use implicit differentiation to find \\( \\frac { d y } { d x } \\) for the equation\n\n\\( \\sqrt { x + y }…

Explanation:

Step1: Differentiate both sides

Differentiate $\sqrt{x + y}$ and $8 + x^{2}y^{2}$ with respect to $x$. For the left - hand side, using the chain rule, if $u=x + y$, then $\frac{d}{dx}(\sqrt{u})=\frac{1}{2\sqrt{u}}\cdot(1+\frac{dy}{dx})=\frac{1 + \frac{dy}{dx}}{2\sqrt{x + y}}$. For the right - hand side, using the product rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = x^{2}$ and $v=y^{2}$, we have $\frac{d}{dx}(8+x^{2}y^{2})=2xy^{2}+2x^{2}y\frac{dy}{dx}$. So, $\frac{1+\frac{dy}{dx}}{2\sqrt{x + y}}=2xy^{2}+2x^{2}y\frac{dy}{dx}$.

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

Multiply both sides by $2\sqrt{x + y}$ to get $1+\frac{dy}{dx}=4xy^{2}\sqrt{x + y}+4x^{2}y\sqrt{x + y}\frac{dy}{dx}$. Rearrange terms: $\frac{dy}{dx}-4x^{2}y\sqrt{x + y}\frac{dy}{dx}=4xy^{2}\sqrt{x + y}-1$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(1 - 4x^{2}y\sqrt{x + y})=4xy^{2}\sqrt{x + y}-1$. Then $\frac{dy}{dx}=\frac{4xy^{2}\sqrt{x + y}-1}{1 - 4x^{2}y\sqrt{x + y}}$.

Answer:

$\frac{4xy^{2}\sqrt{x + y}-1}{1 - 4x^{2}y\sqrt{x + y}}$