use implicit differentiation to find $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$.\n$2x^{2}+y^{2}=1$\n$\frac{d…

Explanation:

Step1: Differentiate both sides with respect to x

Differentiating $2x^{2}+y^{2}=1$ term - by - term. The derivative of $2x^{2}$ with respect to $x$ is $4x$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. For $y^{2}$, by the chain - rule $\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}$. The derivative of the constant 1 is 0. So, $4x + 2y\frac{dy}{dx}=0$.

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

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

Step3: Differentiate $\frac{dy}{dx}$ to find $\frac{d^{2}y}{dx^{2}}$

Using the quotient rule $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$, where $u = - 2x$ and $v = y$. $\frac{du}{dx}=-2$ and $\frac{dv}{dx}=\frac{dy}{dx}=-\frac{2x}{y}$. Then $\frac{d^{2}y}{dx^{2}}=\frac{y(-2)-(-2x)\frac{dy}{dx}}{y^{2}}$. Substitute $\frac{dy}{dx}=-\frac{2x}{y}$ into the above formula: $\frac{d^{2}y}{dx^{2}}=\frac{-2y-(-2x)(-\frac{2x}{y})}{y^{2}}=\frac{-2y-\frac{4x^{2}}{y}}{y^{2}}=\frac{-2y^{2}-4x^{2}}{y^{3}}$. Since $2x^{2}+y^{2}=1$ or $4x^{2}+2y^{2}=2$, then $\frac{d^{2}y}{dx^{2}}=-\frac{2}{y^{3}}$.

Answer:

$\frac{dy}{dx}=-\frac{2x}{y}$, $\frac{d^{2}y}{dx^{2}}=-\frac{2}{y^{3}}$