use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$. $x^{\frac{2}{3}}+y^{\f…

Explanation:

Step1: Recall quotient - rule for differentiation

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$, where $u = y^{\frac{1}{3}}$ and $v = x^{\frac{1}{3}}$.

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

First, find $u^\prime$ and $v^\prime$ using the chain - rule. If $u = y^{\frac{1}{3}}$, then $u^\prime=\frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx}$. If $v = x^{\frac{1}{3}}$, then $v^\prime=\frac{1}{3}x^{-\frac{2}{3}}$. By the quotient - rule: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=-\frac{\frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx}\cdot x^{\frac{1}{3}}-y^{\frac{1}{3}}\cdot\frac{1}{3}x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}\ \end{align*} ] Substitute $\frac{dy}{dx}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=-\frac{\frac{1}{3}y^{-\frac{2}{3}}\left(-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right)\cdot x^{\frac{1}{3}}-y^{\frac{1}{3}}\cdot\frac{1}{3}x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}\ &=-\frac{-\frac{1}{3}y^{-\frac{1}{3}}-\frac{1}{3}y^{\frac{1}{3}}x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}\ &=\frac{\frac{1}{3}y^{-\frac{1}{3}}+\frac{1}{3}y^{\frac{1}{3}}x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}\ &=\frac{1}{3}\cdot\frac{y^{-\frac{1}{3}}x^{\frac{2}{3}} + y^{\frac{1}{3}}}{x^{\frac{4}{3}}y^{\frac{1}{3}}}\ &=\frac{147}{9x^{\frac{4}{3}}y^{\frac{1}{3}}} \end{align*} ] Since $x^{\frac{2}{3}}+y^{\frac{2}{3}} = 147$, we can also simplify in another way. [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{49}{x^{\frac{4}{3}}y^{\frac{1}{3}}} \end{align*} ]

Answer:

$\frac{49}{x^{\frac{4}{3}}y^{\frac{1}{3}}}$