find the slope of the tangent line to the curve (a lemniscate)\n$2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})$\nat the…

Explanation:

Step1: Differentiate both sides with respect to (x)

Using the chain rule and product rule. For the left - hand side: Let (u=x^{2}+y^{2}), then (2u^{2}). The derivative of (2u^{2}) with respect to (x) is (4u\cdot(2x + 2y\frac{dy}{dx})) (by the chain rule (\frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx}) and (\frac{du}{dx}=2x + 2y\frac{dy}{dx})). Substituting (u=x^{2}+y^{2}) back, we get (4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})). For the right - hand side: The derivative of (25(x^{2}-y^{2})) with respect to (x) is (25(2x-2y\frac{dy}{dx})) (by the sum/difference rule (\frac{d}{dx}(f(x)-g(x))=\frac{df}{dx}-\frac{dg}{dx}) and (\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}) by the chain rule). So, (4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})).

Step2: Substitute the point ((-3,1))

First, calculate (x=-3) and (y = 1). Then (x^{2}+y^{2}=(-3)^{2}+1^{2}=9 + 1=10). Substitute into (4(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25(2x - 2y\frac{dy}{dx})): (4\times10\times(-6 + 2\frac{dy}{dx})=25\times(-6-2\frac{dy}{dx})). Expand: (-240+80\frac{dy}{dx}=-150 - 50\frac{dy}{dx}).

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

Add (50\frac{dy}{dx}) to both sides: (-240+(80\frac{dy}{dx}+50\frac{dy}{dx})=-150). (-240 + 130\frac{dy}{dx}=-150). Add (240) to both sides: (130\frac{dy}{dx}=240 - 150=90). Then (\frac{dy}{dx}=\frac{90}{130}=\frac{9}{13}).

Answer:

(\frac{9}{13})