use implicit differentiation to find \\( \\frac { d y } { d x } \\) without first solving for \\( y…

Explanation:

Step1: Differentiate both sides

Differentiate (e^{5xy}) and (y^{8}) with respect to (x). For (e^{5xy}), use the chain rule ((e^{u})^\prime=e^{u}\cdot u^\prime) where (u = 5xy). Then (u^\prime=5y + 5x\frac{dy}{dx}) (using the product rule ((uv)^\prime=u^\prime v+uv^\prime) with (u = 5x) and (v = y)). So ((e^{5xy})^\prime=e^{5xy}(5y + 5x\frac{dy}{dx})). For (y^{8}), use the chain rule ((y^{n})^\prime=ny^{n - 1}\frac{dy}{dx}), so ((y^{8})^\prime=8y^{7}\frac{dy}{dx}). We get (e^{5xy}(5y + 5x\frac{dy}{dx})=8y^{7}\frac{dy}{dx}).

Step2: Expand and collect terms with (\frac{dy}{dx})

Expand the left - hand side: (5ye^{5xy}+5xe^{5xy}\frac{dy}{dx}=8y^{7}\frac{dy}{dx}). Move all terms with (\frac{dy}{dx}) to one side: (5xe^{5xy}\frac{dy}{dx}-8y^{7}\frac{dy}{dx}=- 5ye^{5xy}). Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(5xe^{5xy}-8y^{7})=-5ye^{5xy}).

Step3: Solve for (\frac{dy}{dx})

(\frac{dy}{dx}=\frac{-5ye^{5xy}}{5xe^{5xy}-8y^{7}}=\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}}).

Answer:

(\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}})