If x^2 + y^2 + sin y = 4,then the value of fr | vedclass

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(A) Given the equation: $x^2 + y^2 + \sin y = 4$.Differentiating with respect to $x$:$2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$$2x + (2y + \co

Vedclass · Accepted Answer

$-34$

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(A) Given the equation: $x^2 + y^2 + \sin y = 4$.Differentiating with respect to $x$:$2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$$2x + (2y + \cos y) \frac{dy}{dx} = 0$$\frac{dy}{dx} = -\frac{2x}{2y + \cos y}$.At the point $(-2, 0)$:$\frac{dy}{dx} = -\frac{2(-2)}{2(0) + \cos 0} = \frac{4}{1} = 4$.Differentiating $2x + (2y + \cos y) \frac{dy}{dx} = 0$ with respect to $x$:$2 + (2 \frac{dy}{dx} + \cos y \frac{dy}{dx}) \frac{dy}{dx} + (2y + \cos y) \frac{d^2y}{dx^2} = 0$$2 + (2 + \cos y) \left(\frac{dy}{dx}\right)^2 + (2y + \cos y) \frac{d^2y}{dx^2} = 0$.Substituting $x = -2, y = 0$,and $\frac{dy}{dx} = 4$:$2 + (2 + \cos 0)(4)^2 + (2(0) + \cos 0) \frac{d^2y}{dx^2} = 0$$2 + (2 + 1)(16) + (1) \frac{d^2y}{dx^2} = 0$$2 + 3(16) + \frac{d^2y}{dx^2} = 0$$2 + 48 + \frac{d^2y}{dx^2} = 0$$\frac{d^2y}{dx^2} = -50$.Wait,re-evaluating the derivative of the first equation:$2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$ is correct.Second derivative: $2 + 2(\frac{dy}{dx})^2 + 2y \frac{d^2y}{dx^2} - \sin y (\frac{dy}{dx})^2 + \cos y \frac{d^2y}{dx^2} = 0$.At $(-2, 0)$ with $\frac{dy}{dx} = 4$:$2 + 2(4)^2 + 2(0) \frac{d^2y}{dx^2} - \sin(0)(4)^2 + \cos(0) \frac{d^2y}{dx^2} = 0$$2 + 32 + 0 - 0 + 1 \cdot \frac{d^2y}{dx^2} = 0$$34 + \frac{d^2y}{dx^2} = 0 \Rightarrow \frac{d^2y}{dx^2} = -34$.

If $x^2 + y^2 + \sin y = 4$,then the value of $\frac{d^2y}{dx^2}$ at the point $(-2, 0)$ is

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