The 2nd derivative of $x^2+y^2=4$

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Finding the first derivative of the function $x^2+y^2=4$ gives you$$y'=-\frac{x}{y} $$Then to find the 2nd derivative you apply the quotient rule, which looks like this$$y''=\frac{y(-1)-(-x)y'}{y^2}$$which gives you $$\frac{-y+xy'}{y^2} $$But after looking in the back of the book I realized my answer was wrong.

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3 Answers

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Differentiating $x^2+y^2=4$ twice with respect to $x$ gives$$ 2+2(y')^2+2yy''=0 $$so$$ yy''=-1-(y')^2=-1-\frac{x^2}{y^2}=-\frac{x^2+y^2}{y^2}=-\frac4{y^2}. $$Hence$$ y''=-\frac4{y^3}. $$

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From $$y'=-\frac{x}{y}$$ we get$$y''=-\frac{y-xy'}{y^2}=-\frac{y}{y^2}+\frac{x}{y^2}y'$$ and this is equal to$$y''=-\frac{y}{x^2}+\frac{x}{y^2}\cdot \left(-\frac{x}{y}\right)$$ which can simplified to$$y''=-\frac{1}{y}-\frac{x^2}{y^3}$$

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Let $\color{blue}{x =2\cos\theta}$ and $\color{blue}{y= 2\sin\theta}$ which satisfies the given equation.

$$\frac{dy}{dx} =\frac{{dy}/{d\theta}}{{dx}/{d\theta}} = -\cot\theta $$

$$\frac{d^2y}{dx^2} = \frac{d( -\cot\theta )}{d\theta}\frac{d\theta}{dx} =- \csc^2\theta\frac{1}{2\sin\theta} = \frac{-1}{2\sin^3\theta} = -\frac{4}{y^3}$$

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