The multivariable chain rule
Problem 10.540 · hard
Let \( \displaystyle z = x \cos{\left(y \right)} + y e^{x} \) with \( \displaystyle x = e^{t} \), \( \displaystyle y = 2 t \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x \cos{\left(y \right)} + y e^{x}\right)\\\frac{\partial}{\partial y} \left(x \cos{\left(y \right)} + y e^{x}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x} + \cos{\left(y \right)}\\- x \sin{\left(y \right)} + e^{x}\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).Reviewed
- \[ - 2 e \sin{\left(2 \right)} + 2 e^{e} + e \left(\cos{\left(2 \right)} + 2 e^{e}\right) \]At t = 1, where (x, y) = (E, 2).✓ Proved
Answer \( - 2 e \sin{\left(2 \right)} + 2 e^{e} + e \left(\cos{\left(2 \right)} + 2 e^{e}\right) \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly applies the multivariable chain rule and evaluates the derivatives at the specified point. The intermediate steps are logically sound and the final answer matches the stated result.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies the multivariable chain rule and evaluates the derivatives at the specified point. The intermediate steps are logically sound and the final answer matches the stated result.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies the multivariable chain rule and evaluates the partial derivatives and derivatives of the inner functions at the specified point t=1. The final expression matches the stated answer.
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.