Question #a9319

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

You had:

#-(x+y)^2 - 3(root(3)(xy)) + 8pi = -12#

Well, with the constants, we can ignore those. Their derivatives with respect to any variable is zero. So, this is really going to have the same derivative as

#-(x+y)^2 - 3(root(3)(xy)) = C#,

where #C# is a constant. Rewrite this as

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

by expanding the square. Now the derivative becomes a bit easier. If we had stuck to #(xy)^"1/3"#, we may have gotten something seemingly inseparable...

Taking this as #f(x,y) = C#, and noting the product rule, differentiate with respect to #x# to get:

#-2x - 2(xcdot(dy)/(dx) + y) - 2y(dy)/(dx) - 3*[x^"1/3"1/3y^-"2/3"(dy)/(dx) + y^"1/3"cdot1/3x^-"2/3"] = 0#

Note that when you take the derivative with respect to #x#, since #y# is a function of #x#, we have that

#(d)/(dx)[f(y)] = (df(y))/(dx) = (df(y))/(dy)(dy)/(dx)#.

For instance, #d/(dx)[2y^2] = 4y(dy)/(dx)#, since #(df(y))/(dy) = 4y#.

Simplify first, then try to separate out the #(dy)/(dx)# to one side and everything else to the other side.

#-2x - color(red)(2x(dy)/(dx)) - 2y - color(red)(2y(dy)/(dx)) - color(red)(x^"1/3"y^-"2/3"(dy)/(dx)) - y^"1/3"x^-"2/3" = 0#

#-2x - 2y - y^"1/3"x^-"2/3" = color(red)(2x(dy)/(dx) + 2y(dy)/(dx) + x^"1/3"y^-"2/3"(dy)/(dx))#

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

#=> color(blue)((dy)/(dx) = -(2x + 2y + y^"1/3"x^-"2/3")/(2x + 2y + x^"1/3"y^-"2/3"))#

Wolfram Alpha gives:

#(dy)/(dx) = -(2x(xy)^"2/3"+2y(xy)^"2/3"+y)/(2x(xy)^"2/3"+2y(xy)^"2/3"+x)#

We could verify we got it by matching that; simply multiply by #(xy)^"2/3"/(xy)^"2/3"#:

#(xy)^"2/3"/(xy)^"2/3"[-(2x + 2y + y^"1/3"x^-"2/3")/(2x + 2y + x^"1/3"y^-"2/3")]#

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

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

#= -(2x(xy)^"2/3" + 2y(xy)^"2/3" + y)/(2x(xy)^"2/3" + 2y(xy)^"2/3" + x# #color(blue)(sqrt"")#

Yep, we got it!