find \\( \\frac{d y}{d x} \\) by implicit differentiation\n\\( 2+4 x=\\sin \\left(x y^{4}\\right) \\)\n\\(…

Explanation:

Step1: Differentiate both sides

Differentiate (2 + 4x=\sin(xy^{4})) with respect to (x). The derivative of (2) is (0), the derivative of (4x) is (4). For the right - hand side, use the chain rule. Let (u = xy^{4}), then (\frac{d}{dx}\sin(u)=\cos(u)\cdot\frac{du}{dx}). And (\frac{du}{dx}=y^{4}+x\cdot4y^{3}\frac{dy}{dx}) (using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x) and (v = y^{4})). So, (4=\cos(xy^{4})\left(y^{4}+4xy^{3}\frac{dy}{dx}\right)).

Step2: Expand and solve for (\frac{dy}{dx})

Expand the right - hand side: (4 = y^{4}\cos(xy^{4})+4xy^{3}\cos(xy^{4})\frac{dy}{dx}). Rearrange the terms: (4xy^{3}\cos(xy^{4})\frac{dy}{dx}=4 - y^{4}\cos(xy^{4})). Then (\frac{dy}{dx}=\frac{4 - y^{4}\cos(xy^{4})}{4xy^{3}\cos(xy^{4})}).

Answer:

(\frac{4 - y^{4}\cos(xy^{4})}{4xy^{3}\cos(xy^{4})})