question let y be defined implicitly by the equation 4x^3 + 8y^3 = 71. use implicit differentiation to find…

Explanation:

Step1: Differentiate both sides

Differentiate $4x^{3}+8y^{3}=71$ with respect to $x$. The derivative of $4x^{3}$ with respect to $x$ is $12x^{2}$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. For the term $8y^{3}$, we use the chain - rule. The derivative of $8y^{3}$ with respect to $x$ is $24y^{2}\frac{dy}{dx}$. The derivative of the constant 71 with respect to $x$ is 0. So we have $12x^{2}+24y^{2}\frac{dy}{dx}=0$.

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

First, isolate the term with $\frac{dy}{dx}$: $24y^{2}\frac{dy}{dx}=-12x^{2}$. Then divide both sides by $24y^{2}$ (assuming $y\neq0$) to get $\frac{dy}{dx}=-\frac{12x^{2}}{24y^{2}}$. Simplify the right - hand side: $\frac{dy}{dx}=-\frac{x^{2}}{2y^{2}}$.

Answer:

$-\frac{x^{2}}{2y^{2}}$