The multivariable chain rule
Problem 10.447 · easy
Let \( \displaystyle z = \ln{\left(x^{2} + y^{2} + 1 \right)} \) with \( \displaystyle x = \cos{\left(t \right)} \), \( \displaystyle y = \sin{\left(t \right)} \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = \frac{\pi}{2} \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)}\\\frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)}\end{matrix}\right] = \left[\begin{matrix}\frac{2 x}{x^{2} + y^{2} + 1}\\\frac{2 y}{x^{2} + y^{2} + 1}\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
- \[ 0 \]At t = pi/2, where (x, y) = (0, 1).✓ 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: fail (misleading) — The solution skips the actual evaluation of the partial derivatives and the time derivatives at t=π/2, simply stating 0=0. A student would not see how the chain‑rule computation yields 0, so the explanation is misleading.qwen3.6:27b-mlx: fail (error) — The solution fails to compute the derivatives dx/dt and dy/dt, which are essential for applying the chain rule. It jumps directly to the final answer without showing the intermediate step of multiplying the partial derivatives by the time derivatives of x and y.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to compute the derivatives dx/dt and dy/dt, which are essential for applying the chain rule. It jumps directly to the final answer without showing the intermediate step of multiplying the partial derivatives by the time derivatives of x and y.gpt-oss:20b: fail (misleading) 2026-10-08 — The solution skips the actual evaluation of the partial derivatives and the time derivatives at t=π/2, simply stating 0=0. A student would not see how the chain‑rule computation yields 0, so the explanation is misleading.gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies the chain rule and evaluates the derivatives at the specified point. Although the intermediate step showing the substitution is skipped, the final result is correct and the logic is sound.
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 2026-10-08 with SymPy 1.14.0.