find $\frac{d^{2}y}{dx^{2}}$ implicitly in terms of $x$ and $y$.\n$x^{2}+y^{2}=8$

Explanation:

Step1: Differentiate the given equation

Differentiate $x^{2}+y^{2}=8$ with respect to $x$. Using the power - rule and chain - rule, we have $2x + 2y\frac{dy}{dx}=0$. Solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx}=-2x$, so $\frac{dy}{dx}=-\frac{x}{y}$.

Step2: Differentiate $\frac{dy}{dx}$ to get $\frac{d^{2}y}{dx^{2}}$

Using the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = - x$ and $v = y$. $u'=-1$ and $v'=\frac{dy}{dx}=-\frac{x}{y}$. $\frac{d^{2}y}{dx^{2}}=\frac{-y-(-x)\frac{dy}{dx}}{y^{2}}$. Substitute $\frac{dy}{dx}=-\frac{x}{y}$ into the above formula: $\frac{d^{2}y}{dx^{2}}=\frac{-y + x\left(-\frac{x}{y}\right)}{y^{2}}=\frac{-y^{2}-x^{2}}{y^{3}}$. Since $x^{2}+y^{2}=8$, we can rewrite it as $\frac{d^{2}y}{dx^{2}}=-\frac{8}{y^{3}}$.

Answer:

$-\frac{8}{y^{3}}$