Question

use implicit differentiation to find $\frac{dy}{dx}$. 7xy + y² = x + y $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $7xy + y^{2}=x + y$ with respect to $x$.
For the left - hand side, use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ for $7xy$ (where $u = 7x$ and $v = y$) and the chain rule for $y^{2}$.
The derivative of $7xy$ with respect to $x$ is $7y+7x\frac{dy}{dx}$, and the derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$. The derivative of the right - hand side: the derivative of $x$ with respect to $x$ is $1$, and the derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.
So, $7y + 7x\frac{dy}{dx}+2y\frac{dy}{dx}=1+\frac{dy}{dx}$.

Step2: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side:
$7x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=1 - 7y$.
Factor out $\frac{dy}{dx}$ on the left - hand side:
$\frac{dy}{dx}(7x + 2y-1)=1 - 7y$.

Step3: Solve for $\frac{dy}{dx}$

Divide both sides by $7x + 2y - 1$ to get $\frac{dy}{dx}=\frac{1 - 7y}{7x+2y - 1}$.

Answer:

$\frac{1 - 7y}{7x+2y - 1}$