Implicit differentiation
Problem 2.1161 · medium
The curve \( \displaystyle x^{2} + 3 x y + 3 y^{2} = 1 \) passes through \( \displaystyle (1, 0) \). Find \( \displaystyle \dfrac{dy}{dx} \) by implicit differentiation, and its value at that point.
- \[ 1 \]The point is on the curve.✓ Proved
- Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
- \[ \frac{d}{d x} \left(x^{2} + 3 x Y{\left(x \right)} + 3 Y^{2}{\left(x \right)}\right) = 2 x + \left(3 x + 6 Y{\left(x \right)}\right) \frac{d}{d x} Y{\left(x \right)} + 3 Y{\left(x \right)} \]Every y term picks up a factor dy/dx.✓ Proved
- \[ \frac{- 2 x - 3 y}{3 x + 6 y} = \frac{- \frac{2 x}{3} - y}{x + 2 y} \]Solve for dy/dx: minus F_x over F_y.✓ Proved
- \[ - \frac{2}{3} \]At the point.✓ Proved
Answer \( \frac{dy}{dx} = \frac{- \frac{2 x}{3} - y}{x + 2 y}, \quad \left.\frac{dy}{dx}\right|_{(1,0)} = - \frac{2}{3} \)
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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 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 | sympy.idiff differentiates the curve on its own and agrees |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The derivative calculation in line 3 is incorrect. The term 3xy differentiates to 3y + 3x(dy/dx) via the product rule, but the solution incorrectly groups it as (3x + 6y)(dy/dx) + 3y, effectively treating the 3x term as if it were multiplied by y inside the derivative chain incorrectly or missing the standalone 3y term from the product rule. Specifically, d/dx(3xy) = 3y + 3x y', not part of the coefficient of y' alone without the 3y term outside. The correct derivative of the LHS is 2x + 3y + (3x + 6y)y'. The solution's expression 2x + (3x + 6y)y' + 3y is actually algebraically equivalent to the correct derivative (2x + 3y + (3x+6y)y'), so the algebra in line 3 is technically correct, just written in a grouped way. However, line 4 solves for dy/dx. From 2x + 3y + (3x+6y)y' = 0, we get y' = -(2x+3y)/(3x+6y). The solution gives (-2x-3y)/(3x+6y), which is correct. Wait, let me re-read line 3 carefully. Line 3 says: `2*x + (3*x + 6*Y(x))*Derivative(Y(x), x) + 3*Y(x)`. This is `2x + 3y + (3x+6y)y'`. This is the correct derivative of `x^2 + 3xy + 3y^2`. So line 3 is correct. Line 4 solves `2x + 3y + (3x+6y)y' = 0` for `y'`. `y' = -(2x+3y)/(3x+6y)`. The solution states `(-2*x - 3*y)/(3*x + 6*y)`. This is correct. Line 5 evaluates at (1,0): `-(2(1)+0)/(3(1)+0) = -2/3`. This is correct. Why did I think it was wrong? Let me check the product rule again. d/dx(3xy) = 3(x'y + xy') = 3(1*y + x*y') = 3y + 3xy'. The solution has `3*Y(x)` (which is 3y) and `(3*x)*Derivative(Y(x), x)` (which is 3xy'). It also has `6*Y(x)*Derivative(Y(x), x)` from `3y^2`. So the coefficient of y' is `3x + 6y`. The non-y' terms are `2x + 3y`. The equation is `2x + 3y + (3x+6y)y' = 0`. Solving for y' gives `-(2x+3y)/(3x+6y)`. The solution provides this. The evaluation at (1,0) gives `-2/3`. The solution is actually correct. I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The derivative calculation in line 3 is incorrect. The term 3xy differentiates to 3y + 3x(dy/dx) via the product rule, but the solution incorrectly groups it as (3x + 6y)(dy/dx) + 3y, effectively treating the 3x term as if it were multiplied by y inside the derivative chain incorrectly or missing the standalone 3y term from the product rule. Specifically, d/dx(3xy) = 3y + 3x y', not part of the coefficient of y' alone without the 3y term outside. The correct derivative of the LHS is 2x + 3y + (3x + 6y)y'. The solution's expression 2x + (3x + 6y)y' + 3y is actually algebraically equivalent to the correct derivative (2x + 3y + (3x+6y)y'), so the algebra in line 3 is technically correct, just written in a grouped way. However, line 4 solves for dy/dx. From 2x + 3y + (3x+6y)y' = 0, we get y' = -(2x+3y)/(3x+6y). The solution gives (-2x-3y)/(3x+6y), which is correct. Wait, let me re-read line 3 carefully. Line 3 says: `2*x + (3*x + 6*Y(x))*Derivative(Y(x), x) + 3*Y(x)`. This is `2x + 3y + (3x+6y)y'`. This is the correct derivative of `x^2 + 3xy + 3y^2`. So line 3 is correct. Line 4 solves `2x + 3y + (3x+6y)y' = 0` for `y'`. `y' = -(2x+3y)/(3x+6y)`. The solution states `(-2*x - 3*y)/(3*x + 6*y)`. This is correct. Line 5 evaluates at (1,0): `-(2(1)+0)/(3(1)+0) = -2/3`. This is correct. Why did I think it was wrong? Let me check the product rule again. d/dx(3xy) = 3(x'y + xy') = 3(1*y + x*y') = 3y + 3xy'. The solution has `3*Y(x)` (which is 3y) and `(3*x)*Derivative(Y(x), x)` (which is 3xy'). It also has `6*Y(x)*Derivative(Y(x), x)` from `3y^2`. So the coefficient of y' is `3x + 6y`. The non-y' terms are `2x + 3y`. The equation is `2x + 3y + (3x+6y)y' = 0`. Solving for y' gives `-(2x+3y)/(3x+6y)`. The solution provides this. The evaluation at (1,0) gives `-2/3`. The solution is actually correct. I will mark it as pass.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-28 — The differentiation in step 3 is incorrect. The derivative of the term 3xy with respect to x is 3y + 3x(dy/dx), but the solution incorrectly groups it as (3x + 6y)(dy/dx) + 3y, effectively treating the 3y part as if it were differentiated with respect to y or mixing up product rule terms. The correct derivative of the LHS is 2x + 3y + 3x(dy/dx) + 6y(dy/dx). Consequently, the formula for dy/dx in step 4 is wrong.gpt-oss:20b: pass 2026-09-28
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_differentiation, checked 2026-09-28 with SymPy 1.14.0.