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Higher-order partial derivatives

Problem 10.616 · medium

For \( \displaystyle f(x, y) = \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} \), find \( \displaystyle f_{xx} \), \( \displaystyle f_{yy} \) and \( \displaystyle f_{xy} \), check that \( \displaystyle f_{xy} = f_{yx} \), and decide whether \( \displaystyle f \) is harmonic (\( \displaystyle f_{xx} + f_{yy} = 0 \)).

\[ \frac{\partial^{2}}{\partial x^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \cos{\left(y + \frac{\pi}{4} \right)} \]
f_xx.✓ Proved
\[ \frac{\partial^{2}}{\partial y^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \left(\sin{\left(y \right)} - \cos{\left(y \right)}\right) e^{- x} \]
f_yy.✓ Proved
\[ \frac{\partial^{2}}{\partial y\partial x} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)} \]
f_xy.✓ Proved
\[ \frac{\partial^{2}}{\partial x\partial y} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)} \]
f_yx, the same (Clairaut).✓ Proved
\[ \frac{\partial^{2}}{\partial x^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} + \frac{\partial^{2}}{\partial y^{2}} \left(- \sin{\left(y \right)} + \cos{\left(y \right)}\right) e^{- x} = 0 \]
f_xx + f_yy: 0, so f is harmonic.✓ Proved

Answer \( f_{xx} = \sqrt{2} e^{- x} \cos{\left(y + \frac{\pi}{4} \right)},\ f_{yy} = \left(\sin{\left(y \right)} - \cos{\left(y \right)}\right) e^{- x},\ f_{xy} = \sqrt{2} e^{- x} \sin{\left(y + \frac{\pi}{4} \right)};\ \text{harmonic} \)

✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation. How we verify

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
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	second differences of f at two points

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly computes the partial derivatives, verifies Clairaut's theorem, and confirms the function is harmonic. The algebraic simplifications are correct and verified.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly computes the partial derivatives, verifies Clairaut's theorem, and confirms the function is harmonic. The algebraic forms provided in the equations are correct.
gpt-oss:20b: pass 2026-10-14

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/higher_order_partials, checked 2026-10-17 with SymPy 1.14.0.