Answer:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}[/tex]
Step-by-step explanation:
We have the equation:
[tex]2y^2+2=x^2[/tex]
And we want to find d²y/dx² at the point (-2, -1).
So, let's take the derivative of both sides with respect to x:
[tex]\frac{d}{dx}[2y^2+2]=\frac{d}{dx}[x^2][/tex]
On the left, let's implicitly differentiate:
[tex]4y\frac{dy}{dx}=\frac{d}{dx}[x^2][/tex]
Differentiate normally on the left:
[tex]4y\frac{dy}{dx}=2x[/tex]
Solve for the first derivative. Divide both sides by 4y:
[tex]\frac{dy}{dx}=\frac{x}{2y}[/tex]
Now, let's take the derivative of both sides again:
[tex]\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}][/tex]
We will need to use the quotient rule:
[tex]\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}[/tex]
So:
[tex]\frac{d^2y}{dx^2}=\frac{\frac{d}{dx}[(x)](2y)-x\frac{d}{dx}[(2y)]}{(2y)^2}[/tex]
Differentiate:
[tex]\frac{d^2y}{dx^2}=\frac{(1)(2y)-x(2\frac{dy}{dx})}{4y^2}[/tex]
Simplify:
[tex]\frac{d^2y}{dx^2}=\frac{2y-2x\frac{dy}{dx}}{4y^2}[/tex]
Substitute x/2y for dy/dx. This yields:
[tex]\frac{d^2y}{dx^2}=\frac{2y-2x\frac{x}{2y}}{4y^2}[/tex]
Simplify:
[tex]\frac{d^2y}{dx^2}=\frac{2y-\frac{2x^2}{2y}}{4y^2}[/tex]
Simplify. Multiply both the numerator and denominator by 2y. So:
[tex]\frac{d^2y}{dx^2}=\frac{4y^2-2x^2}{8y^3}[/tex]
Reduce. Therefore, our second derivative is:
[tex]\frac{d^2y}{dx^2}=\frac{2y^2-x^2}{4y^3}[/tex]
We want to find the second derivative at the point (-2, -1).
So, let's substitute -2 for x and -1 for y. This yields:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(-1)^2-(-2)^2}{4(-1)^3}[/tex]
Evaluate:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(1)-(4)}{4(-1)}[/tex]
Multiply:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2-4}{-4}[/tex]
Subtract:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{-2}{-4}[/tex]
Reduce. So, our answer is:
[tex]\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}[/tex]
And we're done!
