Mathematics - Differentiation Question with Solution | TestHub

Given, $x^2+y^2+\sin y=4$ After differentiating the above equation $w$. r. t. $x$ we get $$\begin{array}{rl}& 2x+2y\frac{dy}{dx}+\cos y\frac{dy}{dx}=0\dots (1)\\ & \Rightarrow 2x+(2y+\cos y)\frac{dy}{dx}=0\\ & \Rightarrow \frac{dy}{dx}=\frac{-2x}{2y+\cos y}\\ & \text{ At }(-2,0),{\left(\frac{dy}{dx}\right)}_{(-2,0)}=\frac{-2\times -2}{2\times 0+\cos 0}\\ & \Rightarrow {\left(\frac{dy}{dx}\right)}_{(-2,0)}=\frac{4}{0+1}\\ & \Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,0)}=4\dots (2)\end{array}$$ Again differentiating equation (1) w. r. t to $x$, we get $2+2{\left(\frac{dy}{dx}\right)}^2+2y\frac{d^{2}y}{d{x}^2}-\sin y{\left(\frac{dy}{dx}\right)}^2+\cos \mathrm{y}\frac{d^{2}y}{d{x}^2}=0$ $\Rightarrow 2+(-2\sin y){\left(\frac{dy}{dx}\right)}^2+(2y+\cos \mathrm{y})\frac{d^{2}y}{d{x}^2}=0$ $\Rightarrow (2y+\cos \mathrm{y})\frac{d^{2}y}{d{x}^2}=-2-(2-\sin y){\left(\frac{dy}{dx}\right)}^2$ $\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2-(2-\sin y){\left(\frac{dy}{dx}\right)}^2}{2y+\cos \mathrm{y}}$ So, at $(-2,0)$ $\frac{d^{2}y}{d{x}^2}=\frac{-2-(2-0)\times 4^2}{2\times 0+1}$ $\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2-2\times 16}{1}$ $\Rightarrow \frac{d^{2}y}{d{x}^2}=-34$