Use implicit differentiate to find dy / dx when x² + y³  = 1

To implicitly differentiate the equation \( x^2 + y^3 = 1 \) with respect to \( x \), we will apply the chain rule. Here's the step-by-step process:

### Step 1: Differentiate both sides with respect to \( x \).

\[
\frac{d}{dx}(x^2 + y^3) = \frac{d}{dx}(1)
\]

On the left-hand side, we'll differentiate each term separately:

1. The derivative of \( x^2 \) with respect to \( x \) is:
   \[
   2x
   \]

2. The derivative of \( y^3 \) with respect to \( x \) requires implicit differentiation. Since \( y \) is a function of \( x \), we apply the chain rule:
   \[
   \frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}
   \]

On the right-hand side, the derivative of the constant \( 1 \) is \( 0 \).

### Step 2: Combine the results.

\[
2x + 3y^2 \frac{dy}{dx} = 0
\]

### Step 3: Solve for \( \frac{dy}{dx} \).

Now, isolate \( \frac{dy}{dx} \) on one side of the equation:

\[
3y^2 \frac{dy}{dx} = -2x
\]

\[
\frac{dy}{dx} = \frac{-2x}{3y^2}
\]

### Final Answer:
\[
\frac{dy}{dx} = \frac{-2x}{3y^2}
\]

Would you like any more details or further clarification?

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Here are 5 related questions to expand your understanding:
1. How do you apply implicit differentiation to trigonometric functions?
2. What is the geometric interpretation of \( \frac{dy}{dx} \) when using implicit differentiation?
3. How would the process change if the equation was \( x^3 + y^3 = 1 \)?
4. Can you explain how the chain rule is applied in implicit differentiation?
5. How do we differentiate a product of two functions implicitly?

**Tip**: Always remember to apply the chain rule when differentiating terms involving \( y \) with respect to \( x \).

Use implicit differentiation to find dy/dx when x² + y³ = 1.

Find the derivative of the implicit equation x² + y³ = 1 using implicit differentiation. This solution breaks down the chain rule and shows the step-by-step process to solve for dy/dx, suitable for Grades 10-12 students studying calculus and derivatives.

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