$$\text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. }$$
-$$x ^ { 2 } - y ^ { 3 } = 6$$
A) $\frac { d y } { d x } = \frac { 2 x } { 3 y ^ { 2 } } ; \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 6 y ^ { 3 } - 8 x ^ { 2 } } { 9 y ^ { 5 } }$
B) $\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 2 \mathrm { x } } { 3 \mathrm { y } ^ { 2 } } ; \frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx^{2 } } } = \frac { 5 \mathrm { y } ^ { 3 } - 8 \mathrm { x } ^ { 2 } } { 9 \mathrm { y } ^ { 5 } }$
C) $\frac { d y } { d x } = \frac { 2 x } { 3 y ^ { 2 } } ; \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 6 y ^ { 3 } - 8 x ^ { 2 } } { 9 y ^ { 6 } }$
D) $\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 2 \mathrm { x } } { 3 \mathrm { y } ^ { 2 } } ; \frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } } = \frac { 6 \mathrm { y } ^ { 3 } - 8 \mathrm { x } ^ { 2 } } { 9 \mathrm { y } ^ { 3 } }$

Question

Quizplus · Accepted Answer

Taking the derivative of both sides with respect to $x$ using implicit differentiation:
$2x-3y^2\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{2x}{3y^2}$

Now, taking the derivative of $\frac{dy}{dx}$ using the quotient rule:

\begin{align*}
\frac{d}{dx}\left(\frac{2x}{3y^2}ight)&=\frac{3y^2\cdot2-\frac{d}{dx}(2x)\cdot2\cdot y}{(3y^2)^2}\
\
&=\frac{6y^2-4x}{9y^4}\
\
&=\frac{2y^2-4x}{3y^4}\
\
&=\frac{2(y^3-2x)}{3y^4}\
\
\frac{d^2y}{dx^2}&=\frac{2(y^3-2x)}{3y^4}
\end{align*}

Use implicit differentiation to find dy/dx and d2y/dx2. \text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. } Use implicit differentiation to find dy/dx and d2y/dx2.

$\text { Use implicit differentiation to find } d y / d x \text { and } \mathrm { d } ^ { 2 } y / d x ^ { 2 } \text {. }$