Question

use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$. write the solutions in terms of x and y only.
$y^{2}-2x = 7 - y$
$\frac{dy}{dx}=\frac{2}{2y + 1}$
$\frac{d^{2}y}{dx^{2}}=square$

Explanation:

Step1: Differentiate given equation

Differentiate $y^{2}-2x = 7 - y$ with respect to $x$. Using the chain - rule on $y^{2}$ gives $2y\frac{dy}{dx}$, the derivative of $-2x$ is $-2$, the derivative of the constant $7$ is $0$, and the derivative of $-y$ is $-\frac{dy}{dx}$. So, $2y\frac{dy}{dx}-2=-\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the equation $2y\frac{dy}{dx}-2 = -\frac{dy}{dx}$ to get all terms with $\frac{dy}{dx}$ on one side: $2y\frac{dy}{dx}+\frac{dy}{dx}=2$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y + 1)=2$, then $\frac{dy}{dx}=\frac{2}{2y + 1}$.

Step3: Differentiate $\frac{dy}{dx}$ to get $\frac{d^{2}y}{dx^{2}}$

Use the quotient rule, if $u = 2$ and $v=2y + 1$, then $\frac{d^{2}y}{dx^{2}}=\frac{u'v - uv'}{v^{2}}$. Here, $u' = 0$, and $v'=2\frac{dy}{dx}$. Substitute $\frac{dy}{dx}=\frac{2}{2y + 1}$ into $v'$. So $v' = 2\times\frac{2}{2y+1}=\frac{4}{2y + 1}$. Then $\frac{d^{2}y}{dx^{2}}=\frac{0\times(2y + 1)-2\times\frac{4}{2y + 1}}{(2y + 1)^{2}}=-\frac{8}{(2y + 1)^{3}}$.

Answer:

$-\frac{8}{(2y + 1)^{3}}$