find $\frac{d^{2}y}{dx^{2}}$.\n$6x^{2}+y^{2}=1$\n$\frac{d^{2}y}{dx^{2}}=square$

Explanation:

Step1: Differentiate implicitly once

Differentiate $6x^{2}+y^{2}=1$ with respect to $x$. Using the power - rule and chain - rule, we have $12x + 2y\frac{dy}{dx}=0$. Then solve for $\frac{dy}{dx}$: [ \begin{align*} 2y\frac{dy}{dx}&=- 12x\ \frac{dy}{dx}&=-\frac{6x}{y} \end{align*} ]

Step2: Differentiate $\frac{dy}{dx}$ with respect to $x$

Use the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = - 6x$, $u'=-6$, $v = y$, and $v'=\frac{dy}{dx}$. [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-6y-(-6x)\frac{dy}{dx}}{y^{2}}\ &=\frac{-6y + 6x\frac{dy}{dx}}{y^{2}} \end{align*} ] Substitute $\frac{dy}{dx}=-\frac{6x}{y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-6y+6x\left(-\frac{6x}{y}\right)}{y^{2}}\ &=\frac{-6y^{2}- 36x^{2}}{y^{3}}\ &=-\frac{6y^{2}+36x^{2}}{y^{3}}\ &=-\frac{6(y^{2}+6x^{2})}{y^{3}} \end{align*} ] Since $6x^{2}+y^{2}=1$, then $\frac{d^{2}y}{dx^{2}}=-\frac{6}{y^{3}}$

Answer:

$-\frac{6}{y^{3}}$