ANSWER
[tex]\frac{d^2y}{dx^2}=4[/tex]
EXPLANATION
We want to find the second derivative of the function at the given point:
[tex]2-xy-2y=0[/tex]
First, differentiate the function implicitly:
[tex]0-y-x\frac{dy}{dx}-2\frac{dy}{dx}=0[/tex]
Next, differentiate the function implicitly a second time:
[tex]\begin{gathered} -\frac{dy}{dx}-\frac{dy}{dx}-x\frac{d^2y}{dx^2}-2\frac{d^2y}{dx^2}=0 \\ -2\frac{dy}{dx}-(x+2)\frac{d^2y}{dx^2}=0 \\ -(x+2)\frac{d^2y}{dx^2}=2\frac{dy}{dx} \\ \frac{d^2y}{dx^2}=\frac{2}{-(x+2)}\frac{dy}{dx} \end{gathered}[/tex]
From the first derivative, we have that:
[tex]\begin{gathered} -y-(x+2)\frac{dy}{dx}=0 \\ -(x+2)\frac{dy}{dx}=y \\ \frac{dy}{dx}=\frac{y}{-(x+2)} \end{gathered}[/tex]
Substitute that into the second derivative:
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{2}{-(x+2)}*\frac{y}{-(x+2)} \\ \frac{d^2y}{dx^2}=\frac{2y}{(x+2)^2} \end{gathered}[/tex]
Now, substitute the values of the point given into the equation and simplify:
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{2(2)}{(-1+2)^2} \\ \frac{d^2y}{dx^2}=\frac{4}{1^2} \\ \frac{d^2y}{dx^2}=4 \end{gathered}[/tex]
That is the answer.
