QuestionJuly 14, 2025

If x^2+xy+y^3=1 find the value of y''' at the point where x=1 square

If x^2+xy+y^3=1 find the value of y''' at the point where x=1 square
If x^2+xy+y^3=1 find the value of y''' at the point where x=1 square

Solution
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Answer

2 Explanation 1. Differentiate implicitly Differentiate both sides with respect to x: \frac{d}{dx}(x^2 + xy + y^3) = \frac{d}{dx}(1). 2. Apply the product rule and chain rule The derivative is 2x + (x\frac{dy}{dx} + y) + 3y^2\frac{dy}{dx} = 0. 3. Solve for \frac{dy}{dx} Rearrange: \frac{dy}{dx}(x + 3y^2) = -2x - y. Thus, \frac{dy}{dx} = \frac{-2x - y}{x + 3y^2}. 4. Differentiate \frac{dy}{dx} to find \frac{d^2y}{dx^2} Use quotient rule: \frac{d}{dx}\left(\frac{-2x - y}{x + 3y^2}\right). 5. Apply quotient rule \frac{d^2y}{dx^2} = \frac{(x + 3y^2)(-2 - \frac{dy}{dx}) - (-2x - y)(1 + 6y\frac{dy}{dx})}{(x + 3y^2)^2}. 6. Evaluate at x=1 Substitute x=1 into original equation: 1 + y + y^3 = 1, solve for y: y(y^2 + 1) = 0, so y = 0. 7. Calculate \frac{dy}{dx} at x=1, y=0 \frac{dy}{dx} = \frac{-2(1) - 0}{1 + 3(0)^2} = -2. 8. Substitute values into \frac{d^2y}{dx^2} \frac{d^2y}{dx^2} = \frac{(1 + 3(0)^2)(-2 - (-2)) - (-2(1) - 0)(1 + 6(0)(-2))}{(1 + 3(0)^2)^2} = \frac{0 - (-2)}{1} = 2.

Explanation

1. Differentiate implicitly<br /> Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2 + xy + y^3) = \frac{d}{dx}(1)$.<br /><br />2. Apply the product rule and chain rule<br /> The derivative is $2x + (x\frac{dy}{dx} + y) + 3y^2\frac{dy}{dx} = 0$.<br /><br />3. Solve for $\frac{dy}{dx}$<br /> Rearrange: $\frac{dy}{dx}(x + 3y^2) = -2x - y$. Thus, $\frac{dy}{dx} = \frac{-2x - y}{x + 3y^2}$.<br /><br />4. Differentiate $\frac{dy}{dx}$ to find $\frac{d^2y}{dx^2}$<br /> Use quotient rule: $\frac{d}{dx}\left(\frac{-2x - y}{x + 3y^2}\right)$.<br /><br />5. Apply quotient rule<br /> $\frac{d^2y}{dx^2} = \frac{(x + 3y^2)(-2 - \frac{dy}{dx}) - (-2x - y)(1 + 6y\frac{dy}{dx})}{(x + 3y^2)^2}$.<br /><br />6. Evaluate at $x=1$<br /> Substitute $x=1$ into original equation: $1 + y + y^3 = 1$, solve for $y$: $y(y^2 + 1) = 0$, so $y = 0$.<br /><br />7. Calculate $\frac{dy}{dx}$ at $x=1, y=0$<br /> $\frac{dy}{dx} = \frac{-2(1) - 0}{1 + 3(0)^2} = -2$.<br /><br />8. Substitute values into $\frac{d^2y}{dx^2}$<br /> $\frac{d^2y}{dx^2} = \frac{(1 + 3(0)^2)(-2 - (-2)) - (-2(1) - 0)(1 + 6(0)(-2))}{(1 + 3(0)^2)^2} = \frac{0 - (-2)}{1} = 2$.
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