How do you find dy/dx for #y + 2x - 3xy^3 = 4#?

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How do you find dy/dx for #y + 2x - 3xy^3 = 4#?

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Answer 1 · 2015-05-26T23:42:02

You must remember that #y# is a function of #x# so that to take into account this you have to include #(dy)/(dx)# when deriving any #y# in your function, as:
#1(dy)/(dx)+2-3y^3-9xy^2(dy)/(dx)=0#
Collecting #(dy)/(dx)#:
#(dy)/(dx)=(3y^3-2)/(1-9xy^2)#

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How do you find dy/dx for #y + 2x - 3xy^3 = 4#?

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