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The multivariable chain rule

Problem 10.538 · easy

Let \( \displaystyle z = \sqrt{x^{2} + y^{2} + 4} \) with \( \displaystyle x = \cos{\left(t \right)} \), \( \displaystyle y = \sin{\left(t \right)} \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 0 \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \sqrt{x^{2} + y^{2} + 4}\\\frac{\partial}{\partial y} \sqrt{x^{2} + y^{2} + 4}\end{matrix}\right] = \left[\begin{matrix}\frac{x}{\sqrt{x^{2} + y^{2} + 4}}\\\frac{y}{\sqrt{x^{2} + y^{2} + 4}}\end{matrix}\right] \]
∂z/∂x and ∂z/∂y.✓ Proved
dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
\[ 0 \]
At t = 0, where (x, y) = (1, 0).✓ 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: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to compute the derivatives dx/dt and dy/dt or substitute the values of x and y into the partial derivatives. It jumps directly to the result 0 without showing the necessary chain rule calculation steps.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to compute the derivatives dx/dt and dy/dt or substitute the values of x and y into the partial derivatives. It jumps directly to the result 0 without showing the necessary chain rule calculation steps.
gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to compute the derivatives dx/dt and dy/dt or substitute them into the chain rule formula. It jumps from the formula to the final answer without showing the necessary intermediate steps or evaluating the partial derivatives at the specific point.

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.