find dy/dx by implicit differentiation. 8x³ + x²y - xy³ = 3 dy/dx = need help? read it master it submit answer

Explanation:

Step1: Differentiate each term

Differentiate $8x^{3}+x^{2}y - xy^{3}=3$ term - by - term with respect to $x$. The derivative of $8x^{3}$ with respect to $x$ is $24x^{2}$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$). For $x^{2}y$, use the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u = x^{2}$ and $v = y$. So $\frac{d}{dx}(x^{2}y)=2xy+x^{2}\frac{dy}{dx}$. For $-xy^{3}$, use the product rule and the chain - rule. $\frac{d}{dx}(-xy^{3})=-y^{3}-3xy^{2}\frac{dy}{dx}$. The derivative of the constant 3 with respect to $x$ is 0. So we have $24x^{2}+2xy + x^{2}\frac{dy}{dx}-y^{3}-3xy^{2}\frac{dy}{dx}=0$.

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

Rearrange the terms to get all terms with $\frac{dy}{dx}$ on one side: $x^{2}\frac{dy}{dx}-3xy^{2}\frac{dy}{dx}=y^{3}-24x^{2}-2xy$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(x^{2}-3xy^{2})=y^{3}-24x^{2}-2xy$.

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

Divide both sides by $x^{2}-3xy^{2}$ to solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{y^{3}-24x^{2}-2xy}{x^{2}-3xy^{2}}$.

Answer:

$\frac{y^{3}-24x^{2}-2xy}{x^{2}-3xy^{2}}$