given x² + 2xy - y³ = y + 9, use implicit differentiation to find dy/dx. dy/dx =

Explanation:

Step1: Differentiate each term

Differentiate $x^{2}+2xy - y^{3}=y + 9$ term - by - term with respect to $x$. The derivative of $x^{2}$ with respect to $x$ is $2x$ (using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$). For the term $2xy$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = 2x$ and $v = y$. So $\frac{d}{dx}(2xy)=2y + 2x\frac{dy}{dx}$. The derivative of $-y^{3}$ with respect to $x$ is $-3y^{2}\frac{dy}{dx}$ (using the chain - rule $\frac{d}{dx}(f(y))=f^{\prime}(y)\frac{dy}{dx}$). The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$, and the derivative of the constant 9 with respect to $x$ is 0. So we have $2x+2y + 2x\frac{dy}{dx}-3y^{2}\frac{dy}{dx}=\frac{dy}{dx}+0$.

Step2: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $2x\frac{dy}{dx}-3y^{2}\frac{dy}{dx}-\frac{dy}{dx}=-2x - 2y$. Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(2x-3y^{2}-1)=-2x - 2y$.

Step3: Solve for $\frac{dy}{dx}$

Divide both sides by $(2x - 3y^{2}-1)$ to get $\frac{dy}{dx}=\frac{-2x - 2y}{2x-3y^{2}-1}=\frac{2x + 2y}{3y^{2}+1 - 2x}$.

Answer:

$\frac{2x + 2y}{3y^{2}+1 - 2x}$