# (1-x^4)^½ + (1-y^4)^½ =k(x^2-y^2) then show that dy/dx = x*√(1-y^4)/y*√(1-x^4) ?

Oct 1, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{x}{y}\right) \frac{\sqrt{1 - {y}^{4}}}{\sqrt{1 - {x}^{4}}}$

#### Explanation:

If ${\left(1 - {x}^{4}\right)}^{\frac{1}{2}} + {\left(1 - {y}^{4}\right)}^{\frac{1}{2}} = k \left({x}^{2} - {y}^{2}\right)$ then

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

now

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$ with

f_x = -(2 (x - x^3 y^2 + x sqrt[1 - x^4] sqrt[1 - y^4]))/( sqrt[1 - x^4] (x^2 - y^2)^2)

f_y = (2 (y - x^2 y^3 + sqrt[1 - x^4] y sqrt[1 - y^4]))/((x^2 - y^2)^2 sqrt[ 1 - y^4])

so

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}} = \left(\frac{x}{y}\right) \frac{\sqrt{1 - {y}^{4}}}{\sqrt{1 - {x}^{4}}}$

Note: If you still didn't learn how to calculate partial derivatives, you can proceed as follows.

$g \left(y \left(x\right) , x\right) = \frac{{\left(1 - {x}^{4}\right)}^{\frac{1}{2}} + {\left(1 - y {\left(x\right)}^{4}\right)}^{\frac{1}{2}}}{{x}^{2} - y {\left(x\right)}^{2}} - k = 0$ so

d/dx g(y(x),x) = -((sqrt[1 - x^4] + sqrt[1 - y(x)^4]) (2 x - 2 y(x)y'(x)))/(x^2 - y(x)^2)^2 + -((2 x^3)/sqrt[ 1 - x^4] - (2 y(x)^3 y'(x))/sqrt[1 - y(x)^4])/( x^2 - y(x)^2) = 0

Now solving for $y ' \left(x\right)$ we find

$y ' \left(x\right) = \left(\frac{x}{y \left(x\right)}\right) \frac{\sqrt{1 - y {\left(x\right)}^{4}}}{\sqrt{1 - {x}^{4}}}$