The multivariable chain rule
Problem 10.543 · medium
Let \( \displaystyle z = x \cos{\left(y \right)} + y e^{x} \) with \( \displaystyle x = s^{2} - t^{2} \), \( \displaystyle y = 2 s t \). Find \( \displaystyle \frac{\partial z}{\partial s} \) at \( \displaystyle s = 1 \), \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x \cos{\left(y \right)} + y e^{x}\right)\\\frac{\partial}{\partial y} \left(x \cos{\left(y \right)} + y e^{x}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x} + \cos{\left(y \right)}\\- x \sin{\left(y \right)} + e^{x}\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- ∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s).
- \[ 2 \cos{\left(2 \right)} + 6 \]Substitute.✓ Proved
Answer \( 2 \cos{\left(2 \right)} + 6 \)
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 skips the calculation of the partial derivatives of x and y with respect to s, and fails to substitute the values of x and y at s=1, t=1 into the expressions for ∂z/∂x and ∂z/∂y before combining them. While the final numerical answer happens to be correct, the logical step from line 2 to line 3 is missing the intermediate evaluation of the chain rule components, making the derivation incomplete and potentially misleading if a student assumes the substitution was trivial or handled implicitly without showing the work.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution skips the calculation of the partial derivatives of x and y with respect to s, and fails to substitute the values of x and y at s=1, t=1 into the expressions for ∂z/∂x and ∂z/∂y before combining them. While the final numerical answer happens to be correct, the logical step from line 2 to line 3 is missing the intermediate evaluation of the chain rule components, making the derivation incomplete and potentially misleading if a student assumes the substitution was trivial or handled implicitly without showing the work.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution skips the calculation of the partial derivatives of x and y with respect to s, and fails to evaluate the partial derivatives of z with respect to x and y at the specific point (x=0, y=2) derived from s=1, t=1. It jumps directly to the final numerical answer without showing the intermediate steps required by the chain rule application.
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.