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

Jun 15, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0.6359991$

Explanation:

$F \left(x , y\right) = 2 \frac{{x}^{2} + {y}^{2}}{x} - 3 \frac{{x}^{2} - {y}^{2}}{y} = 0$ is an homogeneous function so susbtituting $y = \lambda x$ we obtain

$F \left(x , \lambda x\right) = \frac{x \left(- 3 + 2 \lambda + 3 {\lambda}^{2} + 2 {\lambda}^{3}\right)}{\lambda} = 0$ 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 $\lambda$.

$\lambda = 0.6359991$.

$F \left(x , y\right) = 0 \equiv y = 0.6359991 x$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = 0.6359991$