Second derivatives by implicit differentiation
Problem 2.2326 · hard
The curve \( \displaystyle x^{2} + 2 y^{2} = 3 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle y'' \) there.
- \[ \left. x^{2} + 2 y^{2} \right|_{\substack{ x=1\\ y=1 }} = 3 \]The point is on the curve.✓ Proved
- \[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + 2 y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + 2 y^{2}\right)} = - \frac{x}{2 y} \]y′ = −F_x/F_y.✓ Proved
- \[ \frac{d}{d x} \left(- \frac{x}{2 Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{2 Y^{2}{\left(x \right)}} - \frac{1}{2 Y{\left(x \right)}} \]Differentiate y′ again, with y a function of x.✓ Proved
- Replace dy/dx in that by y′ itself, and simplify.
- \[ \left. \frac{- x^{2} - 2 y^{2}}{4 y^{3}} \right|_{\substack{ x=1\\ y=1 }} = - \frac{3}{4} \]At (1, 1).✓ Proved
Answer \( y'' = \frac{- x^{2} - 2 y^{2}}{4 y^{3}},\quad y''(1) = - \frac{3}{4} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to substitute the original constraint x^2 + 2y^2 = 3 into the expression for y'' before evaluating at the point, or simply substitutes y' = -x/(2y) into the second derivative formula incorrectly. The derived formula y'' = (-x^2 - 2y^2)/(4y^3) is algebraically incorrect for the given curve; the correct second derivative is y'' = -3/(4y^3). The numerator should be -3 (from the constant on the RHS of the curve equation), not -x^2 - 2y^2, although they are equal at the point, the general formula stated is wrong because it implies y'' depends on x and y in a way that doesn't reflect the constant constraint properly in the general form (usually one substitutes the constraint to simplify). More critically, step 3 computes the derivative of -x/(2y) but step 4 claims to replace dy/dx by y' itself. The expression in step 3 is d/dx(-x/(2y)). Using quotient rule: (-1*2y - (-x)*2y')/(4y^2) = (-2y + 2xy')/(4y^2). Substituting y' = -x/(2y) gives (-2y + 2x(-x/(2y)))/(4y^2) = (-2y - x^2/y)/(4y^2) = (-2y^2 - x^2)/(4y^3). This matches the numerator in step 5. However, since x^2 + 2y^2 = 3, the numerator is -3. The formula presented in the stated answer and step 5 is technically correct as an intermediate step, but usually, one simplifies using the constraint. The main issue is that the stated answer presents a general formula y'' = (-x^2 - 2y^2)/(4y^3) which is not simplified using the curve equation. While not strictly 'wrong' numerically, it is poor practice. But wait, is it wrong? No. Is it misleading? Maybe. Let's look closer. Step 3 is the derivative of y'. Step 4 says 'Replace dy/dx in that by y' itself'. This is confusing phrasing. You replace y' with its expression in terms of x and y. The result in step 5 is correct. The verdict should be pass if the math is correct. The math is correct. The formula is unsimplified but correct. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution fails to substitute the original constraint x^2 + 2y^2 = 3 into the expression for y'' before evaluating at the point, or simply substitutes y' = -x/(2y) into the second derivative formula incorrectly. The derived formula y'' = (-x^2 - 2y^2)/(4y^3) is algebraically incorrect for the given curve; the correct second derivative is y'' = -3/(4y^3). The numerator should be -3 (from the constant on the RHS of the curve equation), not -x^2 - 2y^2, although they are equal at the point, the general formula stated is wrong because it implies y'' depends on x and y in a way that doesn't reflect the constant constraint properly in the general form (usually one substitutes the constraint to simplify). More critically, step 3 computes the derivative of -x/(2y) but step 4 claims to replace dy/dx by y' itself. The expression in step 3 is d/dx(-x/(2y)). Using quotient rule: (-1*2y - (-x)*2y')/(4y^2) = (-2y + 2xy')/(4y^2). Substituting y' = -x/(2y) gives (-2y + 2x(-x/(2y)))/(4y^2) = (-2y - x^2/y)/(4y^2) = (-2y^2 - x^2)/(4y^3). This matches the numerator in step 5. However, since x^2 + 2y^2 = 3, the numerator is -3. The formula presented in the stated answer and step 5 is technically correct as an intermediate step, but usually, one simplifies using the constraint. The main issue is that the stated answer presents a general formula y'' = (-x^2 - 2y^2)/(4y^3) which is not simplified using the curve equation. While not strictly 'wrong' numerically, it is poor practice. But wait, is it wrong? No. Is it misleading? Maybe. Let's look closer. Step 3 is the derivative of y'. Step 4 says 'Replace dy/dx in that by y' itself'. This is confusing phrasing. You replace y' with its expression in terms of x and y. The result in step 5 is correct. The verdict should be pass if the math is correct. The math is correct. The formula is unsimplified but correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies implicit differentiation and the quotient rule. The algebraic simplification in step 4 is valid, and the final evaluation is correct.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-14 — The differentiation in step 3 is incorrect; the derivative of -x/(2y) with respect to x is (-2y - x(2y'))/(4y^2), not the expression shown. The final formula for y'' is also algebraically incorrect.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/implicit_second_derivative, checked 2026-10-18 with SymPy 1.14.0.