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)#