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

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Aicharya · Accepted Answer

Given, $x^2+y^2+\sin y=4$ After differentiating the above equation $w$. r. t. $x$ we get $ \begin{aligned} &2 x+2 y \frac{d y}{d x}+\cos y \frac{d y}{d x}=0 \ldots(1) \ &\Rightarrow 2 x+(2 y+\cos y) \frac{d y}{d x}=0 \ &\Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 y+\cos y} \ &	ext { At }(-2,0),\left(\frac{d y}{d x}ight)_{(-2,0)}=\frac{-2 	imes-2}{2 	imes 0+\cos 0} \ &\Rightarrow\left(\frac{d y}{d x}ight)_{(-2,0)}=\frac{4}{0+1} \ &\Rightarrow\left(\frac{d y}{d x}ight)_{(2,0)}=4 \ldots(2) \end{aligned} $ Again differentiating equation (1) w. r. t to $x$, we get $2+2\left(\frac{d y}{d x}ight)^2+2 y \frac{d^2 y}{d x^2}-\sin y\left(\frac{d y}{d x}ight)^2+\cos \mathrm{y} \frac{d^2 y}{d x^2}=0$ $\Rightarrow 2+(-2 \sin y)\left(\frac{d y}{d x}ight)^2+(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=0$ $\Rightarrow(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=-2-(2-\sin y)\left(\frac{d y}{d x}ight)^2$ $\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-(2-\sin y)\left(\frac{d y}{d x}ight)^2}{2 y+\cos \mathrm{y}}$ So, at $(-2,0)$ $\frac{d^2 y}{d x^2}=\frac{-2-(2-0) 	imes 4^2}{2 	imes 0+1}$ $\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-2 	imes 16}{1}$ $\Rightarrow \frac{d^2 y}{d x^2}=-34$

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

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