Home›Calculus 3›Partial derivatives›Problem 10.126

Partial derivatives

Problem 10.126 · hard

For \( \displaystyle f(x, y) = \ln{\left(x^{2} + 3 y^{2} + 1 \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).

For f_x hold y constant; for f_y hold x constant.
Reviewed
\[ \frac{\partial}{\partial x} \ln{\left(x^{2} + 3 y^{2} + 1 \right)} = \frac{2 x}{x^{2} + 3 y^{2} + 1} \]
f_x.✓ Proved
\[ \frac{\partial}{\partial y} \ln{\left(x^{2} + 3 y^{2} + 1 \right)} = \frac{6 y}{x^{2} + 3 y^{2} + 1} \]
f_y.✓ Proved
\[ \frac{\partial}{\partial y} \frac{2 x}{x^{2} + 3 y^{2} + 1} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \]
f_xy: differentiate f_x with respect to y.✓ Proved
\[ \frac{\partial}{\partial x} \frac{6 y}{x^{2} + 3 y^{2} + 1} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \]
Clairaut: f_yx is the same.✓ Proved

Answer \( f_x = \frac{2 x}{x^{2} + 3 y^{2} + 1},\quad f_y = \frac{6 y}{x^{2} + 3 y^{2} + 1},\quad f_{xy} = - \frac{12 x y}{\left(x^{2} + 3 y^{2} + 1\right)^{2}} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. 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	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: pass
qwen3.6:27b-mlx: pass — The solution correctly computes the partial derivatives and the mixed partial derivative. The application of Clairaut's theorem is valid since the function is smooth.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly computes the partial derivatives and the mixed partial derivative. The application of Clairaut's theorem is valid since the function is smooth.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly computes the partial derivatives and verifies Clairaut's theorem. The setup and calculations are sound.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the partial derivatives and the mixed partial derivative. The application of Clairaut's theorem is valid since the function is smooth, and the algebraic results are correct.
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/partial_derivatives, checked 2026-09-28 with SymPy 1.14.0.