How do you determine dy/dx given #x^2y+y=3#?

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Answer 1 · 2016-11-03T13:29:48

#(dy)/(dx)=-(2xy)/(x^2+1)#

#d/dx(x^2y+y=3)#

#d/(dx)(x^2y)+d/(dx)(y)=d/(dx)(3)#

using the product rule on the first term and differentiating as normal for the rest.

#2xy+x^2(dy)/(dx)+(dy)/(dx)=0#

rearranging for #(dy)/(dx)#

#x^2(dy)/(dx)+(dy)/(dx)=-2xy#

#(dy)/(dx)(x^2+1)=-2xy#

#(dy)/(dx)=-(2xy)/(x^2+1)#

Answer 2 · 2016-11-03T14:06:32

Unless I was asked to use implicit differentiation, I would solve for #y# first.

#x^2y+y=3# is equivalent to

#y = 3/(x^2+1) = 3(x^2+1)^-1#

So,

#dy/dx = -3(x^2+1)^-2 * [2x]# #" "# (Use the chain rule)

# = (-6x)/(x^2+1)^2#