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

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Answer 1 · 2016-11-04T17:47:06

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

$5=4x^3-4y^2$

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

$8ydy/dx=12x^2$

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

To find the second derivative use the quotient rule

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

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

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

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

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

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

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

How do you find (d^2y)/(dx^2)(d^2y)/(dx^2) for 5=4x^3-4y^25=4x^3-4y^2?