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The multivariable chain rule

Problem 10.490 · easy

Let \( \displaystyle z = \ln{\left(x^{2} + y^{2} + 1 \right)} \) with \( \displaystyle x = e^{t} \), \( \displaystyle y = 2 t \). Use the chain rule to find \( \displaystyle \frac{dz}{dt} \) at \( \displaystyle t = 0 \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} \ln{\left(x^{2} + y^{2} + 1 \right)}\\\frac{\partial}{\partial y} \ln{\left(x^{2} + y^{2} + 1 \right)}\end{matrix}\right] = \left[\begin{matrix}\frac{2 x}{x^{2} + y^{2} + 1}\\\frac{2 y}{x^{2} + y^{2} + 1}\end{matrix}\right] \]
∂z/∂x and ∂z/∂y.✓ Proved
dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
\[ 1 \]
At t = 0, where (x, y) = (1, 0).✓ Proved

Answer \( 1 \)

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, which are required by the chain rule formula stated in step 2. It jumps directly to the final numerical answer without showing the substitution or calculation.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to compute the derivatives dx/dt and dy/dt, which are required by the chain rule formula stated in step 2. It jumps directly to the final numerical answer without showing the substitution or calculation.
gpt-oss:20b: pass 2026-10-10
gpt-oss:20b: fail (style) 2026-10-10 — The final sentence merely states "1 = 1" without showing the actual computation of dz/dt; it does not clearly convey that the derivative equals 1 at t=0, which could confuse a reader.
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution skips the calculation of dx/dt and dy/dt, which are required to apply the chain rule formula stated in line 2. Without showing these derivatives, the jump to the final answer is unjustified and hides the actual computation.

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.