Gradient and directional derivatives
Problem 10.247 · medium
Find the directional derivative of \( \displaystyle f(x, y) = x y^{2} + 2 x \) at \( \displaystyle (-1, 1) \) in the direction of \( \displaystyle \langle 1, 1 \rangle \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x y^{2} + 2 x\right)\\\frac{\partial}{\partial y} \left(x y^{2} + 2 x\right)\end{matrix}\right] = \left[\begin{matrix}y^{2} + 2\\2 x y\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}3\\-2\end{matrix}\right] \]∇f at the point.✓ Proved
- \[ \left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]The unit direction u.✓ Proved
- \[ \frac{\sqrt{2}}{2} \]D_u f = ∇f · u.✓ Proved
Answer \( \frac{\sqrt{2}}{2} \)
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 |
| 4 | ✓ 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 | f differenced along the unit direction agrees |
Reviewers
gpt-oss:20b: fail (misleading) — The final line claims the directional derivative is a vector (sqrt(2)/2, sqrt(2)/2) instead of the scalar sqrt(2)/2 obtained from the dot product. This would mislead a student into thinking the directional derivative is a vector.qwen3.6:27b-mlx: pass — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product. The final answer is correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product. The final answer is correct.gpt-oss:20b: fail (misleading) 2026-09-29 — The final line claims the directional derivative is a vector (sqrt(2)/2, sqrt(2)/2) instead of the scalar sqrt(2)/2 obtained from the dot product. This would mislead a student into thinking the directional derivative is a vector.qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product. The final answer matches the stated answer.gpt-oss:20b: pass 2026-09-29
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/gradient_directional, checked 2026-09-29 with SymPy 1.14.0.