What is the implicit derivative of #4= xy^2-y^3+xy #?

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Answer 1 · 2016-10-26T09:50:11

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

Formally is quite easy. Calling #f(x,y) = x y^2 - y^3 + x y - 4=0# we have:

#df=f_x dx + f_y dy = 0# so

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

but suppose you need #dy/dx# at #x_0=1#. In this case you have to solve first

#f(1,y)=y^2-y^3+y-4=0# giving

#y_0=1/3 (1 - 4 (2/(97 - 9 sqrt[113]))^(1/3) - (1/2 (97 - 9 sqrt[113]))^( 1/3)) = -1.48558#
as the real solution, and then

#dy/dx)_(x_0,y_0) = 0.0839585#