math 2110 (calc find $$ \frac{d y}{d x} $$ 4) $$ y^{2} x^{2}+x y=7 $$

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We start with the equation
$$
x^2 y^2 + xy = 7.
$$
Differentiate both sides with respect to $$ x $$:

1. Differentiating the first term $$ x^2 y^2 $$ using the product rule:
   $$
   \frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(x^2) \cdot y^2 + x^2 \cdot \frac{d}{dx}(y^2).
   $$
   Note that
   $$
   \frac{d}{dx}(x^2) = 2x \quad \text{and} \quad \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.
   $$
   Thus,
   $$
   \frac{d}{dx}(x^2 y^2) = 2x y^2 + 2x^2 y \frac{dy}{dx}.
   $$

2. Differentiating the second term $$ xy $$ using the product rule:
   $$
   \frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}.
   $$

3. The derivative of the constant $$ 7 $$ is $$ 0 $$.

Putting both differentiated parts together, we have:
$$
2xy^2 + 2x^2y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0.
$$

4. Group the terms containing $$\frac{dy}{dx}$$:
$$
(2x^2y + x) \frac{dy}{dx} + (2xy^2 + y) = 0.
$$

5. Isolate the term with $$\frac{dy}{dx}$$:
$$
(2x^2y + x) \frac{dy}{dx} = - (2xy^2 + y).
$$

6. Solve for $$\frac{dy}{dx}$$:
$$
\frac{dy}{dx} = \frac{ - (2xy^2 + y) }{ 2x^2y + x }.
$$

7. Factor common factors in the numerator and denominator:
   - In the numerator, factor $$ y $$: $$ 2xy^2 + y = y(2xy + 1) $$.
   - In the denominator, factor $$ x $$: $$ 2x^2y + x = x(2xy + 1) $$.
   
Thus,
$$
\frac{dy}{dx} = \frac{ - y(2xy + 1) }{ x(2xy + 1) }.
$$

8. Cancel the common factor $$ 2xy + 1 $$ (assuming $$ 2xy + 1 \neq 0 $$):
$$
\frac{dy}{dx} = -\frac{y}{x}.
$$

The final derivative is
$$
\boxed{\frac{dy}{dx} = -\frac{y}{x}}.
$$
$$ \frac{dy}{dx} = -\frac{y}{x} $$

math 2110 (calc find \( \frac{d y}{d x} \) 4) \( y^{2} x^{2}+x y=7 \)

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