`x^4 + y^4 = a^4` Find y'' by implicit differentiation.

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Chapter 3, 3.5 - Problem 38 - 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^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)

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