use implicit differentiation to find an equation of the tangent line to the curve at the given point…

Explanation:

Step1: Differentiate both sides

Differentiate $x^{2/3}+y^{2/3}=4$ with respect to $x$. Using the power - rule and chain - rule, we have $\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}y' = 0$.

Step2: Solve for $y'$

First, factor out $\frac{2}{3}$: $\frac{2}{3}(x^{-1/3}+y^{-1/3}y') = 0$. Then, solve for $y'$: [ \begin{align*} x^{-1/3}+y^{-1/3}y'&=0\ y^{-1/3}y'&=-x^{-1/3}\ y'&=-\left(\frac{y}{x}\right)^{1/3} \end{align*} ]

Step3: Find the slope at the given point

Substitute $x = - 3\sqrt{3}$ and $y = 1$ into $y'$: [ \begin{align*} y'&=-\left(\frac{1}{-3\sqrt{3}}\right)^{1/3}\ &=\left(\frac{1}{3\sqrt{3}}\right)^{1/3} \end{align*} ] The slope of the tangent line $m = \sqrt{3}$.

Step4: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$. Here, $x_1=-3\sqrt{3}$, $y_1 = 1$ and $m=\sqrt{3}$. [ \begin{align*} y - 1&=\sqrt{3}(x + 3\sqrt{3})\ y - 1&=\sqrt{3}x+9\ y&=\sqrt{3}x + 10 \end{align*} ]

Answer:

$y=\sqrt{3}x + 10$