`x^3 + y^3 = 1` Find y'' by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 37 - 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^3) + (y^3) = 1

Differentiating both sides w.r.t 'x' we get;

3(x^2) + 3(y^2)*(dy/dx) = 0

or, (x^2) + (y^2)*(dy/dx) = 0 .........(1)

or, dy/dx = -(x^2)/(y^2)..........(2)

Differentiating (1) again w.r.t 'x' we get

2x  + 2y*{(dy/dx)^2} + [(y^2)*y"] = 0..........(3)

Putting the value of dy/dx from (2) in (3) we get

2x + 2{(x^4)/(y^3)} + [(y^2)*y"] = 0

or, y" = -[2{x(y^3) + (x^4)}/(y^4)

 

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