Finally, we factor out the $$ y^{\prime} $$ and divide to conclude that the derivative $$ \frac{d y}{d x} $$ is as follows. $$ \frac{d y}{d x}=\square $$

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Assume that after differentiating an equation in $$ x $$ and $$ y $$ we obtained an expression of the form

$$
A(x,y)\,y' + B(x,y) = 0.
$$

The next step is to isolate $$ y' $$. To do this, we first factor out $$ y' $$ from the terms that contain it (if not already factored) so that the equation appears as

$$
y'\,A(x,y) = -B(x,y).
$$

Then, we divide both sides by the coefficient $$ A(x,y) $$—assuming $$ A(x,y) \neq 0 $$—to solve for $$ y' $$:

$$
y' = \frac{-B(x,y)}{A(x,y)}.
$$

Thus, the derivative is

$$
\frac{dy}{dx} = -\frac{B(x,y)}{A(x,y)}.
$$

Without additional context, this is the general form that results from factoring out $$ y' $$ and then dividing by its coefficient.
$$ \frac{dy}{dx} = -\frac{B(x,y)}{A(x,y)} $$

Finally, we factor out the and divide to conclude that the derivative is as follows.

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