find $\frac{d^{2}y}{dx^{2}}$.\n$15x^{2}+y^{2}=6$\n$\frac{d^{2}y}{dx^{2}}=square$

Explanation:

Step1: Differentiate implicitly once

Differentiate $15x^{2}+y^{2}=6$ with respect to $x$. Using the power - rule and chain - rule, we have $30x + 2y\frac{dy}{dx}=0$. Then solve for $\frac{dy}{dx}$: [ \begin{align*} 2y\frac{dy}{dx}&=- 30x\ \frac{dy}{dx}&=-\frac{15x}{y} \end{align*} ]

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

Using the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u=-15x$ and $v = y$. $u'=-15$ and $v'=\frac{dy}{dx}$. [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-15y-(-15x)\frac{dy}{dx}}{y^{2}}\ &=\frac{-15y + 15x\frac{dy}{dx}}{y^{2}} \end{align*} ] Substitute $\frac{dy}{dx}=-\frac{15x}{y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{-15y+15x\left(-\frac{15x}{y}\right)}{y^{2}}\ &=\frac{-15y^{2}-225x^{2}}{y^{3}}\ &=-\frac{15(y^{2}+15x^{2})}{y^{3}} \end{align*} ] Since $15x^{2}+y^{2}=6$, then $\frac{d^{2}y}{dx^{2}}=-\frac{15\times6}{y^{3}}=-\frac{90}{y^{3}}$

Answer:

$-\frac{90}{y^{3}}$