use implicit differentiation to find $\\frac{dy}{dx}$ for the equation $\\sqrt{x + y}=8 + x^{2}y^{2}$…

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})). So (\frac{d}{dx}(\sqrt{x + y})=\frac{1 + \frac{dy}{dx}}{2\sqrt{x + y}}). For the right - hand side, using the sum rule and the product rule. The derivative of a constant (8) is (0). For (x^{2}y^{2}), by the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = x^{2}), (u^\prime=2x) and (v = y^{2}), (v^\prime = 2y\frac{dy}{dx}). So (\frac{d}{dx}(x^{2}y^{2})=2xy^{2}+2x^{2}y\frac{dy}{dx}). We get (\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}})