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

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(B) Given the equation $x^2+y^2+\sin y=4$. At $x=-2$,we have $(-2)^2+y^2+\sin y=4$,which simplifies to $4+y^2+\sin y=4$,so $y^2+\sin y=0$. Since $y=0$

Vedclass · Accepted Answer

$-34$

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(B) Given the equation $x^2+y^2+\sin y=4$. At $x=-2$,we have $(-2)^2+y^2+\sin y=4$,which simplifies to $4+y^2+\sin y=4$,so $y^2+\sin y=0$. Since $y=0$ is a solution,we have $y=0$ at $x=-2$. Differentiating $x^2+y^2+\sin y=4$ with respect to $x$,we get $2x+2y\frac{dy}{dx}+\cos y \frac{dy}{dx}=0$. Substituting $x=-2$ and $y=0$,we get $2(-2)+2(0)\frac{dy}{dx}+\cos(0)\frac{dy}{dx}=0$,which gives $-4+\frac{dy}{dx}=0$,so $\frac{dy}{dx}=4$. Differentiating $2x+(2y+\cos y)\frac{dy}{dx}=0$ again with respect to $x$,we get $2+(2\frac{dy}{dx}-\sin y \frac{dy}{dx})\frac{dy}{dx}+(2y+\cos y)\frac{d^2y}{dx^2}=0$. Substituting $x=-2, y=0, \frac{dy}{dx}=4$,we get $2+(2(4)-\sin(0)(4))(4)+(2(0)+\cos(0))\frac{d^2y}{dx^2}=0$. This simplifies to $2+(8)(4)+(1)\frac{d^2y}{dx^2}=0$,so $2+32+\frac{d^2y}{dx^2}=0$. Thus,$\frac{d^2y}{dx^2}=-34$.

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

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