find \\( \\frac{d y}{d x} \\) by implicit differentiation.\n\nanswer: \\( \\frac{d y}{d x}= \\)\n\n\\( 4+4…

Explanation:

Step1: Differentiate both sides

Differentiate (4 + 4x=\sin(xy^{2})) with respect to (x). Using the sum rule ((u + v)^\prime=u^\prime+v^\prime), ((4)^\prime = 0), ((4x)^\prime=4). For the right - hand side, use the chain rule ((\sin(u))^\prime=\cos(u)\cdot u^\prime), where (u = xy^{2}). Then ((\sin(xy^{2}))^\prime=\cos(xy^{2})\cdot(xy^{2})^\prime). Now, use the product rule ((uv)^\prime = u^\prime v+uv^\prime) for ((xy^{2})^\prime), where (u = x), (u^\prime=1), (v = y^{2}), (v^\prime = 2y\frac{dy}{dx}). So ((xy^{2})^\prime=y^{2}+2xy\frac{dy}{dx}). The equation becomes (4=\cos(xy^{2})\left(y^{2}+2xy\frac{dy}{dx}\right)).

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

Expand the right - hand side: (4 = y^{2}\cos(xy^{2})+2xy\cos(xy^{2})\frac{dy}{dx}). Subtract (y^{2}\cos(xy^{2})) from both sides: (4 - y^{2}\cos(xy^{2})=2xy\cos(xy^{2})\frac{dy}{dx}). Then (\frac{dy}{dx}=\frac{4 - y^{2}\cos(xy^{2})}{2xy\cos(xy^{2})}).

Answer:

(\frac{4 - y^{2}\cos(xy^{2})}{2xy\cos(xy^{2})})