sqrt(1-x^(8))+sqrt(1-y^(8))=a[x^(4)-y^(8)]

If sqrt(1-x^(4))+sqrt(1-y^(4))=k(x^(2)-y^(2)), prove that (dy)/(dx)=(x sqrt(1-y^(4)))/(y sqrt(1-x^(4)))

If sqrt(1-x^4) + sqrt(1-y^4) =k(x^2 - y^2) then show that dy/dx = {x sqrt(1-y^4)}/{y sqrt(1-x^4)}

y=Tan^(-1)((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))) then find (8ddy)/(8ddx)

lim_(x rarr 0) (x sqrt(y^(2) - (y - x)^(2)))/((sqrt(8xy - 4x^(2)) + sqrt(8xy))^(3)) equals :

lim_( x to 0)(xsqrt(y^(2)-(y-x)^(2)))/({sqrt((8xy-4x^(2))+sqrt(8xy))}^(3))=

lim_(xrarr0)(xy(sqrt(y^(2)-(y-x)^(2))))/((sqrt(8xy-4x^(2))+sqrt(8xy))^(3))" equals"

The following are the steps involved in finding the value of x-y from (sqrt(8)-sqrt(5))/(sqrt(8)+sqrt(5))=x-ysqrt(40) . Arrange them in sequential order. (A) (13-2sqrt(40))/(8-5)=x-ysqrt(40) (B) ((sqrt(8))^(2)+(sqrt(5))^(2)-2(sqrt(8))(sqrt(5)))/((sqrt(8))^(2)-(sqrt(5))^(2))=x-ysqrt(40) (C) x-y=(11)/(3) (D) x=(13)/(3) and y=(2)/(3) (E) ((sqrt(8)-sqrt(5))(sqrt(8)-sqrt(5)))/((sqrt(8)+sqrt(5))(sqrt(8)-sqrt(5)))=x-ysqrt(40)

(3+sqrt(8))^(x^(2)-2x+1)+(3-sqrt(8))^(x^(2)-2x-1)=((2)/(3-sqrt(8))) then the difference between the greatest value and the least possible value of x is:

The shortest distance between line y-x=1 and curve is : (1)(sqrt(3))/(4) (2) (3sqrt(2))/(8) (3) (8)/(3sqrt(2))(4)(4)/(sqrt(3))