homework7: problem 2\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point…

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (4x^{2}-xy + 4y^{3}) term - by - term. Using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), (\frac{d}{dx}(4x^{2})=8x). For the term (-xy), use the product rule ((uv)^\prime=u^\prime v+uv^\prime), where (u=-x) and (v = y). So (\frac{d}{dx}(-xy)=-y - x\frac{dy}{dx}). For the term (4y^{3}), use the chain rule (\frac{d}{dx}(f(g(x)))=f^\prime(g(x))\cdot g^\prime(x)). Let (u = y), then (\frac{d}{dx}(4y^{3})=12y^{2}\frac{dy}{dx}). Differentiating the right - hand side (\frac{d}{dx}(264) = 0). So, (8x-y - x\frac{dy}{dx}+12y^{2}\frac{dy}{dx}=0).

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

Group the terms with (\frac{dy}{dx}) together: ((-x + 12y^{2})\frac{dy}{dx}=y - 8x). Then (\frac{dy}{dx}=\frac{y - 8x}{-x + 12y^{2}}).

Step3: Substitute (x = 2) and (y = 4)

Substitute (x = 2) and (y = 4) into (\frac{dy}{dx}=\frac{y - 8x}{-x + 12y^{2}}). (\frac{dy}{dx}=\frac{4-8\times2}{-2 + 12\times4^{2}}=\frac{4 - 16}{-2+192}=\frac{-12}{190}=-\frac{6}{95}).

Answer:

(-\frac{6}{95})