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

Question

William Abdalla · Accepted Answer

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

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))#

Samantha Adkison · Answer

undefined To differentiate the equation $ \sqrt{1-x^4} + \sqrt{1-y^4} = k(x^2-y^2) $ with respect to $ x $ and $ y $, you can follow these steps:

1. Square both sides of the equation.
2. Differentiate both sides with respect to $ x $ and $ y $ separately.
3. Solve for $ \frac{dy}{dx} $ and $ \frac{dx}{dy} $.

This process involves some algebraic manipulations and differentiation techniques such as the chain rule and implicit differentiation.