use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$. write the solutions…

Explanation:

Step1: Differentiate both sides with respect to x

Differentiating $y^{2}-5x = 4 - 7y$ term - by - term. For the left - hand side, using the chain rule, the derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$, and the derivative of $-5x$ is $-5$. For the right - hand side, the derivative of $4$ is $0$ and the derivative of $-7y$ is $-7\frac{dy}{dx}$. So we have $2y\frac{dy}{dx}-5=-7\frac{dy}{dx}$.

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

Rearrange the equation $2y\frac{dy}{dx}-5=-7\frac{dy}{dx}$ to get all terms with $\frac{dy}{dx}$ on one side: $2y\frac{dy}{dx}+7\frac{dy}{dx}=5$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y + 7)=5$. Then $\frac{dy}{dx}=\frac{5}{7 + 2y}$.

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

Using the quotient rule, if $u = 5$ and $v=7 + 2y$, then $\frac{du}{dx}=0$ and $\frac{dv}{dx}=2\frac{dy}{dx}$. The quotient rule states that $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Substituting the values, we have $\frac{d^{2}y}{dx^{2}}=\frac{(7 + 2y)\times0-5\times2\frac{dy}{dx}}{(7 + 2y)^{2}}$. Substitute $\frac{dy}{dx}=\frac{5}{7 + 2y}$ into the above formula: $\frac{d^{2}y}{dx^{2}}=\frac{- 10\times\frac{5}{7 + 2y}}{(7 + 2y)^{2}}=-\frac{50}{(7 + 2y)^{3}}$.

Answer:

$\frac{dy}{dx}=\frac{5}{7 + 2y}$, $\frac{d^{2}y}{dx^{2}}=-\frac{50}{(7 + 2y)^{3}}$