Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 36

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^(1/2) + y^(1/2) = 1
Differentiating both sides w.r.t 'x' we get;
(1/2)*x^(-1/2) + (1/2)*{y^(-1/2)}*(dy/dx)
or, x^(-1/2) + {y^(-1/2)}(dy/dx) = 0.........(1)
or, dy/dx = -{(x/y)^(-1/2)}..........(2)
Differentiating (1) again w.r.t 'x' we get
-(1/2)*x^(-3/2) - (1/2)*y^(-3/2)*{(dy/dx)^2} + [{y^(-1/2)}*y"]=0.......(3)
Putting the value of dy/dx from (2) in (3) we get
-(1/2)*x^(-3/2) - (1/2)*x*{y^(-1/2)}+ [{y^(-1/2)}*y"] = 0
or, y" = (1/2)*[x^(-3/2) + x*{y^(-1/2)}]/{y^(-1/2)}