find the slope of the tangent line to the curve\n\n$2x^{2}-4xy - 2y^{3}=104$\n\nat the point…

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (2x^{2}-4xy - 2y^{3}) term - by - term. Using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), (\frac{d}{dx}(2x^{2})=4x). For the term (-4xy), use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u=-4x) and (v = y). So (\frac{d}{dx}(-4xy)=-4y-4x\frac{dy}{dx}). For the term (-2y^{3}), use the chain rule (\frac{d}{dx}(f(y))=f^\prime(y)\frac{dy}{dx}). So (\frac{d}{dx}(-2y^{3})=-6y^{2}\frac{dy}{dx}). Differentiating the right - hand side (\frac{d}{dx}(104) = 0). The derivative of the entire equation is: (4x-4y-4x\frac{dy}{dx}-6y^{2}\frac{dy}{dx}=0)

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

Group the terms with (\frac{dy}{dx}) together: (-4x\frac{dy}{dx}-6y^{2}\frac{dy}{dx}=4y - 4x) Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(-4x - 6y^{2})=4y - 4x) Then (\frac{dy}{dx}=\frac{4y - 4x}{-4x - 6y^{2}}=\frac{2y - 2x}{-2x - 3y^{2}})

Step3: Substitute (x=-2) and (y = - 4) into (\frac{dy}{dx})

Substitute (x=-2) and (y=-4) into (\frac{dy}{dx}=\frac{2y - 2x}{-2x - 3y^{2}}) (\frac{dy}{dx}=\frac{2(-4)-2(-2)}{-2(-2)-3(-4)^{2}}) First, calculate the numerator: (2(-4)-2(-2)=-8 + 4=-4) Then, calculate the denominator: (-2(-2)-3(-4)^{2}=4-3\times16=4 - 48=-44)

Answer:

(\frac{1}{11})