Question

Find the second derivative at the point (1,2), given the function below. y^2-2=2x^3

Answer 1

Solution:

Given:

$y^2-2=2x^3$

Lets First Differentiate the given equation with respect to x

$\frac{d}{dx} ( y^2 - 2 ) = \frac{d}{dx} 2x^3$

$2y \cdot \frac{dy}{dx} - 0 = 6x^2$

$\frac{dy}{dx} = \frac{6x^2}{2y}$

$\frac{dy}{dx} = \frac{3x^2}{y}$-----------------------(1)

this can be rewritten as

$\frac{dy}{dx} =3x^2y^{-1}$

Now differentiating again with respect to x

$\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{dx}{dy}$

Now substituting (1) we get

$\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{3x^2}{y}$

$\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{3x^2}{y^2}) \cdot \frac{3x^2}{y}$

$\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{9x^4}{y^3})$

At(1,2) $\frac{d^2y}{dx^2} = \frac{6(1)^2}{2} + ( \frac{9(1)^4}{(2)^3})$

$\frac{d^2y}{dx^2} = \frac{6}{2} + ( \frac{9}{8})$

$\frac{d^2y}{dx^2} = 3 + 1.125$

$\frac{d^2y}{dx^2} = 4.125$