if (x + 3y^{\frac{1}{3}}=y), what is (\frac{dy}{dx}) at the point ((2,8))?\na (\frac{1}{3})\nb…

Explanation:

Step1: Differentiate both sides with respect to x

Differentiating $x + 3y^{\frac{1}{3}}$ with respect to $x$ gives $1+3\times\frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx}$, and differentiating $y$ with respect to $x$ gives $\frac{dy}{dx}$. So, $1 + y^{-\frac{2}{3}}\frac{dy}{dx}=\frac{dy}{dx}$.

Step2: Rearrange the equation for $\frac{dy}{dx}$

Move terms involving $\frac{dy}{dx}$ to one - side: $\frac{dy}{dx}-y^{-\frac{2}{3}}\frac{dy}{dx}=1$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(1 - y^{-\frac{2}{3}})=1$. Then $\frac{dy}{dx}=\frac{1}{1 - y^{-\frac{2}{3}}}$.

Step3: Substitute the point $(2,8)$

Substitute $y = 8$ into the derivative formula. Since $y^{-\frac{2}{3}}=8^{-\frac{2}{3}}=(2^{3})^{-\frac{2}{3}}=2^{-2}=\frac{1}{4}$, then $\frac{dy}{dx}=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$.

Answer:

D. $\frac{4}{3}$