How do I find y" of 9x^2+y^2=9 using implicit differentiation?

Answer 1

9x² + y² = 9

By the chain rule, 18x + 2y(dy/dx) = 0

2y(dy/dx) = -18x

dy/dx = -9x/y

So,by the quotient rule, d²y / dx² = [-9y + 9x(dy/dx)] / y²

d²y/dx² = [-9y + 9x(-9x/y)] / y²

= [-9y - 81x²/y] / y²

= -9/y - 81x² / y³

= -9(y² + 9x²) / y³

Answer 2

if 9x² + y² = 9 then y² = 9 - 9x² then 2yy' = -18x or y' = -18x/2y = -9/SQRT(9-9x²)

going further 2(y')² + 2yy" = -18 Then (y')² + yy" = -9

So y" = (-9 -(y')²)/y where y = SQRT(9-9x²) and y' is shown above. Make the substitutions and

you have d²y/dx^{2 in terms of x}