Question

How do you differentiate `sqrt (1-x^4) + sqrt (1-y^4)= k(x^2-y^2)`?

Answer 1

`dy/dx=(kxsqrt(1-x^4)+x^3)/(kysqrt(1-y^4)-y^3)sqrt((1-y^4)/(1-x^4))`

Explanation:

Writing as:

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

When we differentiate this, we will use the chain rule frequently. Recall that, as an example, the derivative with respect to `x` of `y^3` is not `3y^2` but `3y^2*dy/dx`.

Differentiating:

`1/2(1-x^4)^(-1/2)(-4x^3)+1/2(1-y^4)^(-1/2)(-4y^3dy/dx)=k(2x-2ydy/dx)`

Expanding and rewriting:

`(-2x^3)/sqrt(1-x^4)+(-2y^3)/sqrt(1-y^4)(dy/dx)=2kx-2ky(dy/dx)`

Rewriting to solve for `dy/dx`, the derivative:

`2ky(dy/dx)-(2y^3)/sqrt(1-y^4)(dy/dx)=2kx+(2x^3)/sqrt(1-x^4)`

Factoring `dy/dx` and dividing through by `2`:

`dy/dx(ky-y^3/sqrt(1-y^4))=kx+x^3/sqrt(1-x^4)`

`dy/dx=(kx+x^3/sqrt(1-x^4))/(ky-y^3/sqrt(1-y^4))`

Personally, this is as far as I want to go, but we can "simplify" by getting common denominators:

`dy/dx=((kxsqrt(1-x^4)+x^3)/sqrt(1-x^4))/((kysqrt(1-y^4)-y^3)/sqrt(1-y^4))`

`dy/dx=(kxsqrt(1-x^4)+x^3)/(kysqrt(1-y^4)-y^3)sqrt((1-y^4)/(1-x^4))`