`sqrt(x) + sqrt(y) = 1` Find y'' by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 36 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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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)}

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