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

# 2x^2-3y^2=4 #

Differentiating implicitly we get;
# 4x-6ydy/dx=0 #
# :. dy/dx=(4x)/(6y) #
# :. dy/dx=(2x)/(3y) #

So, using the quotient rule
# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
the second derivative is given by:

# (d^2y)/(dx^2) = ((3y)(d/dx2x)-(2x)(d/dx3y))/(3y)^2#
# :. (d^2y)/(dx^2) = (6y-(2x)(dy/dxd/dy3y))/(9y^2)#
# :. (d^2y)/(dx^2) = (6y-(2x)(3dy/dx))/(9y^2)#
# :. (d^2y)/(dx^2) = (6y-6xdy/dx)/(9y^2)#
# :. (d^2y)/(dx^2) = (6y-6x((2x)/(3y)))/(9y^2)#
# :. (d^2y)/(dx^2) = (6y-4x^2/y)/(9y^2)#
# :. (d^2y)/(dx^2) = (6y^2-4x^2)/(9y^3)#
# :. (d^2y)/(dx^2) = 2(3y^2-2x^2)/(9y^3)#

But # 2x^2-3y^2=4 => 3y^2-2x^2=-4 #
# :. (d^2y)/(dx^2) = 2(-4)/(9y^3)#
# :. (d^2y)/(dx^2) = -8/(9y^3)#

Computing the second implicit differentiation is determined by computing the first implicit derivative .

#(2dx^2)/dx-(3dy^2)/dx=(d4)/dx#
#rArr2(2x)-3xx2yxx(dy)/dx=0#
#rArr4x-6yxx(dy)/dx=0#
#rArr-6yxx(dy)/dx=-4x#
#rArr(dy)/dx=(-4x)/(-6y)#

#rArrcolor(blue)((dy)/dx=(2x)/(3y))#

#color(red)((d^2y)/dx^2=??)#

#color(red)((d^2y)/dx^2)=((d(2x))/(dx)xx3y-3((dy)/dx)xx2x)/(3y)^2#

#color(red)((d^2y)/dx^2)=(2xx3y-6x((dy)/dx))/(3y)^2#

#color(red)((d^2y)/dx^2)=(6y-6x((dy)/dx))/(9y^2)#

#color(red)((d^2y)/dx^2)=(2y-2x((dy)/dx))/(3y^2)#