Question

`(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)` ?

Answer 1

`(dy)/(dx) =(x/y) ( sqrt[1 - y^4])/(sqrt[1 - x^4] )`

Explanation:

If `(1-x^4)^(1/2) + (1-y^4)^(1/2) =k(x^2-y^2)` then

`f(x,y) = ((1-x^4)^(1/2) + (1-y^4)^(1/2))/(x^2-y^2)=k`

now

`df = f_x dx + f_y 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

`(dy)/(dx) = - (f_x)/(f_y) =(x/y) ( 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(y(x),x) = ((1-x^4)^(1/2) + (1-y(x)^4)^(1/2))/(x^2-y(x)^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'(x)` we find

`y'(x) = (x/(y(x)))sqrt(1-y(x)^4)/sqrt(1-x^4)`