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

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Answer 1 · 2016-11-07T12:51:13

Please see the explanation section below.

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

Differentiate w.r.t. #x# to get

#-4y dy/dx = 3#

So, #" "# #dy/dx = -3/4 y^-1#

Differentiate again w.r.t. #x# and we find

#(d^2y)/dx^2 =3/4y^-2 dy/dx#

Replace #dy/dx = -3/4y^-1# and simplify

#(d^2y)/dx^2 = -9/(16y^3)#