differentiate implicitly to find dy/dx. then find the slope of the curve at the given point.\n\n$x^{2}y^{2}=4…

Explanation:

Step1: Differentiate both sides with respect to (x)

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x^{2}) and (v = y^{2}). Differentiating (x^{2}y^{2}) gives (2xy^{2}+2x^{2}y\frac{dy}{dx}), and differentiating (49) gives (0). So, (2xy^{2}+2x^{2}y\frac{dy}{dx}=0).

Step2: Solve for (\frac{dy}{dx})

First, factor out (2xy) from the left - hand side: (2xy(y + x\frac{dy}{dx})=0). Since (x = 1) and (y = 7) (non - zero values at the given point ((1,7))), we can solve (2xy^{2}+2x^{2}y\frac{dy}{dx}=0) for (\frac{dy}{dx}) directly. Subtract (2xy^{2}) from both sides: (2x^{2}y\frac{dy}{dx}=- 2xy^{2}). Then divide both sides by (2x^{2}y) ((x\neq0,y\neq0)): (\frac{dy}{dx}=-\frac{y}{x}).

Step3: Find the slope at the point ((1,7))

Substitute (x = 1) and (y = 7) into (\frac{dy}{dx}=-\frac{y}{x}). (\frac{dy}{dx}\big|_{(1,7)}=-\frac{7}{1}=-7).

Answer:

(\frac{dy}{dx}=-\frac{y}{x}), and the slope at the point ((1,7)) is (-7).