`9x^2 + y^2 = 9` Find y'' by implicit differentiation.

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Chapter 3, 3.5 - Problem 35 - 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 :-

9(x^2) + (y^2) = 9

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

18x + 2y(dy/dx) = 0

or, 9x + y(dy/dx) = 0 .........(1)

or, dy/dx = -(9x)/(y)..........(2)

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

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

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

9 + 81{(x^2)/(y^2)} + [(y)*y"] = 0

or, y" = -[(9{1 + 9(x^2)/(y^2)}/(y)

or, y" = -[9{(y^2) + 9(x^4)}]/(y^3)

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