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find the second derivative of 2x^3-3y^2=8​

Sagot :

Answer:

[tex]\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}[/tex]

Step-by-step explanation:

Find the first implicit derivative using implicit differentiation

[tex]2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}[/tex]

Use the substitution of dy/dx to find the second derivative (d²y/dx²)

[tex]\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}[/tex]

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