if ( x^{3}+y^{3}=19 ), find the value of ( \frac{d^{2} y}{d x^{2}} ) at the point ( (-2,3) ).
the value of ( \frac{d^{2} y}{d x^{2}} ) at the point ( (-2,3) ) is (type a simplified fraction.)

Question

Accepted Answer

# Explanation:
## Step1: Differentiate implicitly for first - derivative
Differentiate $x^{3}+y^{3}=19$ with respect to $x$.
Using the power rule $(u^{n})^\prime = nu^{n - 1}u^\prime$, we have $3x^{2}+3y^{2}\frac{dy}{dx}=0$.
Solve for $\frac{dy}{dx}$:
$$
\begin{align*}
3y^{2}\frac{dy}{dx}&=- 3x^{2}\
\frac{dy}{dx}&=-\frac{x^{2}}{y^{2}}
\end{align*}
$$
## Step2: Differentiate $\frac{dy}{dx}$ for second - derivative
Use the quotient rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=-x^{2}$, $u^\prime=-2x$, $v = y^{2}$, $v^\prime = 2y\frac{dy}{dx}$.
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{(-2x)y^{2}-(-x^{2})(2y\frac{dy}{dx})}{y^{4}}\
&=\frac{-2xy^{2}+2x^{2}y\frac{dy}{dx}}{y^{4}}\
&=\frac{-2x}{y^{2}}+\frac{2x^{2}\frac{dy}{dx}}{y^{3}}
\end{align*}
$$
Substitute $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$ into the above formula:
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{-2x}{y^{2}}+\frac{2x^{2}(-\frac{x^{2}}{y^{2}})}{y^{3}}\
&=\frac{-2x}{y^{2}}-\frac{2x^{4}}{y^{5}}
\end{align*}
$$
## Step3: Substitute $x=-2$, $y = 3$
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{-2(-2)}{3^{2}}-\frac{2(-2)^{4}}{3^{5}}\
&=\frac{4}{9}-\frac{32}{243}\
&=\frac{4\times27 - 32}{243}\
&=\frac{108 - 32}{243}\
&=\frac{76}{243}
\end{align*}
$$

# Answer:
$\frac{76}{243}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Differentiate implicitly for first - derivative

Step2: Differentiate \(\frac{dy}{dx}\) for second - derivative

Step3: Substitute \(x=-2\), \(y = 3\)

Answer: