Solution:
Given:
[tex]y^2-2=2x^3[/tex]
Lets First Differentiate the given equation with respect to x
[tex]\frac{d}{dx} ( y^2 - 2 ) = \frac{d}{dx} 2x^3[/tex]
[tex]2y \cdot \frac{dy}{dx} - 0 = 6x^2[/tex]
[tex]\frac{dy}{dx} = \frac{6x^2}{2y}[/tex]
[tex]\frac{dy}{dx} = \frac{3x^2}{y}[/tex]-----------------------(1)
this can be rewritten as
[tex]\frac{dy}{dx} =3x^2y^{-1}[/tex]
Now differentiating again with respect to x
[tex]\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{dx}{dy}[/tex]
Now substituting (1) we get
[tex]\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{3x^2}{y}[/tex]
[tex]\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{3x^2}{y^2}) \cdot \frac{3x^2}{y}[/tex]
[tex]\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{9x^4}{y^3})[/tex]
At(1,2) [tex]\frac{d^2y}{dx^2} = \frac{6(1)^2}{2} + ( \frac{9(1)^4}{(2)^3})[/tex]
[tex]\frac{d^2y}{dx^2} = \frac{6}{2} + ( \frac{9}{8})[/tex]
[tex]\frac{d^2y}{dx^2} = 3 + 1.125[/tex]
[tex]\frac{d^2y}{dx^2} = 4.125[/tex]
