The multivariable chain rule
Problem 10.343 · easy
Let \( \displaystyle z = x^{2} + 3 x y - 3 y^{2} \) with \( \displaystyle x = t^{2} \), \( \displaystyle y = t^{3} \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 3 x y - 3 y^{2}\right)\\\frac{\partial}{\partial y} \left(x^{2} + 3 x y - 3 y^{2}\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 3 y\\3 x - 6 y\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
- \[ 1 \]At t = 1, where (x, y) = (1, 1).✓ Proved
Answer \( 1 \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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: fail (error) — The solution skips the essential computation of dx/dt and dy/dt and jumps directly to the conclusion 1=1 at t=1, which is not justified. The chain‑rule step is correct, but the missing intermediate calculations make the reasoning invalid.qwen3.6:27b-mlx: fail (error) — The solution states the chain rule formula but fails to compute the derivatives dx/dt and dy/dt or substitute the values of x, y, and t. It jumps from the formula to the final answer '1' without showing the necessary intermediate calculation steps.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution states the chain rule formula but fails to compute the derivatives dx/dt and dy/dt or substitute the values of x, y, and t. It jumps from the formula to the final answer '1' without showing the necessary intermediate calculation steps.gpt-oss:20b: fail (error) 2026-10-04 — The solution skips the essential computation of dx/dt and dy/dt and jumps directly to the conclusion 1=1 at t=1, which is not justified. The chain‑rule step is correct, but the missing intermediate calculations make the reasoning invalid.qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to compute the derivatives dx/dt and dy/dt, nor does it evaluate the partial derivatives at the correct point (x=1, y=1). It simply asserts the final answer is 1 without showing the chain rule calculation.gpt-oss:20b: fail (error) 2026-10-04 — The solution skips the essential steps of computing dx/dt and dy/dt, and it does not actually evaluate the chain‑rule expression at t=1. The conclusion that dz/dt=1 is unsupported by the given work.
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 2026-10-04 with SymPy 1.14.0.