find y by implicit differentiation. simplify where possible.\nx^2 + 2y^2 = 2\ny =

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $x^{2}+2y^{2}=2$ term - by - term. The derivative of $x^{2}$ with respect to $x$ is $2x$. For $2y^{2}$, using the chain - rule $\frac{d}{dx}(2y^{2}) = 4y\frac{dy}{dx}$. The derivative of the constant 2 is 0. So, $2x + 4y\frac{dy}{dx}=0$. $$2x + 4y\frac{dy}{dx}=0$$

Step2: Solve for $\frac{dy}{dx}$

Subtract $2x$ from both sides: $4y\frac{dy}{dx}=-2x$. Then divide both sides by $4y$ (assuming $y\neq0$) to get $\frac{dy}{dx}=-\frac{x}{2y}$. $$\frac{dy}{dx}=-\frac{x}{2y}$$

Step3: Differentiate $\frac{dy}{dx}$ with respect to x to find $y''$

Using the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = - x$ and $v = 2y$. So, $u'=-1$ and $v' = 2\frac{dy}{dx}$. [ \begin{align*} y''&=\frac{-1\times(2y)-(-x)\times(2\frac{dy}{dx})}{(2y)^{2}}\ &=\frac{-2y + 2x\frac{dy}{dx}}{4y^{2}} \end{align*} ]

Step4: Substitute $\frac{dy}{dx}=-\frac{x}{2y}$ into the expression for $y''$

[ \begin{align*} y''&=\frac{-2y+2x\left(-\frac{x}{2y}\right)}{4y^{2}}\ &=\frac{-2y-\frac{x^{2}}{y}}{4y^{2}}\ &=\frac{\frac{-2y^{2}-x^{2}}{y}}{4y^{2}}\ &=\frac{-2y^{2}-x^{2}}{4y^{3}} \end{align*} ] Since $x^{2}+2y^{2}=2$, then $x^{2}=2 - 2y^{2}$. Substitute $x^{2}=2 - 2y^{2}$ into the above expression: [ \begin{align*} y''&=\frac{-2y^{2}-(2 - 2y^{2})}{4y^{3}}\ &=\frac{-2y^{2}-2 + 2y^{2}}{4y^{3}}\ &=-\frac{1}{2y^{3}} \end{align*} ]

Answer:

$-\frac{1}{2y^{3}}$