How does $\frac{d^2y}{dx^2} = \frac{-1}{4y^3}$ if $x-y^2 = 1$? I'm getting conflicting answers on all the online calculators.
2026-03-28 22:24:30.1774736670
second derivative implicit differentiation.
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you have $$x-y^2=1$$ and $$y=y(x)$$ thus we get $$1-2yy'=0$$ or $$y'=\frac{1}{2}y^{-1}$$ from here we get $$y''=\frac{1}{2}(-1)y^{-2}y'$$ and you can plug $$y'$$ in this equation $$y''=-\frac{1}{4y^3}$$