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

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.)

Accepted Answer

# Explanation:
## Step1: Differentiate $x^{3}+y^{3}=19$ implicitly
Differentiate both sides with respect to $x$:
$3x^{2}+3y^{2}\frac{dy}{dx}=0$
Solve for $\frac{dy}{dx}$:
$\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$
## Step2: Differentiate $\frac{dy}{dx}$ to find $\frac{d^{2}y}{dx^{2}}$
Use the quotient rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = -x^{2}$, $u^\prime=-2x$, $v = y^{2}$, $v^\prime = 2y\frac{dy}{dx}$
$\frac{d^{2}y}{dx^{2}}=\frac{-2x\cdot y^{2}-(-x^{2})\cdot2y\frac{dy}{dx}}{y^{4}}$
Substitute $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$ into the above formula:
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{-2xy^{2}+2x^{2}y\cdot\frac{x^{2}}{y^{2}}}{y^{4}}\
&=\frac{-2xy^{3}+2x^{4}}{y^{5}}
\end{align*}
$$
## Step3: Substitute $x = - 2$, $y = 3$
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{-2(-2)(3)^{3}+2(-2)^{4}}{3^{5}}\
&=\frac{4\times27 + 2\times16}{243}\
&=\frac{108 + 32}{243}\
&=\frac{140}{243}
\end{align*}
$$
# Answer:
$\frac{140}{243}$

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QUESTION IMAGE

Question

Explanation:

Step1: Differentiate \(x^{3}+y^{3}=19\) implicitly

Step2: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^{2}y}{dx^{2}}\)

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

Answer: