Find the second derivative at the point (1,1), given the function below. 2x^9+1=3y^3

Question

Find the second derivative at the point (1,1), given the function below. 2x^9+1=3y^3

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Raelynn 3 weeks 2021-11-16T04:30:04+00:00 1 Answer 0 views 0

Answers ( )

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    2021-11-16T04:31:11+00:00

    The value at point ( 1, 1 ) is 8.

    Explanation:

    Given

    3y³ = 2x⁹ + 1

    y³ = 2/3x⁹ + 1/3

    y = ∛2/3 x⁹ + 1/3

    We need to first find dy/dx which is the first derivative and then d²y/dx² which is the second derivative

    \frac{dy}{dx} = \frac{2x^8}{y^2}

    \frac{d^{2}y }{dx^2} = \frac{16x^7}{y^2} - \frac{8x^1^6}{y^5}

    Put the value of (1,1) : x = 1 and y = 1 in d²y/dx² equation

    \frac{d^2y}{dx^2}  = \frac{16 X 1^7}{1^2} - \frac{8 X 1^1^6}{1^5}

    \frac{d^2y}{dx^2} = 16 - 8\\\\\frac{d^2y}{dx^2}  = 8

    Therefore, the value is 8.

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45:7+7-4:2-5:5*4+35:2 =? ( )