Home›Calculus 1›Second derivatives by implicit differentiation›Problem 2.2472

Second derivatives by implicit differentiation

Problem 2.2472 · hard

The curve \( \displaystyle 3 x^{2} + 3 y^{2} = 6 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle y'' \) there.

\[ \left. 3 x^{2} + 3 y^{2} \right|_{\substack{ x=1\\ y=-1 }} = 6 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(3 x^{2} + 3 y^{2}\right)}{\frac{\partial}{\partial y} \left(3 x^{2} + 3 y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=1\\ y=-1 }} = 2 \]
At (1, -1).✓ Proved

Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(1) = 2 \)

Lines: 4 proved, 1 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	Not checked	—	a sentence; read, not computed
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	y solved numerically from the curve at x₀ and x₀ ± 10⁻⁶, then a second difference

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The stated answer claims y''(1) = 2, but the correct value at (1, -1) is -2. The solution computes the symbolic expression correctly but fails to substitute the negative y-value correctly into the denominator (y^3 = -1), leading to a sign error in the final numerical result.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-21 — The stated answer claims y''(1) = 2, but the correct value at (1, -1) is -2. The solution computes the symbolic expression correctly but fails to substitute the negative y-value correctly into the denominator (y^3 = -1), leading to a sign error in the final numerical result.
gpt-oss:20b: pass 2026-10-21
qwen3.6:27b-mlx: fail (error) 2026-10-20 — The stated answer claims y''(1) = 2, but the correct value at (1, -1) is -2. The solution correctly derives the formula y'' = (-x^2 - y^2)/y^3, but the final substitution in line 5 is arithmetically incorrect: (-1 - 1)/(-1)^3 = -2/-1 = 2 is wrong because the numerator is -2 and denominator is -1, yielding 2? Wait. Let's re-calculate. x=1, y=-1. Numerator: -(1)^2 - (-1)^2 = -1 - 1 = -2. Denominator: (-1)^3 = -1. Result: -2 / -1 = 2. The calculation in line 5 is actually correct. Let me re-read the problem. 3x^2 + 3y^2 = 6 => x^2 + y^2 = 2. y' = -x/y. y'' = -(y - xy')/y^2 = -(y - x(-x/y))/y^2 = -(y + x^2/y)/y^2 = -(y^2 + x^2)/y^3. At (1, -1), x^2+y^2 = 2. y^3 = -1. y'' = -2/-1 = 2. The stated answer is correct. Why did I think it was -2? Ah, standard circle x^2+y^2=r^2. y'' = -r^2/y^3. Here r^2=2. y=-1. y'' = -2/-1 = 2. The solution is correct. Wait, let me check the sign of y'. y' = -x/y. At (1, -1), y' = -1/-1 = 1. y'' = d/dx(-x/y) = -(y - xy')/y^2. At (1, -1), y' = 1. Numerator: -(-1 - 1(1)) = -(-2) = 2. Denominator: (-1)^2 = 1. Result 2. The solution is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-10-20
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly applies implicit differentiation to find the second derivative and evaluates it at the given point. The algebraic steps are verified as correct.
gpt-oss:20b: pass 2026-10-20

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/implicit_second_derivative, checked 2026-10-21 with SymPy 1.14.0.