To find the implicit derivative of the equation $y^2(6 - x) = x^3$, follow these steps:

Step 1: Differentiate both sides with respect to $x$.

Apply the product rule on the left-hand side, where $y^2$ is one factor and $(6 - x)$ is the other factor. Also, differentiate the right-hand side directly.

$\frac{d}{dx} \left( y^2(6 - x) \right) = \frac{d}{dx}(x^3)$

Step 2: Apply the product rule.

For the left-hand side, use the product rule: $\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)$, where $u(x) = y^2$ and $v(x) = (6 - x)$.

• The derivative of $y^2$ with respect to $x$ is $2y \cdot \frac{dy}{dx}$ (using the chain rule).
• The derivative of $(6 - x)$ with respect to $x$ is $-1$.

So, the derivative of the left-hand side is: $2y \cdot \frac{dy}{dx} \cdot (6 - x) + y^2 \cdot (-1)$

Now differentiate the right-hand side: $\frac{d}{dx}(x^3) = 3x^2$

Step 3: Combine the results.

Equating both sides, we get: $2y \cdot \frac{dy}{dx} \cdot (6 - x) - y^2 = 3x^2$

Step 4: Solve for $\frac{dy}{dx}$.

Rearrange the equation to isolate $\frac{dy}{dx}$:

$2y \cdot \frac{dy}{dx} \cdot (6 - x) = 3x^2 + y^2$

$\frac{dy}{dx} = \frac{3x^2 + y^2}{2y(6 - x)}$

Final Answer:

The implicit derivative is: $\frac{dy}{dx} = \frac{3x^2 + y^2}{2y(6 - x)}$

Would you like more details or have any questions?

Here are some related questions to consider:

1. What is implicit differentiation used for in calculus?
2. Can you apply implicit differentiation to higher-order derivatives?
3. How does the chain rule affect implicit differentiation?
4. How can we use implicit derivatives to find tangent lines to curves?
5. What are common mistakes to avoid when applying the product rule in implicit differentiation?

Tip: Always apply the chain rule when differentiating terms involving $y$, as $y$ is a function of $x$.