We are tasked with finding the equation of the tangent line to the curve given by $x^{2/3} + y^{2/3} = 4$ at the point $(-3, 1)$. To do this, we will use implicit differentiation.

Step-by-step Solution:

1. Differentiate implicitly the given equation with respect to $x$: $x^{2/3} + y^{2/3} = 4$

   Taking the derivative of both sides with respect to $x$, we use the chain rule for the term involving $y$, as $y$ is a function of $x$:

   $\frac{d}{dx}\left( x^{2/3} \right) + \frac{d}{dx}\left( y^{2/3} \right) = 0$

2. Differentiate each term:

   • For $x^{2/3}$, the derivative is: $\frac{d}{dx}\left( x^{2/3} \right) = \frac{2}{3}x^{-1/3}$
   • For $y^{2/3}$, applying the chain rule: $\frac{d}{dx}\left( y^{2/3} \right) = \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx}$

   So the equation becomes: $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$: First, subtract $\frac{2}{3}x^{-1/3}$ from both sides: $\frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3}$

   Cancel the $\frac{2}{3}$ terms: $y^{-1/3} \cdot \frac{dy}{dx} = -x^{-1/3}$

   Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3}$

4. Substitute the point $(-3, 1)$ to find the slope of the tangent line at that point: $\frac{dy}{dx} = -\left(\frac{1}{-3}\right)^{1/3} = -(-\frac{1}{3})^{1/3}$

   Since the cube root of $-\frac{1}{3}$ is still $-\frac{1}{3}$, we have: $\frac{dy}{dx} = \frac{1}{3}$

5. Equation of the tangent line: The slope of the tangent line at the point $(-3, 1)$ is $\frac{1}{3}$, and we use the point-slope form of the line equation: $y - y_1 = m(x - x_1)$ Substituting $m = \frac{1}{3}$, $(x_1, y_1) = (-3, 1)$: $y - 1 = \frac{1}{3}(x + 3)$ Simplifying: $y - 1 = \frac{1}{3}x + 1$ $y = \frac{1}{3}x + 2$

Final Answer:

The equation of the tangent line at the point $(-3, 1)$ is: $y = \frac{1}{3}x + 2$

Would you like any further clarifications or details?

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Relative Questions:

1. What is implicit differentiation and when is it used?
2. How does the chain rule apply in implicit differentiation?
3. What is the general form of the equation for a tangent line?
4. How do you differentiate terms like $x^{n}$ and $y^{n}$ implicitly?
5. How does the cube root impact the differentiation of a function?

Tip: Always remember to apply the chain rule when differentiating implicitly, as it accounts for dependent variables like $y(x)$.