### A problem asked by Rudra

Given $\sqrt{1-x^4} + \sqrt{1 -y^4} =k(x^2 - y^2)$, find $\frac{dy}{dx}$.

Solution:

Differentiate both sides with respect to x and get

$\frac{x^3}{\sqrt{1-x^4}} + \frac{y^3y'}{\sqrt{1-y^4}} = k(yy' - x)$

or, $yy'\left(k - \frac{y^2}{\sqrt{1-y^4}}\right) = x\left(k + \frac{x^2}{\sqrt{1-x^4}}\right)$

or, $y' = \frac{x}{y} \frac{\sqrt{1-y^4}}{\sqrt{1-x^4}} \frac{ k\sqrt{1-x^4} + x^2 }{k\sqrt{1-y^4} - y^2}$

or, $y' = \frac{x}{y} \frac{\sqrt{1-y^4}}{\sqrt{1-x^4}} \frac{ k(x^2-y^2)(\sqrt{1-x^4} + x^2(x^2-y^2) }{k(x^2-y^2)\sqrt{1-y^4} - y^2(x^2-y^2)}$

or, $y' = \frac{x}{y} \frac{\sqrt{1-y^4}}{\sqrt{1-x^4}} \frac{ 1- x^4 +\sqrt{1-x^4}\sqrt{1-x^4} + x^2(x^2-y^2) }{1+\sqrt{1-x^4}\sqrt{1-y^4} - y^2(x^2-y^2}$

or, $y' = \frac{x}{y} \frac{\sqrt{1-y^4}}{\sqrt{1-x^4}} \frac{ 1 +\sqrt{1-x^4}\sqrt{1-x^4} -x^2y^2) }{1+\sqrt{1-x^4}\sqrt{1-y^4} - x^2y^2}$

or, $y' = \frac{x}{y} \frac{\sqrt{1-y^4}}{\sqrt{1-x^4}}$