6. find \\( \\frac { d y } { d x } \\) by implicit differentiation \\( x ^ { 4 } + x ^ { 2 } y ^ { 2 } + y ^…

Explanation:

Step1: Differentiate each term

Differentiate (x^{4}+x^{2}y^{2}+y^{3}) with respect to (x). For (x^{4}), using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1}), we have (\frac{d}{dx}(x^{4}) = 4x^{3}). For (x^{2}y^{2}), use the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = x^{2}) and (v=y^{2}). (\frac{d}{dx}(x^{2}y^{2})=2xy^{2}+x^{2}\cdot2y\frac{dy}{dx}). For (y^{3}), using the chain rule (\frac{d}{dx}(y^{n})=ny^{n - 1}\frac{dy}{dx}), we get (\frac{d}{dx}(y^{3})=3y^{2}\frac{dy}{dx}). The derivative of the constant (5) is (0). So, (4x^{3}+2xy^{2}+2x^{2}y\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0).

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

Group the terms with (\frac{dy}{dx}) together: ((2x^{2}y + 3y^{2})\frac{dy}{dx}=-4x^{3}-2xy^{2}). Then (\frac{dy}{dx}=\frac{-4x^{3}-2xy^{2}}{2x^{2}y + 3y^{2}}=\frac{-2x(2x^{2}+y^{2})}{y(2x^{2}+3y)}).

Answer:

(\frac{dy}{dx}=\frac{-2x(2x^{2}+y^{2})}{y(2x^{2}+3y)})