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

Question

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.

Answer 1

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 $$y''=-\frac{1}{y}-\frac{x^2}{y^3}$$

Answer 2

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 $$ y''=-\frac4{y^3}. $$

Answer 3

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

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

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