use implicit differentiation to find \\( \\frac { d y } { d x } \\).\n\\( \\sqrt { x ^ { 4 } + y ^ { 2 } } =…

Explanation:

Step1: Differentiate both sides

Differentiate $\sqrt{x^{4}+y^{2}}$ and $3x + 9y^{3}$ with respect to $x$. For the left - hand side, use the chain rule. Let $u=x^{4}+y^{2}$, then $\frac{d}{dx}(\sqrt{u})=\frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$. $\frac{du}{dx}=4x^{3}+2y\frac{dy}{dx}$, so $\frac{d}{dx}(\sqrt{x^{4}+y^{2}})=\frac{4x^{3}+2y\frac{dy}{dx}}{2\sqrt{x^{4}+y^{2}}}=\frac{2x^{3}+y\frac{dy}{dx}}{\sqrt{x^{4}+y^{2}}}$. For the right - hand side, $\frac{d}{dx}(3x + 9y^{3})=3+27y^{2}\frac{dy}{dx}$. So we have $\frac{2x^{3}+y\frac{dy}{dx}}{\sqrt{x^{4}+y^{2}}}=3 + 27y^{2}\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Multiply both sides by $\sqrt{x^{4}+y^{2}}$ to get $2x^{3}+y\frac{dy}{dx}=(3 + 27y^{2}\frac{dy}{dx})\sqrt{x^{4}+y^{2}}$. Expand the right - hand side: $2x^{3}+y\frac{dy}{dx}=3\sqrt{x^{4}+y^{2}}+27y^{2}\sqrt{x^{4}+y^{2}}\frac{dy}{dx}$. Group the terms with $\frac{dy}{dx}$: $y\frac{dy}{dx}-27y^{2}\sqrt{x^{4}+y^{2}}\frac{dy}{dx}=3\sqrt{x^{4}+y^{2}}-2x^{3}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(y - 27y^{2}\sqrt{x^{4}+y^{2}})=3\sqrt{x^{4}+y^{2}}-2x^{3}$. Then $\frac{dy}{dx}=\frac{3\sqrt{x^{4}+y^{2}}-2x^{3}}{y - 27y^{2}\sqrt{x^{4}+y^{2}}}$.

Answer:

$\frac{3\sqrt{x^{4}+y^{2}}-2x^{3}}{y - 27y^{2}\sqrt{x^{4}+y^{2}}}$