified and incorrect.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-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 terms at the specified point (s=2, t=0). Instead, it provides a tautology '0=0' as the substitution step, which is mathematically unjustified and incorrect.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-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 chain rule expression at the given point (s=2, t=0). It merely states the formula and asserts '0=0' without showing the calculation that leads to the result.
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.