Partial derivatives
Problem 10.204 · hard
For \( \displaystyle f(x, y) = \frac{x + y}{x - y + 1} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
- For f_x hold y constant; for f_y hold x constant.Reviewed
- \[ \frac{\partial}{\partial x} \frac{x + y}{x - y + 1} = \frac{1 - 2 y}{\left(x - y + 1\right)^{2}} \]f_x.✓ Proved
- \[ \frac{\partial}{\partial y} \frac{x + y}{x - y + 1} = \frac{2 x + 1}{\left(x - y + 1\right)^{2}} \]f_y.✓ Proved
- \[ \frac{\partial}{\partial y} \frac{1 - 2 y}{\left(x - y + 1\right)^{2}} = \frac{- 2 x - 2 y}{\left(x - y + 1\right)^{3}} \]f_xy: differentiate f_x with respect to y.✓ Proved
- \[ \frac{\partial}{\partial x} \frac{2 x + 1}{\left(x - y + 1\right)^{2}} = \frac{- 2 x - 2 y}{\left(x - y + 1\right)^{3}} \]Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = \frac{1 - 2 y}{\left(x - y + 1\right)^{2}},\quad f_y = \frac{2 x + 1}{\left(x - y + 1\right)^{2}},\quad f_{xy} = \frac{2 \left(- x - y\right)}{\left(x - y + 1\right)^{3}} \)
Lines: 4 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ 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 | difference quotients of f at (0.6, 0.4) agree with f_x and f_y |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the quotient rule and chain rule for partial derivatives. The setup, calculations, and verification via Clairaut's theorem are all correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the quotient rule and chain rule for partial derivatives. The setup, calculations, and verification via Clairaut's theorem are all correct.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution claims to find f_xy but presents a matrix containing f_x, f_y, and f_xy. The problem asks for three specific scalar values, not a matrix representation. Additionally, the label 'Clairaut' in step 5 is misapplied; while Clairaut's theorem justifies equality, the step itself is just a verification calculation, not the application of the theorem to derive the result.gpt-oss:20b: pass 2026-09-28
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/partial_derivatives, checked 2026-09-28 with SymPy 1.14.0.