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Brilliant_brown [7]
3 years ago
12

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

Mathematics
1 answer:
lyudmila [28]3 years ago
6 0

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

<u>Explanation:</u>

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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