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Gradient and directional derivatives

Problem 10.10 · medium Video

Find the directional derivative of \( \displaystyle f(x, y) = x e^{y} \) at \( \displaystyle (-1, 1) \) in the direction of \( \displaystyle \langle 1, 1 \rangle \).

\[ \left[\begin{matrix}\frac{\partial}{\partial x} x e^{y}\\\frac{\partial}{\partial y} x e^{y}\end{matrix}\right] = \left[\begin{matrix}e^{y}\\x e^{y}\end{matrix}\right] \]
∇f.✓ Proved
\[ \left[\begin{matrix}e\\- e\end{matrix}\right] \]
∇f at the point.
\[ \left[\begin{matrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{matrix}\right] \]
The unit direction u.
\[ 0 \]
D_u f = ∇f · u.

Answer \( 0 \)

A contour map of f(x, y) = x·e^(y): each curve joins points where f has the same value. At (-1, 1) the gradient arrow points straight uphill, across the contours, and the direction of travel is drawn beside it.

Lines: 1 proved, 3 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. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
3	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
4	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
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: pass
qwen3.6:27b-mlx: fail (error) — The gradient calculation in step 2 is incorrect; f_x(-1, 1) is -e, not e. Consequently, the final dot product is -sqrt(2)e, not 0.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The gradient calculation in step 2 is incorrect; f_x(-1, 1) is -e, not e. Consequently, the final dot product is -sqrt(2)e, not 0.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The dot product in step 4 is calculated incorrectly. The gradient is <E, -E> and the unit vector is <1/sqrt(2), 1/sqrt(2)>, so the dot product is E/sqrt(2) - E/sqrt(2) = 0. Wait, the calculation is actually correct. Let me re-read. Gradient at (-1, 1): f_x = e^y = e^1 = E. f_y = x e^y = -1 * e^1 = -E. So grad f = <E, -E>. Direction vector v = <1, 1>. Magnitude |v| = sqrt(1^2 + 1^2) = sqrt(2). Unit vector u = <1/sqrt(2), 1/sqrt(2)>. Directional derivative D_u f = grad f . u = E * (1/sqrt(2)) + (-E) * (1/sqrt(2)) = E/sqrt(2) - E/sqrt(2) = 0. The solution is actually correct. Why did I think it was wrong? Let me double check the problem statement. f(x,y) = x e^y. Point (-1, 1). Direction <1, 1>. f_x = e^y. At (-1, 1), f_x = e. f_y = x e^y. At (-1, 1), f_y = -e. u = <1, 1> / sqrt(2). Dot product: e * (1/sqrt(2)) + (-e) * (1/sqrt(2)) = 0. The stated answer is 0. The solution derives 0. The steps are correct. The algebra is marked as proved. The sentences are just labels. Wait, is there a subtle error? Step 1: Gradient formula. Correct. Step 2: Evaluation. Correct. Step 3: Unit vector. Correct. Step 4: Dot product result. Correct. It seems the solution is sound. I will mark it as pass.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly computes the gradient, normalizes the direction vector, and calculates the dot product, which results in 0.
gpt-oss:20b: pass 2026-09-26

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 2026-10-13 with SymPy 1.14.0.