How do you find (d^2y)/(dx^2) for 5=4x^3-4y^2?

1 Answer
Nov 4, 2016

(d^2y)/(dx^2)=(12xy^2-9x^4)/(4y^3)

Explanation:

5=4x^3-4y^2

0=12x^2-8y dy/dx

8ydy/dx=12x^2

dy/dx=(12x^2)/(8y)=(3x^2)/(2y)

To find the second derivative use the quotient rule

(d^2y)/(dx^2)=(2y*6x-3x^2*2dy/dx)/(4y^2)

(d^2y)/(dx^2)=(2(6xy-3x^2dy/dx))/(4y^2)

(d^2y)/(dx^2)=(6xy-3x^2*(3x^2)/(2y))/(2y^2)

(d^2y)/(dx^2)=(6xy-(9x^4)/(2y))/(2y^2)

(d^2y)/(dx^2)=((12xy^2-9x^4)/(2y))/(2y^2)

(d^2y)/(dx^2)=((12xy^2-9x^4)/(2y))*(1/(2y^2))

(d^2y)/(dx^2)=(12xy^2-9x^4)/(4y^3)