Partial derivatives
Problem 10.8 · hard
For \( \displaystyle f(x, y) = x e^{2 y} + y^{2} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
- For f_x hold y constant; for f_y hold x constant.
- \[ \frac{\partial}{\partial x} \left(x e^{2 y} + y^{2}\right) = e^{2 y} \]f_x.✓ Proved
- \[ \frac{\partial}{\partial y} \left(x e^{2 y} + y^{2}\right) = 2 x e^{2 y} + 2 y \]f_y.✓ Proved
- \[ \frac{d}{d y} e^{2 y} = 2 e^{2 y} \]f_xy: differentiate f_x with respect to y.✓ Proved
- \[ \frac{\partial}{\partial x} \left(2 x e^{2 y} + 2 y\right) = 2 e^{2 y} \]Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = e^{2 y},\quad f_y = 2 x e^{2 y} + 2 y,\quad f_{xy} = 2 e^{2 y} \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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 |
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 2026-09-26 with SymPy 1.14.0.