The multivariable chain rule
Problem 10.439 · easy
Let \( \displaystyle z = \ln{\left(x^{2} + y^{2} + 1 \right)} \) with \( \displaystyle x = t^{2} \), \( \displaystyle y = t^{3} \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)}\\\frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)}\end{matrix}\right] = \left[\begin{matrix}\frac{2 x}{x^{2} + y^{2} + 1}\\\frac{2 y}{x^{2} + y^{2} + 1}\end{matrix}\right] \]∂z/∂x and ∂z/∂y.✓ Proved
- dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
- \[ \frac{10}{3} \]At t = 1, where (x, y) = (1, 1).✓ Proved
Answer \( \frac{10}{3} \)
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 (style) — The solution skips the explicit computation of dx/dt, dy/dt and the substitution of t=1, so the conclusion 10/3 is not justified from the preceding lines.qwen3.6:27b-mlx: fail (error) — The solution skips the calculation of dx/dt and dy/dt and fails to substitute the values of x, y, dx/dt, and dy/dt into the chain rule formula to demonstrate how 10/3 is derived. It merely asserts the final numerical equality without showing the intermediate step required by the instruction 'Use the chain rule'.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution skips the calculation of dx/dt and dy/dt and fails to substitute the values of x, y, dx/dt, and dy/dt into the chain rule formula to demonstrate how 10/3 is derived. It merely asserts the final numerical equality without showing the intermediate step required by the instruction 'Use the chain rule'.gpt-oss:20b: fail (style) 2026-10-08 — The solution skips the explicit computation of dx/dt, dy/dt and the substitution of t=1, so the conclusion 10/3 is not justified from the preceding lines.gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies the chain rule and the final numerical evaluation is correct.
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.