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 $x$.