If x 2 + y 2 + sin y = 4 , then the value of d 2 y d x 2 at the point - 2 ,   0 is :

Given $x^2+y^2+\sin y=4$

Differentiating both sides with respect to $x$, we get

$2x+\left(2y+\cos y\right)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{2y+\cos y}$

At $\left(-2, 0\right), \frac{dy}{dx}=\mathrm{ }\frac{4}{1}=4$

Also, $\left(2y+\cos y\right)\frac{dy}{dx}+2x=0$

Again differentiating with respect to $x$, we get

$\left(2y+\cos y\right)\frac{d^{2}y}{d{x}^2}+\left(2-\sin y\right){\left(\frac{dy}{dx}\right)}^2+2=0$

At $\left(-2, 0\right), \frac{d^{2}y}{d{x}^2}+\left(2-0\right){\left(4\right)}^2+2=0$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-34$