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. $y^{2}-2x = 7 - y$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $y^{2}-2x = 7 - y$ with respect to $x$. Using the chain - rule for $y^{2}$ and $y$ terms. The derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$, the derivative of $-2x$ is $-2$, the derivative of $7$ is $0$, and the derivative of $-y$ is $-\frac{dy}{dx}$. So we have $2y\frac{dy}{dx}-2=0 - \frac{dy}{dx}$.

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

Add $\frac{dy}{dx}$ to both sides and add $2$ to both sides: $2y\frac{dy}{dx}+\frac{dy}{dx}=2$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2y + 1)=2$. Then $\frac{dy}{dx}=\frac{2}{2y + 1}$.

Answer:

$\frac{2}{2y + 1}$