The multivariable chain rule
Problem 10.498 · easy
Let \( \displaystyle z = x y^{2} - x \) with \( \displaystyle x = e^{t} \), \( \displaystyle y = 2 t \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x y^{2} - x\right)\\\frac{\partial}{\partial y} \left(x y^{2} - x\right)\end{matrix}\right] = \left[\begin{matrix}y^{2} - 1\\2 x y\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
- \[ 11 e \]At t = 1, where (x, y) = (E, 2).✓ Proved
Answer \( 11 e \)
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 explicitly calculate or state the values of dx/dt and dy/dt at t=1, which are required to apply the chain rule formula stated in line 2. It jumps directly to the final numerical result without showing the intermediate substitution steps.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly calculate or state the values of dx/dt and dy/dt at t=1, which are required to apply the chain rule formula stated in line 2. It jumps directly to the final numerical result without showing the intermediate substitution steps.gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly calculate or state the derivatives dx/dt and dy/dt, which are required to apply the chain rule formula stated in line 2. It jumps directly to the final numerical answer without showing the intermediate step of multiplying the partial derivatives by the time derivatives.
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 2026-10-10 with SymPy 1.14.0.