How do you implicitly differentiate #2(x^2+y^2)/x = 3(x^2-y^2)/y#?

Question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-15T15:02:16

#dy/dx = 0.6359991#

#F(x,y)=2 (x^2 + y^2)/x - 3 (x^2 - y^2)/y = 0# is an homogeneous function so susbtituting #y = lambda x# we obtain

#F(x, lambda x) = (x (-3 + 2 lambda+ 3 lambda^2 + 2lambda^3))/lambda = 0# with one real solution which is

#lambda = 1/2 ((63 + 2 sqrt[993])^(1/3)/3^(2/3) - 1/(3 (63 + 2 sqrt[993]))^(1/3)-1)#

discarding #x=0,lambda=0# and considering only real values for #lambda#.

#lambda =0.6359991#.

#F(x,y)=0 equiv y = 0.6359991 x#

so #dy/dx = 0.6359991#