use implicit differentiation to find \\( \\frac { d y } { d x } \\).\n\n\\( x ^ { 4 } + \\sin y = x ^ { 3 }…

Explanation:

Step1: Differentiate both sides

Differentiate (x^{4}+\sin y) and (x^{3}y^{5}) 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 (\sin y), using the chain - rule (\frac{d}{dx}(f(g(x)))=f^{\prime}(g(x))\cdot g^{\prime}(x)), where (f(u)=\sin u) and (u = y), we get (\cos y\frac{dy}{dx}). For (x^{3}y^{5}), using the product rule (\frac{d}{dx}(uv)=u^{\prime}v + uv^{\prime}) (where (u=x^{3}), (u^{\prime}=3x^{2}) and (v = y^{5}), (v^{\prime}=5y^{4}\frac{dy}{dx})), we have (3x^{2}y^{5}+5x^{3}y^{4}\frac{dy}{dx}). So, (4x^{3}+\cos y\frac{dy}{dx}=3x^{2}y^{5}+5x^{3}y^{4}\frac{dy}{dx}).

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

Rearrange the terms: (\cos y\frac{dy}{dx}-5x^{3}y^{4}\frac{dy}{dx}=3x^{2}y^{5}-4x^{3}). Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(\cos y - 5x^{3}y^{4})=3x^{2}y^{5}-4x^{3}). Then (\frac{dy}{dx}=\frac{3x^{2}y^{5}-4x^{3}}{\cos y - 5x^{3}y^{4}}).

Answer:

(\frac{3x^{2}y^{5}-4x^{3}}{\cos y - 5x^{3}y^{4}})