Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x2/3 + y2/3 = 4
(−3

3
, 1)
(astroid)

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 \(y(x)\).

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x^(2/3) + y^(2/3) = 4 (−3, 1) (astroid)

Discover how to find the tangent line to the astroid curve x^(2/3) + y^(2/3) = 4 at the point (-3, 1) using implicit differentiation. This problem involves using the chain rule, implicit function theorem, and point-slope form, suitable for advanced high school or introductory college calculus students.

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