Implicit differentiation with partial derivatives
Problem 10.410 · medium
The equation \( \displaystyle y z + e^{x z} = e^{-1} + 1 \) defines \( \displaystyle z \) as a function of \( \displaystyle x \) and \( \displaystyle y \) near \( \displaystyle (1, -1, -1) \). Find \( \displaystyle \frac{\partial z}{\partial x} \) there.
- \[ e^{-1} + 1 \]The point is on the surface.✓ Proved
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(y z + e^{x z}\right)\\\frac{\partial}{\partial z} \left(y z + e^{x z}\right)\end{matrix}\right] = \left[\begin{matrix}z e^{x z}\\x e^{x z} + y\end{matrix}\right] \]F_x and F_z.✓ Proved
- \[ \frac{1}{e \left(-1 + e^{-1}\right)} = \frac{1}{1 - e} \]∂z/∂x = −F_x/F_z.✓ Proved
Answer \( \frac{1}{1 - e} \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 | solved the equation numerically near the point and differenced |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the implicit function theorem formula ∂z/∂x = -F_x/F_z. The partial derivatives are computed correctly, and the evaluation at the point (1, -1, -1) yields the correct result.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the implicit function theorem formula ∂z/∂x = -F_x/F_z. The partial derivatives are computed correctly, and the evaluation at the point (1, -1, -1) yields the correct result.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution incorrectly identifies the partial derivative with respect to x as z*exp(x*z), omitting the term from the yz product. The correct F_x is exp(x*z), leading to a different final 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/implicit_partials, checked 2026-10-07 with SymPy 1.14.0.