Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 38

Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number
2) If y = u*v ; where both u & v are functions of 'x' , then
dy/dx = u*(dv/dx) + v*(du/dx)
3) If y = k ; where 'k' = constant ; then dy/dx = 0
Now, the given function is :-
(x^4) + (y^4) = a^4
Differentiating both sides w.r.t 'x' we get;
4(x^3) + 4(y^3)*(dy/dx) = 0 .........(1)
or, dy/dx = -(x^3)/(y^3)..........(2)
Differentiating (1) again w.r.t 'x' we get
12(x^2) + {12(y^2)*(dy/dx)^2} + [(y^3)*y"] = 0..........(3)
Putting the value of dy/dx from (2) in (3) we get
12(x^2) + 12{(x^6)/(y^4)} + [(y^3)*y"] = 0
or, y" = -[(12(x^2)*{1 + (x^4)/(y^4)}/(y^3)
or, y" = -[(12(x^2)*{(y^4) + (x^4)}]/(y^7)