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Semmy [17]
3 years ago
15

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

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
5 0

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

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