Question

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

Answer 1

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.