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Find y'' by implicit differentiation: 9x^2+y^2 = 9
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messed up the quotient rule...sigh
...
\[9x^2 +y^2 =9 \] \[\text{ } 18x +2y(dy/dx) = 0\] \[dy/dx=-(18x/2y)\] \[18 +2(dy/dx) + 2y(d^2y/dx^2) =0 \] \[d^2y/dx^2 = -(18 + 2(dy/dx))/2y\] now sub in dy/dx \[d^2y/dx^2 = -(18 - 2(18x/2y))/2y = 36(x - y)/(2y)^2\]
no need for the quotient rule with that method
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