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

##### 1 Answer

#### 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

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

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

Factoring

#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))#