If x^2y + y^3 = 2,then the value of frac{d^2y | vedclass

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(B) Given the equation $x^2y + y^3 = 2$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(2)$ $x

Vedclass · Accepted Answer

$-\frac{3}{8}$

Vedclass · Answer

(B) Given the equation $x^2y + y^3 = 2$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(2)$ $x^2 \frac{dy}{dx} + 2xy + 3y^2 \frac{dy}{dx} = 0$ At the point $(1, 1)$: $(1)^2 \frac{dy}{dx} + 2(1)(1) + 3(1)^2 \frac{dy}{dx} = 0$ $\frac{dy}{dx} + 2 + 3 \frac{dy}{dx} = 0$ $4 \frac{dy}{dx} = -2 \implies \frac{dy}{dx} = -\frac{1}{2}$ Now,differentiate $x^2 \frac{dy}{dx} + 2xy + 3y^2 \frac{dy}{dx} = 0$ again with respect to $x$: $\frac{d}{dx}(x^2 \frac{dy}{dx}) + \frac{d}{dx}(2xy) + \frac{d}{dx}(3y^2 \frac{dy}{dx}) = 0$ $(x^2 \frac{d^2y}{dx^2} + 2x \frac{dy}{dx}) + (2x \frac{dy}{dx} + 2y) + (3y^2 \frac{d^2y}{dx^2} + 6y (\frac{dy}{dx})^2) = 0$ Substitute $x=1, y=1, \frac{dy}{dx} = -\frac{1}{2}$: $(1)^2 \frac{d^2y}{dx^2} + 2(1)(-\frac{1}{2}) + 2(1)(-\frac{1}{2}) + 2(1) + 3(1)^2 \frac{d^2y}{dx^2} + 6(1)(-\frac{1}{2})^2 = 0$ $\frac{d^2y}{dx^2} - 1 - 1 + 2 + 3 \frac{d^2y}{dx^2} + 6(\frac{1}{4}) = 0$ $4 \frac{d^2y}{dx^2} + \frac{3}{2} = 0$ $4 \frac{d^2y}{dx^2} = -\frac{3}{2} \implies \frac{d^2y}{dx^2} = -\frac{3}{8}$

If $x^2y + y^3 = 2$,then the value of $\frac{d^2y}{dx^2}$ at the point $(1, 1)$ is:

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