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

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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 $dy/dx = 0.6359991$

How do you implicitly differentiate 2(x^2+y^2)/x = 3(x^2-y^2)/y2(x^2+y^2)/x = 3(x^2-y^2)/y?